math 152 calculus ii chapter 4

8112019 Math 152 Calculus II Chapter 4

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Contents

1 Chapter 1 1

2 Chapter 1 1

3 Chapter 1 1

4 Integration 2

41 Antiderivatives 2411 Definition of an Antiderivative 2412 Antiderivatives of Trigonometric Functions 3413 Antiderivatives of Other Functions 4

42 Area 6421 Sigma Notation 6422 Expanding Summations 6

423 Area of Plane Regions 943 Riemann Sum 13431 Definition of a Riemann Sum 14

44 Definite Integrals 20441 Fundamental Theorem of Calculus 20442 Evaluating Integrals 20443 Integrals of Absolute Value Functions 21444 Mean Value Theorem 22445 The Average Value of a Function 22

45 Integration by Substitution 25451 Review of the Chain Rule 25

452 Basic Substitutions 2546 The Natural Logarithm Function 2947 Antiderivative of the Inverse Trigonometric Functions 31

471 Basic Antiderivatives of the Inverse Trigonometric Functions 31472 Integration by Substitution 31

48 Simpsonrsquos Rule and the Trapezoid Rule 35481 The Trapezoid Rule 35482 Simpsonrsquos Rule 35

1 Chapter 1

See Math 151 Notes

2 Chapter 1

See Math 151 Notes

3 Chapter 1

See Math 151 Notes

1



4 Integration

41 Antiderivatives

411 Definition of an Antiderivative

A function F is an antiderivative of f on an interval I if F prime(x) = f (x) for all x on I

The integral or antiderivative of a function is essentially the reverse process of differentiationTo start will look at the most basic antiderivative which the power rule for antiderivativesThe power rule for antiderivatives is the reverse process of the power rule for derivatives

Power Rule for Antiderivatives

int cxndx = c

n+1xn+1 + C

Now letrsquos try doing so examples that use the power rule

Example 1

Evaluate the following integralint

x2dx

Solution

int x2dx = 1

2+1x2+1 + C = 1

3x3 + C

Example 2


(x4 + 3x2 + 3x)dx

Solution

int (x4 + 3x2 + 3x)dx = 1

4+1x4+1 + 3

2+1x2+1 + 3

1+1x1+1 + C = 1

5x5 + x3 + 3

2x2 + C

Example 3


(x3 minus 5x2 + 6)dx

Solution

int (x3 minus 5x2 + 6x0)dx = 1

3+1x3+1 minus 5

2+1x2+1 + 6

1+0x0+1 + C = 1

4x4 minus 5

3x3 + 6x + C

2



Example 4


5x4dx

Solutionint 5x4dx = 5

4+1x4+1 + C = x5 + C

412 Antiderivatives of Trigonometric Functions

Basic Integration Rules

int cos(x)dx = sin(x) + C

int sin(x)dx = minuscos(x) + C

int sec2(x)dx = tan(x) + C

int csc2(x)dx = minuscot(x) + C

int sec(x)tanxdx = sec(x) + C

int csc(x)cotxdx = minuscsc(x) + C

Here are some example of antiderivatives of trigonometric functions

Example 5


(sec2x minus csc2x)dx

Solution

int (sec2(x) minus csc2(x))dx = tan(x) + cot(x) + C

Example 6


(sin(x) + 3x2)dx

Solution

int (sin(x) + 3x2)dx = minuscos(x) + 3

2+1x2+1 + C = minuscos(x) + x3 + C

3



Example 7



Solution

int (cos(x) + 4)dx = sin(x) + 4

0+1x0+1 + C = sin(x) + 4x + C

413 Antiderivatives of Other Functions

Antiderivatives of exponential and logarithmic functions

int 1x

dx = ln|x| + C

int exdx = ex + C

Example 8


exdx

Solution

int ex = ex + C

Example 9


( 1x

+ x4)dx

Solution

int ( 1

x + x4) = ln|x| + 1

4+1x4+1 + C = 1

5x5 + C

Example 10


x3

2 dx

Solution

int x

3

2 dx = 13

2+1

x3

2+1 + C = 1

5

2

x5

2 + C = 25

x5

2 + C

4



Example 11


1x3

dx

Solution

int 1x3

dxint

xminus3dx = 1minus3+1

xminus3+1 + C = minus12

xminus2 + C = minus 12x2

+ C

Example 12


( 1x2

+ cos(x))dx

Solution

int ( 1

x2 + cos(x))dx =

int (xminus2 + cos(x))dx = 1

minus2+1xminus2+1 + sin(x) + C = minusxminus1 + sin(x) + C =

minus 1x

+ sin(x) + C

5



42 Area

421 Sigma Notation

The sum of the terms a1 a2 a3 a4an isnsum

i=1

ai = a1 + a2 + a3 + a4 + + an

where i is the index of a summation ai is the ith term of the sum and the upper and lowersum of the summation are 1 and n

Here are some example of summations

Example 1

Evaluate6sum

i=1

i

Solution

6sumi=1

i = 1 + 2 + 3 + 4 + 5 + 6 = 21

Example 2

Evaluate4sum

i=0

(2i + 1)

Solution4sum

i=0

(2i + 1) = 2(0) + 1 + 2(1) + 1 + 2(2) + 1 + 2(3) + 1 + 2(4) + 1 = 1 + 3 + 5 + 7 + 9 = 25

422 Expanding Summations

Example 3

Evaluate4sum

j=0

j2

Solution4sum

j=0

j2 = 12 + 22 + 32 + 42 = 1 + 4 + 9 + 16 = 30

6



Example 4

Evaluate6sum

k=3

k(k minus 2)

Solution6sum

k=3

k(kminus2) = 3(3minus2)+4(4minus2)+5(5minus2)+6(6minus2) = 3middot1+4middot2+5middot3+6middot4 = 3+8+15+24 = 50

Example 5

Evaluate5sum

j=3

1

j

Solution

5sum j=3

1

j =

1

3 +

1

4 +

1

5 =

47

60

Example 6

Evaluate4

sum j=1

[(i

minus1)2 + (i + 1)3]

Solution4sum

j=1

[(i minus 1)2 + (i + 1)3] = (1 minus 1)2 + ( 1 + 1 )3 + (2 minus 1)2 + ( 2 + 1 )3 + (3 minus 1)2 + ( 3 + 1 )3 + (4 minus

1)2 +(4+1)3 = 02 + 23 + 12 + 33 + 22 + 43 + 32 + 53 = 0 + 8 + 1 + 27 + 4 + 64 + 9 + 125 = 237

Example 7

Express the following series of terms into a expression using summation notation 5

1+1 + 5

1+2 +

51+3

+ + 51+12

Solution

Since the series of terms is adding one to each successive term and then the dividing thistotal into 5 the summation would be written as follows12sum

j=1

[ 5

1 + i]

7



Example 8

Express the following series of terms into a expression using summation notation [1 minus (14

)2] +

[1 minus (24

)2] + 1 minus [(34

)2] + [1 minus (44

)2]

SolutionSince the series of terms is dividing each successive term by 4 squaring the result and addingone the summation would be written as follows4sum

j=1

[1 + (i

4)2]

Theorem 42 Summation Formulas

1usum

j=1

c = cn

2usum

j=1

i = n(n+1)

2

3usum

j=1

i2 = n(n+1)(2n+1)

6

4usum

j=1

i3 = n

2(n+1)2

4

Now letrsquos use the four properties above to expand some examples of summations

Example 9

Use the properties of Theorem 42 to expand out the following summation

15sum

j=1

(2i minus 3)

Solution15sum

j=1

(2i minus 3) =15sum

j=1

2i minus15sum

j=1

3 = 2 middot15sum

j=1

i minus15sum

j=1

3 = 2 middot 15(16)

2 minus 15(3) = 240 minus 45 = 195

8



Example 10


15

sum j=1

i(i2 + 1)

Solution10sum

j=1

i(i2 + 1) =10sum

j=1

i3 + i

= n

2(n+1)2

4 + n(n+1)

2 (where n=10)

= 102(10+1)2

4 + 10(10+1)

2

= 100middot1214 + 10middot11

2= 25

middot121 + 5

middot11

= 3025 + 55 = 3080

423 Area of Plane Regions

Now we are prepared to use summation to find areas of plane regions We are going toapproximate areas under the curve by dividing the region under the curve into smallerrectangles and summing up the rectangles as shown in the next diagram

Theorem 43 Limits of Upper and Lower Sums

Let f be continuous and nonnegative on the interval [a b] The limits as n rarrinfin of both thelower and upper sums exist and are equal to each other

limhrarrinfin

S (n) = limhrarrinfin

f (m j)x = limhrarrinfin

f (M j)x = S (n) where x = bminusa

n

9



Example 10

Use upper sums and lower to approximate the area of the function f (x) =radic

x below boundedby the x-axis y = 0 and the values x = 0 and x = 1

Solution

First find x (The length of each rectangle)

a = 0 and b = 1 rArr x = 1minus04

= 14

Next compute the left and right endpoints

Left Endpoint mi = a + (i minus 1)x = 0 + (i minus 1) middot 14

= iminus14

Right Endpoint M i = a + ix = 0 + i middot 14

= i

4

Computing the Lower Sums

S (n) =4sum

j=1

f (mi)x

=4

sumi=1

f (iminus 1

4 ) middot (

1

4)

=4sum

i=1

radic iminus 1

4 middot 1

4

=4sum

i=1

radic iminus 1

2 middot 1

4

=4sum

i=1

radic iminus 1

8

=radic 1minus18 +

radic 2minus18 +

radic 3minus18 +

radic 4minus18

=

radic 0

8 +

radic 1

8 +

radic 2

8 +

radic 3

8 = 0 + 125 + 177 + 217 asymp 519

10



Computing the Upper Sums

S (n) =4sum

j=1

f (M i)x

=4sum

i=1

f (i

4) middot (

1

4)

=4sum

i=1

radic i

4middot 1

4

=

4

sumi=1

radic i

2 middot1

4

=4sum

i=1

radic i

8

=radic 18

+radic 28

+radic 38

+radic 48

= 125 + 177 + 217 + 25 asymp 769

Example 11

Find the limit of S (n) as n

rarrinfinof the summation S (n) = 64

n3[n(n+1)(2n+1)

6 ]

Solution

limhrarrinfin


64

n3[n(n + 1)(2n + 1)

6 ]

= limhrarrinfin

64

n3[(n2 + n))(2n + 1)

6 ]

= limhrarrinfin

64

n3[(2n3 + n2 + 2n2 + n

6 ]

= limhrarrinfin

64[2n3 + 3n2 + n

6n3 ]

= limhrarrinfin

128n3

+ 192n2

+ 64n6n3

11



= limhrarrinfin

128n3

6n3 +

192n2

6n3 +

64n

6n3

= limhrarrinfin

64

3 +

32

n +

32

3n2

= 643

+ 0 + 0 = 643

12



43 Riemann Sum

Example 1 A Partition with Unequal Widths

Consider the region bounded by the graph of f (x) =radic

x and the x-axis for 0

lexi

le1

Now letrsquos evaluate the limit limhrarrinfin

nsumi=1

f (ci)xi

where ci is the right endpoint of the portion given by ci = i2

n2 and xi is the width of theith interval

Let xi = right endpoint - left endpoint

xi = i2

n2 minus (iminus1)2n2

= i2

n2 minus i2minus2i+1

n2

= i2minus(i2+2iminus1)

n2

= i2minusi

2minus2i+1)n2

= minus2i+1n2

Now we can evaluate the following limit

limhrarrinfin

nsumi=1

f (ci)xi = limhrarrinfin

nsumi=1

f ( i2

n2) middot (

2iminus 1

n2 )

= limhrarrinfin

nsumi=1

radic i2

n2 middot (

2iminus 1

n2 )

= limhrarrinfin

nsumi=1

i

nmiddot 2iminus 1

n2

= limhrarrinfin

nsumi=1

2i2

minusi

n3

13



= 1

n3 limhrarrinfin

nsumi=1

(2i2 minus i)

= 1

n3 limhrarrinfin

[2(n(n + 1)(2n + 1)

6 ) minus n(n + 1)

2 ]

= 1

n3 limhrarrinfin(2n(n + 1)(2n + 1)

6 minus n2 + n

2 ]

= 1

n3 limhrarrinfin

((2n2 + 2n)(2n + 1)

6 minus n2 + n

2 ]

= 1

n3 limhrarrinfin

(4n3 + 6n2 + 2n

6 minus n2 + n

2 ]

= 1

n3 limhrarrinfin

4n3 + 6n2 + 2n minus 3n2 minus 3n

6

= 1

n3 limhrarrinfin

4n3 + 3n2 minus n

6n3

= limhrarrinfin

[4n3 + 3n2 minus n

6n3

]

= limhrarrinfin

4n3

6n3 +

3n2

6n3 +

n

6n3

= limhrarrinfin

2

3 +

1

2n +

1

6n2 =

2

3 + 0 + 0 =

2

3

431 Definition of a Riemann Sum

Let f be defined on a closed interval [a b] and let be a partition of [a b] given by a =x0 lt x1 lt x2 lt x3 lt lt xnminus1 lt xn = b where xi is the width of the ith subinterval If

ci is any point in ith subintervalnsum

i=1

f (ci)xi then the sum is called the Riemann sum

Definition of a Definite Integral

If f defined on a closed interval [a b] and the limit lim||rarrinfin

nsumi=1

f (ci)xi exist as described

above then f is integrable on [a b] and the limit is lim||rarrinfin

nsumi=1

f (ci)xi =int

b

a f (x)dx

14



Example 2

Evaluate the following definite integral by the limit definition

Solution

int 3minus2

xdx

Solution

a = minus2 b = 3 x = bminusa

n = 3minus(minus2)

n

Now rewrite the integral as a summationint 3minus2 xdx = lim

nrarrinfin

nsum

i=1

f (ci)xi

= limnrarrinfin

nsum

i=1

1048616minus 2 +

5i

n

1048617middot 5

n

= limnrarrinfin

nsum

i=1

1048616minus10

n +

25i

n2

1048617

= limnrarrinfin

nsum

i=1

1048616minus10

n +

25

n2 middot n(n + 1)

2

1048617

= limnrarrinfin

nsumi=1

1048616minus10n

+ 25n2

+ 25n2n2

1048617

= limnrarrinfin

nsum

i=1

1048616minus10

n +

25n2

2n2 +

25n

2n21048617

= limnrarrinfin

nsum

i=1

1048616minus10

n +

25

2 +

25

2n

1048617 = minus0 + 125 + 0 = 125

Example 3


Solution

15



int 31 3x2dx

Solution

a = 1 b = 3 x = 3minus1n

= 2n

ci = 1 + i middot 2i = 1 + 2i

n

Now rewrite the integral as a summationint 31 xdx = lim

nrarrinfin

nsum

i=1

f (ci)xi

= limnrarrinfin

nsum

i=1

f (1 + 2i

n ) middot

2

n

= limnrarrinfin

nsum

i=1

3(1 + 2i

n )2 middot

2

n

= limnrarrinfin

nsum

i=1

3(4i2

n2 +

4i

n + 1) middot

2

n

= limnrarrinfin

nsum

i=1

3( 8i2

n3 + 8i

n2 + 2

n)

= limnrarrinfin

nsum

i=1

(24i2

n3 +

24i

n2 +

6

n)

= limnrarrinfin

nsum

i=1

104861624(n(n + 1)(2n + 1))

6n3 +

24n(n + 1)

2n2 +

6n

n

1048617

= limnrarrinfin

nsum

i=1

104861624(2n3 + 3n2 + n)

6n3 +

24n2 + 24n

2n2 +

6n

n

1048617

= limnrarrinfin

n

sumi=1

104861648n3

6n3 + 72n2

6n3 + 24n

6n3 + 24n2

2n2 + 24n

n2 + 61048617

16



=1048616

8 + 12

n +

4

n2 + 12 +

12

n + 6

1048617 = 26

Example 4

Evaluate the following definite integral using the limit definition of a Riemann Sumint 104 6dx

Solution

a = 4 b = 10 x = 10minus4n

= 6n

ci = a + i middot x = 4 + i middot 6n

= 4 + 6in

f (x) = 6

Now rewrite the integral as a summationint 104 6dx = lim

nrarrinfin

nsumi=1

f (ci)xi

= limnrarrinfin

nsumi=1

f 983080

4 + 6i

n

983081middot 6

n

= limnrarrinfin

nsumi=1

(6) middot 6

n

= limn

rarrinfin(6n)

middot

6

n = 36

17



Example 5

Evaluate the following integral using the limit definition of an integralint 21 (x2 + 1)dx

a = 1 b = 2 x = 2minus1n

= 1n


= 1 + i

n

f (x) = x2 + 1

Now rewrite the integral as a summation

int 21 (x2 + 1)dx = lim

nrarrinfin

nsumi=1

f (ci)xi

= limnrarrinfin

n

sumi=1 f

9830801 +

i

n983081 middot

1

n

= limnrarrinfin

nsumi=1

9830809830801 +

i

n

9830812+ 1

983081middot 1

n

= limnrarrinfin

nsumi=1

9830809830801 +

i

n

9830819830801 +

i

n

983081+ 1

983081middot 1

n

= limnrarrinfin

nsumi=1

98308010486161 +

2i

n +

i2

n2

983081+ 1

983081middot 1

n

= limnrarrinfin

n

sumi=1

9830802 + 2i

n

+ i2

n2983081 middot

1

n

= limnrarrinfin

nsumi=1

9830802

n +

2i

n2 +

i2

n3

983081

= limnrarrinfin

983080 2

n middot (n) +

2

n2 middot n(n + 1)

2 +

1

n3 middot n(n + 1)(2n + 1)

6

983081

= limnrarrinfin

9830802 +

2n2 + 2n

2n2 +

(n2 + n)(2n + 1)

6n3

983081

= limnrarrinfin

9830802 +

2n2 + 2n

2n2 +

2n3 + 3n2 + n

6n3

983081

= limnrarrinfin983080

2 + 2n2

2n2

+ 2n

2n2

+ 2n3

6n3

+ 3n2

6n3

+ n

6n3983081

= limnrarrinfin

9830802 + 1 +

1

n +

1

3 +

1

2n +

1

6n2

983081 = 2 + 1 + 0 +

1

3 + 0 + 0 =

10

3

18



Example 6

Evaluate using the limit definition of an integralint 30 5xdx

a = 0 b = 3

x = 3minus0

n = 3

n


= 3n

f (x) = 5x


int 3

0 5xdx = limnrarrinfin

n

sumi=1

f (ci)

xi

= limnrarrinfin

nsumi=1

f 9830803i

n

983081middot 3

n

= limnrarrinfin

nsumi=1

59830803i

n

983081middot 3

n

= limnrarrinfin

nsumi=1

45i

n2

= limnrarrinfin

nsum

i=1

45

n2 middot n(n + 1)

2

= limnrarrinfin

nsumi=1

98308045n(n + 1)

2n2

983081

= limnrarrinfin

nsumi=1

98308045n2 + 45n

2n2

983081

= limnrarrinfin

nsumi=1

98308045n2

2n2 +

45n

2n2

983081

= limnrarrinfin

nsumi=1

98308045

2 +

45

2n

983081 =

45

2 + 0 =

45

2

19



44 Definite Integrals

441 Fundamental Theorem of Calculus

If a function f is continuous on a closed interval [a b] and F is an the antiderivative of f onthe interval [a b] then

int b

a f (x)dx = F (b) minus F (a)

When we evaluating a definite integral we first find the antiderivative and then substitutethe limits into the resulting antiderivative

442 Evaluating Integrals

Example 1

Evaluate the following definite integralint 20 4xdx

Solution

int 20

4xdx = 41+1

x220

= 2x2|20 = 2(2)2 minus 2(0)2 = 8 minus 0 = 8

Example 2

Evaluate the following definite integralint 20 x3 + x2dx

Solution

int 21

x3 + x2dx =1048616

13+1

x3+1 + 12+1

x2+1)21

=1048616 14

x4 + 13

x3)21

=104861614

24 + 13

23) minus (14

14 + 13

13)

= 4 + 83 minus 1048616

14

+ 13

)

= 203 minus 1

12

= 7912

Example 3

Evaluate the following definite integralint 20 6x

2

dx

Solution

int 20 6x2dx = 6

2+1x2+1

21

= 2x3|21= 2 middot 23 minus 2 middot 03

= 16 minus 0= 16

20



Example 4

Evaluate the following definite integralint 20

exdx

Solution

int 20

exdx = ex|20= e2 minus e0

= e2 minus 1 asymp 639

Example 5


x+radic x

2 dx

= [1

2 middot x2

2 + 1

2 middot 1

2 middot x

3

2

]10

= [x2

4 + 1

3 middot x

3

2 ]10

= [12

4 + 1

3 middot 1

3

2 ] minus [02

4 + 1

3 middot 0

3

2 ]14 minus 1

3 minus 0 = 7

12

Solution

int 10

x+radic x

2 dx =

int 10

x2

+radic x

2 dx =

443 Integrals of Absolute Value Functions

Example 6

Evaluate the following definite integralint 20 |2x minus 3|dx

Solution

Note

f (x) = 2

xminus 3

xge

3

2

minus(2x minus 3) x lt 32

int 20 |2x minus 3|dx =

int 32

0 minus(2x minus 3)dx +

int 23

2

(2x minus 3)dx

=int 3

2

0 (minus2x + 3)dx +int 2

3

2

(2x minus 3)dx

= (minusx2 + 3x)|3

2

0 + (x2 minus 3x)|232

= [((32

)2 + 3 middot 32

) minus ((minus0)2 + 3 middot 0)] + [(22 minus 3(2)) minus ((32

)2 minus 3 middot 32

)]= 10

4

21



444 Mean Value Theorem

If f is a continuous function on the closed interval [a b] then there exist a number c in theclosed interval [a b] such that

int ba f (x)dx = f (c)(b minus a)

445 The Average Value of a Function

If f is integrable on the closed interval [a b] the average value of the function f is given bythe integral

1bminusa

int b

a f (x)dx

In the next examples we will average value of a function on a given interval

Example 7

Find the average value of the function f (x) = x2 + 2 on the interval [minus2 2]

Solution

AV = 12minus(minus2)

int 2minus2 (x2 + 2)dx

= 14


= 1

4 middot (1

3x3

+ x)2minus2

= 14 middot [1

3(2)3 + 2] minus 1

3 middot [1

3(minus2)3 minus 2]

= 14 middot 14

3 minus 1

3 middot minus14

3

= 76

+ 76

= 146

= 73

Example 8

Find the average value of the function f (x) = cos(x) on the interval [0 π

2

]

Solution

AV = 1π

2minus0int π

2

0 cos(x)dx

= 2π

int π2

0 cos(x)dx

= 2π middot 1048616 sin(x))|

π

2

0

= 2π middot [sin(π

2) minus sin(0)]

= 2π middot [1 minus 0]

= 2π middot 1

= 2

π

22



Example 9

Find the area of the region bounded by the following graphs of equations

y = x2 + 2 y = 0 x = 0 x = 2

Solution

Area=int 20 (x2 + 2)dx

=1048616 x3

3 + 2x)

2

0

= [23

3 + 2(2)] minus [03

3 + 2(0)]= [8

3 + 4] minus [0 + 0]

= 203 minus 0

= 203

23



Example 10


y = ex y = 0 x = 0 x = 2

Solution

Area=int 20 (ex)dx

= [ex]|20

= e2 minus e0

= e2 minus 1

Example 11

Find the value of the transcendental functionint π

0 (1 + cos(x))dx

Solution

int π0 (1 + cos(x))dx =

1048616 x + sin(x))|

π

0

= (π + sin(π)) minus (0 + sin(0))= π + 0 minus 0 = π

24



45 Integration by Substitution

451 Review of the Chain Rule

Recall that with the chain rule we used a basic substitution to find the derivative

For example letrsquos look at the following derivative

Example 1

Find the derivative of the following function f (x) = (x2 + 4x)6

Solution

f (x) = (x2 + 4x)6

let f (x) = u6 where u = x2 + 4x rArr du = 2x + 4Using the chain rule we get the follow derivativef prime(x) = 6u5 middot du

f (x) = 6(x2 + 4x)5(2x + 4)Now we will reverse the process of the chain rule which is integration using substitutionLetrsquos start by examining the following example

452 Basic Substitutions

Example 2

Evaluateint

(2x + 3)(x2 + 3x)6dx

Solution

Let u = x2 + 3x rArr du = (2x + 3)dx

int (2x + 3)(x2 + 3x)6dx =

int u6 middot du

Now find the antiderivativeint u6 middot du = 1

6+1u6+1 + C = 1

7u7 + C = 1

7(x2 + 3x)7 + C

Example 3

Evaluateint

3(1 + 3x)3dx

Solution

Let u = 1 + 3x rArr du = 3dx

int 3(1 + 3x)3dx =

int u3 middot du

25




3+1u3+1 + C = 1

4u4 + C = 1

4(1 + 3x)4 + C

Example 4

Evaluateint

x2(2 + x3)4dx

Solution

Let u = 2 + x3 rArr du = 3x2dx rArr 13

du = x2dxint x2(2 + x3)4dx =

int 13

u4 middot du

Now find the antiderivative

int 13

u4 middot du = 13 middot 1

5u5 + C = 1

15u5 + C = 1

15(2 + x3)5 + C

26



Example 5

Evaluateint

2tradic

t2 + 3dt

Solution

Let u = t2 + 3 rArr du = 2tdtint 2t

radic t2 + 3dt =

int radic u middot du =

int u

1

2 middot du

Now find the antiderivativeint u

1

2 middot du = 23 middot u

3

2 + C = 23

(t2 + 3)3

2 + C = 23

(radic

t2 + 3)3 + C

Example 6

Evaluateint

πcos(πx)dx

Solution

Let u = πx rArr du = πdxint πcosπxdx =

int cos(u)du


int cos(u)du = sin(u) + C = sin(πx) + C

Example 7

Evaluateint

xradic 1minus4x2 dx

Solution

Let u = 1 minus 4x2 rArr du = minus8xdx rArr minusdu8

= xdx

int x

radic 1minus4x2

dx = int minus1

8 middot du

radic uNow find the antiderivativeint

minus18 middot uminus

1

2 = minus18 middot 2u

1

2 + C = minus14

u1

2 + C = minus14

(1 minus 4x2)1

2 + C = minus14

radic 1 minus 4x2 + C

27



Example 8

Evaluateint

2xe2x2

dx

Solution

Let u = 2x2 rArr du = 4xdx rArr du2

= 2xdx

int 2xe2x

2

dx =int

12

eudu

Now find the antiderivativeint 12

eudu = 12

eu + C = 12

e2x2

+ C

Example 9

Evaluateint

2xsin(2x2)dx

Solution

Let u = 2x2 rArr du = 4xdx rArr 12

du = 2xdx

int 2xsin2x2dx =

int 12

sin(u)du


int 12

sin(u)du =minus

1

2cos(u) + C =

minus1

2cos(2x2) + C

28



46 The Natural Logarithm Function

Definition of the antiderivative of the natural function

Let u be a differentiable function of x

1int 1x dx = ln|x| + C

2int 1u

du = ln|x| + C

Example 1

Evaluateint

13x

dx

Solution

Let u = 3x rArr du = 3dx rArr 1

3du = dxint 13x

dx =int

13 middot 1

u du


int 13 middot 1

u du = 1

3ln|u| + C = 1

3ln|3x| + C

Example 2

Evaluateint

15x+3

dx

Solution

Let u = 5x + 3 rArr du = 5dx rArr 15

du = dx

int 13x

dx =int

15 middot 1

u du


int 15 middot 1

u du = 1

5ln|u| + C = 1

5ln|5x + 3| + C

29



Example 3

Evaluateint

x2

3minusx3 dx

Solution

Let u = 3 minus x3 rArr du = minus3x2dx rArr minus13

du = x2dx

int x

2

3minusx3 dx =int

minus13 middot 1

u du


int minus1

3 middot 1

u du = minus1

3ln|u| + C = minus1

3ln|3 minus x3| + C

Example 4

Evaluateint

sinθ

cosθ dθ

Solution

Let u = cosθ rArr du = minussinθ rArr minusdu = sinθint sinθ

cosθ dθ =

int minus 1

u du


int minus 1

udu = minusln|u| + C = minusln|cosθ| + C

Example 5

Evaluateint

cos(t)3+sin(t)

dt

Solution

Let u = 3 + sin(t)

rArr du = cos(t)dt

int cos(t)3+sin(t)

dθ =int

1u

du


int 1u

du = ln|u| + C = ln|3 + sin(t)| + C

30



47 Antiderivative of the Inverse Trigonometric Functions

471 Basic Antiderivatives of the Inverse Trigonometric Functions

Theorem 48

Integrals involving inverse trigonometric functions

Let u be a differentiable function of x and let a gt 0

1int

duradic a2minusu2

= arcsin u

a +C

2int

du

a2+u2 = 1a

arctan u

a +C

3int

du

uradic u2minusa2

= 1a

arcsec |u|a

+C


Here are some example of integration that results in inverse trigonometric functions

Example 1

Evaluateint

41+9x2 middotdx

Solution


3du = dx

Now integrate by substitution

int 41+9x2 middotdx

=int

41+(3x)2

middotdx

= 13

int 4du1+u2

= 43

int du

1+u2

= 4

3arctan(u) + C

= 43

arctan(3x) + C

Example 2

Evaluateint

dx

4+(x+1)2

Solution

Let u = x + 1 rArr du = dx


31



int dx

4+(x+1)2 =int

du

22+u2

= 12

arctan(u2

) + C

= 12

arctan(x+12

) + C

Example 3

Evaluateint

7radic 4minusx2

middotdx

Solution

Let u = x rArr du = dx


int 7

radic 4minusx2

= int 7du

radic 22

minusu2

= 7 middot arcsin(u2

) + C

= 7arcsin(x2

) + C

Example 4

Evaluateint

dx

xmiddotradic x2minus9

Solution



int dx


=int

du

uradic u2minus9

= 13

arcsec(u3

) + C

= 13

arcsec( |x|3

) + C

32





Example 8

Evaluateint

x+5radic 9+(xminus3)2

Solution

Let u = x minus 3 rArr du = dx

Also let v = 9 minus (x minus 3)2

rArr v = 9 minus (x2 minus 6x + 9)rArr v = 6x minus x2

rArr dv = (minus2x + 6)dx

rArr minus12

du = (x minus 3)dx

Now substitute

int x+5radic 9+(xminus3)2

=int

xminus3radic 9+(xminus3)2+int

8radic 9+(xminus3)2

= minus12

int dvradic

v + 8 middot int duradic

9+(u)2

= 12

int vminus 1

2 dv + 8 middot int duradic 9+(u)2

= minus12

(2v1

2 ) + 8arcsin(u3

) + C

= minus(6x minus x2)1

2 + 8arcsin(xminus33

) + C

= 8arcsin(xminus33 ) minus radic 6x minus x2 + C

34



48 Simpsonrsquos Rule and the Trapezoid Rule

481 The Trapezoid Rule

Let be continuous of [a b] The Trapezoid rule approximating the integral

int b

a f (x)dx is

given by the functionint ba

f (x)dx = bminusa

2n [f (x0) + 2f (x1) + 2f (x2) + 2f (x3) + + 2f (xnminus1 + f (xn)]

Example 1

Use the Trapezoid Rule to the approximate value of the definite integral for the given valueof n

int 20

2x2dx n = 4

Solution

x = 2minus04

= 24

= 12

xi = a + i middot x = 0 + i middot 12

= i2

int 20 2x2dx = 2minus0

2(4)[f (0) + 2f (1

2) + 2f (1) + 2f (3

2) + f (2)]

= 28

[2(0)2 + 2(2(12

)2) + 2(2(1)2) + 2(2( 32

)2) + 2(22)]

= 1

4

[0 + 2(1

2

) + 2(2) + 2( 18

4

) + 8]= 1

4[0 + 1 + 4 + 9 + 8]

= 14

[22]= 55

482 Simpsonrsquos Rule

Let be continuous of [a b] The Simpsonrsquos rule approximating the integralint b

a f (x)dx is

given by the function

int b

a f (x)dx = bminusa3n [f (x0) + 4f (x1) + 2f (x2) + 4f (x3) + + 4f (xnminus1 + f (xn)]

Example 2

Use the Trapezoid Rule to the approximate value of the definite integral for the given valueof n (Same definite integral as Example 1)

int 2

0 2x2dx n = 4

Solution

35



x = 2minus04

= 24

= 12


= i2

int 20

2x2dx = 2minus03(4)

[f (0) + 4f (12

) + 2f (1) + 4f (32

) + f (2)]

= 212

[2(0)2 + 4(2(12

)2) + 2(2(1)2) + 4(2(32

)2) + 2(22)]= 1

6[0 + 4(1

2) + 2(2) + 4( 18

4 ) + 8]

= 16

[0 + 2 + 4 + 18 + 8]

= 16

[32]= 53333

Now lets find the exact value of the same definite integral in Example 1 and Example 2 useintegration

int 2

0 2x2dx = 22+1

x2+120

= 23

x320

= 23

(2)3 minus 23

(0)3

= 23 middot 8 minus 0

= 163

Example 3

Use the Trapezoid Rule and Simpsonrsquos Rule to approximate value of the define integral usingn = 4

int π0 sin(x)dx

Solution

x = πminus04

= π4

xi = a + i middot x = 0 + i middot π4

= iπ4int π

0 sin(x)dx = πminus02(4)

[f (0) + 2f (π4

) + 2f (2π4

) + 2f (3π4

) + f (π)]

= π8

[sin(0) + 2sin(π4

) + 2sin(π2

) + 2sin(3π4

) + sin(π)]

= π8 [0 + 2(radic 22 ) + 2(1) + 2(radic 22 ) + 0]

= π8

[2radic

2 + 2]asymp 2003

Now find the value of the integral using Simpsonrsquos Rule

int π0 sin(x)dx

Solution

x = πminus04

= π4

36





= 112

[111 + 4(0947) + 2(0816) + 4(0771) + 0625] asymp 086

Now compare the solutions to value of the definite integral

Let u = x + 2 rArr du = dxint 21

1(x+2)2

dx =int 2

11u2

du =int 2

1 uminus2du = = minusuminus1|21= minus(x + 2)minus1|21= minus 1

x+2

21

= minus 12+2

minus (minus 11+2

)

= minus14

+ 13

= 112



4 Integration

41 Antiderivatives

411 Definition of an Antiderivative

A function F is an antiderivative of f on an interval I if F prime(x) = f (x) for all x on I

The integral or antiderivative of a function is essentially the reverse process of differentiationTo start will look at the most basic antiderivative which the power rule for antiderivativesThe power rule for antiderivatives is the reverse process of the power rule for derivatives

Power Rule for Antiderivatives

int cxndx = c

n+1xn+1 + C

Now letrsquos try doing so examples that use the power rule

Example 1


x2dx

Solution

int x2dx = 1

2+1x2+1 + C = 1

3x3 + C

Example 2


(x4 + 3x2 + 3x)dx

Solution

int (x4 + 3x2 + 3x)dx = 1

4+1x4+1 + 3

2+1x2+1 + 3

1+1x1+1 + C = 1

5x5 + x3 + 3

2x2 + C

Example 3


(x3 minus 5x2 + 6)dx

Solution

int (x3 minus 5x2 + 6x0)dx = 1

3+1x3+1 minus 5

2+1x2+1 + 6

1+0x0+1 + C = 1

4x4 minus 5

3x3 + 6x + C

2



Example 4


5x4dx


4+1x4+1 + C = x5 + C










Example 5



Solution


Example 6


(sin(x) + 3x2)dx

Solution


2+1x2+1 + C = minuscos(x) + x3 + C

3



Example 7



Solution


0+1x0+1 + C = sin(x) + 4x + C



int 1x

dx = ln|x| + C

int exdx = ex + C

Example 8


exdx

Solution

int ex = ex + C

Example 9


( 1x

+ x4)dx

Solution

int ( 1

x + x4) = ln|x| + 1

4+1x4+1 + C = 1

5x5 + C

Example 10


x3

2 dx

Solution

int x

3

2 dx = 13

2+1

x3

2+1 + C = 1

5

2

x5

2 + C = 25

x5

2 + C

4



Example 11


1x3

dx

Solution

int 1x3

dxint




+ C

Example 12


( 1x2

+ cos(x))dx

Solution

int ( 1

x2 + cos(x))dx =



minus 1x

+ sin(x) + C

5



42 Area

421 Sigma Notation


i=1

ai = a1 + a2 + a3 + a4 + + an



Example 1

Evaluate6sum

i=1

i

Solution

6sumi=1

i = 1 + 2 + 3 + 4 + 5 + 6 = 21

Example 2

Evaluate4sum

i=0

(2i + 1)

Solution4sum

i=0

(2i + 1) = 2(0) + 1 + 2(1) + 1 + 2(2) + 1 + 2(3) + 1 + 2(4) + 1 = 1 + 3 + 5 + 7 + 9 = 25


Example 3

Evaluate4sum

j=0

j2

Solution4sum

j=0

j2 = 12 + 22 + 32 + 42 = 1 + 4 + 9 + 16 = 30

6



Example 4

Evaluate6sum

k=3

k(k minus 2)

Solution6sum

k=3


Example 5

Evaluate5sum

j=3

1

j

Solution

5sum j=3

1

j =

1

3 +

1

4 +

1

5 =

47

60

Example 6

Evaluate4

sum j=1

[(i

minus1)2 + (i + 1)3]

Solution4sum

j=1


1)2 +(4+1)3 = 02 + 23 + 12 + 33 + 22 + 43 + 32 + 53 = 0 + 8 + 1 + 27 + 4 + 64 + 9 + 125 = 237

Example 7


1+1 + 5

1+2 +

51+3

+ + 51+12

Solution


j=1

[ 5

1 + i]

7



Example 8


)2] +

[1 minus (24

)2] + 1 minus [(34

)2] + [1 minus (44

)2]


j=1

[1 + (i

4)2]


1usum

j=1

c = cn

2usum

j=1

i = n(n+1)

2

3usum

j=1

i2 = n(n+1)(2n+1)

6

4usum

j=1

i3 = n

2(n+1)2

4


Example 9


15sum

j=1

(2i minus 3)

Solution15sum

j=1

(2i minus 3) =15sum

j=1

2i minus15sum

j=1

3 = 2 middot15sum

j=1

i minus15sum

j=1

3 = 2 middot 15(16)

2 minus 15(3) = 240 minus 45 = 195

8



Example 10


15

sum j=1

i(i2 + 1)

Solution10sum

j=1

i(i2 + 1) =10sum

j=1

i3 + i

= n

2(n+1)2

4 + n(n+1)

2 (where n=10)

= 102(10+1)2

4 + 10(10+1)

2


2= 25

middot121 + 5

middot11

= 3025 + 55 = 3080





limhrarrinfin




n

9



Example 10



Solution



= 14



= iminus14


= i

4


S (n) =4sum

j=1

f (mi)x

=4

sumi=1

f (iminus 1

4 ) middot (

1

4)

=4sum

i=1

radic iminus 1

4 middot 1

4

=4sum

i=1

radic iminus 1

2 middot 1

4

=4sum

i=1

radic iminus 1

8

=radic 1minus18 +

radic 2minus18 +

radic 3minus18 +

radic 4minus18

=

radic 0

8 +

radic 1

8 +

radic 2

8 +

radic 3

8 = 0 + 125 + 177 + 217 asymp 519

10




S (n) =4sum

j=1

f (M i)x

=4sum

i=1

f (i

4) middot (

1

4)

=4sum

i=1

radic i

4middot 1

4

=

4

sumi=1

radic i

2 middot1

4

=4sum

i=1

radic i

8

=radic 18

+radic 28

+radic 38

+radic 48

= 125 + 177 + 217 + 25 asymp 769

Example 11



n3[n(n+1)(2n+1)

6 ]

Solution

limhrarrinfin


64

n3[n(n + 1)(2n + 1)

6 ]

= limhrarrinfin

64

n3[(n2 + n))(2n + 1)

6 ]

= limhrarrinfin

64

n3[(2n3 + n2 + 2n2 + n

6 ]

= limhrarrinfin

64[2n3 + 3n2 + n

6n3 ]

= limhrarrinfin

128n3

+ 192n2

+ 64n6n3

11



= limhrarrinfin

128n3

6n3 +

192n2

6n3 +

64n

6n3

= limhrarrinfin

64

3 +

32

n +

32

3n2

= 643

+ 0 + 0 = 643

12



43 Riemann Sum




lexi

le1


nsumi=1

f (ci)xi




xi = i2


= i2


n2


n2

= i2minusi

2minus2i+1)n2

= minus2i+1n2


limhrarrinfin

nsumi=1


nsumi=1

f ( i2

n2) middot (

2iminus 1

n2 )

= limhrarrinfin

nsumi=1

radic i2

n2 middot (

2iminus 1

n2 )

= limhrarrinfin

nsumi=1

i

nmiddot 2iminus 1

n2

= limhrarrinfin

nsumi=1

2i2

minusi

n3

13



= 1

n3 limhrarrinfin

nsumi=1

(2i2 minus i)

= 1

n3 limhrarrinfin

[2(n(n + 1)(2n + 1)

6 ) minus n(n + 1)

2 ]

= 1


6 minus n2 + n

2 ]

= 1

n3 limhrarrinfin

((2n2 + 2n)(2n + 1)

6 minus n2 + n

2 ]

= 1

n3 limhrarrinfin

(4n3 + 6n2 + 2n

6 minus n2 + n

2 ]

= 1

n3 limhrarrinfin


6

= 1

n3 limhrarrinfin

4n3 + 3n2 minus n

6n3

= limhrarrinfin

[4n3 + 3n2 minus n

6n3

]

= limhrarrinfin

4n3

6n3 +

3n2

6n3 +

n

6n3

= limhrarrinfin

2

3 +

1

2n +

1

6n2 =

2

3 + 0 + 0 =

2

3




i=1




nsumi=1



nsumi=1

f (ci)xi =int

b

a f (x)dx

14



Example 2


Solution

int 3minus2

xdx

Solution


n = 3minus(minus2)

n


nrarrinfin

nsum

i=1

f (ci)xi

= limnrarrinfin

nsum

i=1

1048616minus 2 +

5i

n

1048617middot 5

n

= limnrarrinfin

nsum

i=1

1048616minus10

n +

25i

n2

1048617

= limnrarrinfin

nsum

i=1

1048616minus10

n +

25

n2 middot n(n + 1)

2

1048617

= limnrarrinfin

nsumi=1

1048616minus10n

+ 25n2

+ 25n2n2

1048617

= limnrarrinfin

nsum

i=1

1048616minus10

n +

25n2

2n2 +

25n

2n21048617

= limnrarrinfin

nsum

i=1

1048616minus10

n +

25

2 +

25

2n

1048617 = minus0 + 125 + 0 = 125

Example 3


Solution

15



int 31 3x2dx

Solution

a = 1 b = 3 x = 3minus1n

= 2n

ci = 1 + i middot 2i = 1 + 2i

n


nrarrinfin

nsum

i=1

f (ci)xi

= limnrarrinfin

nsum

i=1

f (1 + 2i

n ) middot

2

n

= limnrarrinfin

nsum

i=1

3(1 + 2i

n )2 middot

2

n

= limnrarrinfin

nsum

i=1

3(4i2

n2 +

4i

n + 1) middot

2

n

= limnrarrinfin

nsum

i=1

3( 8i2

n3 + 8i

n2 + 2

n)

= limnrarrinfin

nsum

i=1

(24i2

n3 +

24i

n2 +

6

n)

= limnrarrinfin

nsum

i=1

104861624(n(n + 1)(2n + 1))

6n3 +

24n(n + 1)

2n2 +

6n

n

1048617

= limnrarrinfin

nsum

i=1

104861624(2n3 + 3n2 + n)

6n3 +

24n2 + 24n

2n2 +

6n

n

1048617

= limnrarrinfin

n

sumi=1

104861648n3

6n3 + 72n2

6n3 + 24n

6n3 + 24n2

2n2 + 24n

n2 + 61048617

16



=1048616

8 + 12

n +

4

n2 + 12 +

12

n + 6

1048617 = 26

Example 4


Solution

a = 4 b = 10 x = 10minus4n

= 6n


= 4 + 6in

f (x) = 6


nrarrinfin

nsumi=1

f (ci)xi

= limnrarrinfin

nsumi=1

f 983080

4 + 6i

n

983081middot 6

n

= limnrarrinfin

nsumi=1

(6) middot 6

n

= limn

rarrinfin(6n)

middot

6

n = 36

17



Example 5


a = 1 b = 2 x = 2minus1n

= 1n


= 1 + i

n

f (x) = x2 + 1


int 21 (x2 + 1)dx = lim

nrarrinfin

nsumi=1

f (ci)xi

= limnrarrinfin

n

sumi=1 f

9830801 +

i

n983081 middot

1

n

= limnrarrinfin

nsumi=1

9830809830801 +

i

n

9830812+ 1

983081middot 1

n

= limnrarrinfin

nsumi=1

9830809830801 +

i

n

9830819830801 +

i

n

983081+ 1

983081middot 1

n

= limnrarrinfin

nsumi=1

98308010486161 +

2i

n +

i2

n2

983081+ 1

983081middot 1

n

= limnrarrinfin

n

sumi=1

9830802 + 2i

n

+ i2

n2983081 middot

1

n

= limnrarrinfin

nsumi=1

9830802

n +

2i

n2 +

i2

n3

983081

= limnrarrinfin

983080 2

n middot (n) +

2

n2 middot n(n + 1)

2 +

1

n3 middot n(n + 1)(2n + 1)

6

983081

= limnrarrinfin

9830802 +

2n2 + 2n

2n2 +

(n2 + n)(2n + 1)

6n3

983081

= limnrarrinfin

9830802 +

2n2 + 2n

2n2 +

2n3 + 3n2 + n

6n3

983081


2 + 2n2

2n2

+ 2n

2n2

+ 2n3

6n3

+ 3n2

6n3

+ n

6n3983081

= limnrarrinfin

9830802 + 1 +

1

n +

1

3 +

1

2n +

1

6n2

983081 = 2 + 1 + 0 +

1

3 + 0 + 0 =

10

3

18



Example 6


a = 0 b = 3

x = 3minus0

n = 3

n


= 3n

f (x) = 5x


int 3


n

sumi=1

f (ci)

xi

= limnrarrinfin

nsumi=1

f 9830803i

n

983081middot 3

n

= limnrarrinfin

nsumi=1

59830803i

n

983081middot 3

n

= limnrarrinfin

nsumi=1

45i

n2

= limnrarrinfin

nsum

i=1

45

n2 middot n(n + 1)

2

= limnrarrinfin

nsumi=1

98308045n(n + 1)

2n2

983081

= limnrarrinfin

nsumi=1

98308045n2 + 45n

2n2

983081

= limnrarrinfin

nsumi=1

98308045n2

2n2 +

45n

2n2

983081

= limnrarrinfin

nsumi=1

98308045

2 +

45

2n

983081 =

45

2 + 0 =

45

2

19






int b




Example 1


Solution

int 20

4xdx = 41+1

x220

= 2x2|20 = 2(2)2 minus 2(0)2 = 8 minus 0 = 8

Example 2


Solution

int 21

x3 + x2dx =1048616

13+1

x3+1 + 12+1

x2+1)21

=1048616 14

x4 + 13

x3)21

=104861614

24 + 13

23) minus (14

14 + 13

13)

= 4 + 83 minus 1048616

14

+ 13

)

= 203 minus 1

12

= 7912

Example 3


2

dx

Solution

int 20 6x2dx = 6

2+1x2+1

21


= 16 minus 0= 16

20



Example 4


exdx

Solution

int 20



Example 5


x+radic x

2 dx

= [1

2 middot x2

2 + 1

2 middot 1

2 middot x

3

2

]10

= [x2

4 + 1

3 middot x

3

2 ]10

= [12

4 + 1

3 middot 1

3

2 ] minus [02

4 + 1

3 middot 0

3

2 ]14 minus 1

3 minus 0 = 7

12

Solution

int 10

x+radic x

2 dx =

int 10

x2

+radic x

2 dx =


Example 6


Solution

Note

f (x) = 2

xminus 3

xge

3

2



int 32


int 23

2

(2x minus 3)dx

=int 3

2


3

2

(2x minus 3)dx

= (minusx2 + 3x)|3

2

0 + (x2 minus 3x)|232

= [((32

)2 + 3 middot 32



)]= 10

4

21








1bminusa

int b

a f (x)dx


Example 7


Solution



= 14


= 1

4 middot (1

3x3

+ x)2minus2

= 14 middot [1

3(2)3 + 2] minus 1

3 middot [1

3(minus2)3 minus 2]

= 14 middot 14

3 minus 1

3 middot minus14

3

= 76

+ 76

= 146

= 73

Example 8


2

]

Solution

AV = 1π

2minus0int π

2

0 cos(x)dx

= 2π

int π2

0 cos(x)dx

= 2π middot 1048616 sin(x))|

π

2

0


2) minus sin(0)]


= 2π middot 1

= 2

π

22



Example 9


y = x2 + 2 y = 0 x = 0 x = 2

Solution


=1048616 x3

3 + 2x)

2

0

= [23

3 + 2(2)] minus [03

3 + 2(0)]= [8

3 + 4] minus [0 + 0]

= 203 minus 0

= 203

23



Example 10


y = ex y = 0 x = 0 x = 2

Solution

Area=int 20 (ex)dx

= [ex]|20

= e2 minus e0

= e2 minus 1

Example 11


0 (1 + cos(x))dx

Solution


1048616 x + sin(x))|

π

0


24







Example 1


Solution

f (x) = (x2 + 4x)6




Example 2

Evaluateint

(2x + 3)(x2 + 3x)6dx

Solution


int (2x + 3)(x2 + 3x)6dx =

int u6 middot du


6+1u6+1 + C = 1

7u7 + C = 1

7(x2 + 3x)7 + C

Example 3

Evaluateint

3(1 + 3x)3dx

Solution


int 3(1 + 3x)3dx =

int u3 middot du

25




3+1u3+1 + C = 1

4u4 + C = 1

4(1 + 3x)4 + C

Example 4

Evaluateint

x2(2 + x3)4dx

Solution



int 13

u4 middot du


int 13


5u5 + C = 1

15u5 + C = 1

15(2 + x3)5 + C

26



Example 5

Evaluateint

2tradic

t2 + 3dt

Solution


radic t2 + 3dt =


int u

1

2 middot du


1


3

2 + C = 23

(t2 + 3)3

2 + C = 23

(radic

t2 + 3)3 + C

Example 6

Evaluateint

πcos(πx)dx

Solution


int cos(u)du



Example 7

Evaluateint

xradic 1minus4x2 dx

Solution


= xdx

int x

radic 1minus4x2

dx = int minus1

8 middot du



1


1

2 + C = minus14

u1

2 + C = minus14

(1 minus 4x2)1

2 + C = minus14


27



Example 8

Evaluateint

2xe2x2

dx

Solution


= 2xdx

int 2xe2x

2

dx =int

12

eudu


eudu = 12

eu + C = 12

e2x2

+ C

Example 9

Evaluateint

2xsin(2x2)dx

Solution


du = 2xdx

int 2xsin2x2dx =

int 12

sin(u)du


int 12

sin(u)du =minus

1

2cos(u) + C =

minus1

2cos(2x2) + C

28







2int 1u

du = ln|x| + C

Example 1

Evaluateint

13x

dx

Solution


3du = dxint 13x

dx =int

13 middot 1

u du


int 13 middot 1

u du = 1

3ln|u| + C = 1

3ln|3x| + C

Example 2

Evaluateint

15x+3

dx

Solution


du = dx

int 13x

dx =int

15 middot 1

u du


int 15 middot 1

u du = 1

5ln|u| + C = 1

5ln|5x + 3| + C

29



Example 3

Evaluateint

x2

3minusx3 dx

Solution


du = x2dx

int x

2

3minusx3 dx =int

minus13 middot 1

u du


int minus1

3 middot 1

u du = minus1

3ln|u| + C = minus1

3ln|3 minus x3| + C

Example 4

Evaluateint

sinθ

cosθ dθ

Solution


cosθ dθ =

int minus 1

u du


int minus 1


Example 5

Evaluateint

cos(t)3+sin(t)

dt

Solution

Let u = 3 + sin(t)

rArr du = cos(t)dt

int cos(t)3+sin(t)

dθ =int

1u

du


int 1u

du = ln|u| + C = ln|3 + sin(t)| + C

30





Theorem 48



1int

duradic a2minusu2

= arcsin u

a +C

2int

du

a2+u2 = 1a

arctan u

a +C

3int

du

uradic u2minusa2

= 1a

arcsec |u|a

+C



Example 1

Evaluateint

41+9x2 middotdx

Solution


3du = dx


int 41+9x2 middotdx

=int

41+(3x)2

middotdx

= 13

int 4du1+u2

= 43

int du

1+u2

= 4

3arctan(u) + C

= 43

arctan(3x) + C

Example 2

Evaluateint

dx

4+(x+1)2

Solution



31



int dx

4+(x+1)2 =int

du

22+u2

= 12

arctan(u2

) + C

= 12

arctan(x+12

) + C

Example 3

Evaluateint

7radic 4minusx2

middotdx

Solution



int 7

radic 4minusx2

= int 7du

radic 22

minusu2


) + C

= 7arcsin(x2

) + C

Example 4

Evaluateint

dx


Solution



int dx


=int

du

uradic u2minus9

= 13

arcsec(u3

) + C

= 13

arcsec( |x|3

) + C

32





Example 8

Evaluateint


Solution





rArr minus12

du = (x minus 3)dx

Now substitute


=int


8radic 9+(xminus3)2

= minus12

int dvradic


9+(u)2

= 12

int vminus 1


= minus12

(2v1

2 ) + 8arcsin(u3

) + C



) + C


34






int b

a f (x)dx is


f (x)dx = bminusa


Example 1


int 20

2x2dx n = 4

Solution

x = 2minus04

= 24

= 12


= i2


2(4)[f (0) + 2f (1

2) + 2f (1) + 2f (3

2) + f (2)]

= 28

[2(0)2 + 2(2(12

)2) + 2(2(1)2) + 2(2( 32

)2) + 2(22)]

= 1

4

[0 + 2(1

2

) + 2(2) + 2( 18

4

) + 8]= 1

4[0 + 1 + 4 + 9 + 8]

= 14

[22]= 55



a f (x)dx is


int b


Example 2


int 2

0 2x2dx n = 4

Solution

35



x = 2minus04

= 24

= 12


= i2

int 20

2x2dx = 2minus03(4)

[f (0) + 4f (12

) + 2f (1) + 4f (32

) + f (2)]

= 212

[2(0)2 + 4(2(12

)2) + 2(2(1)2) + 4(2(32

)2) + 2(22)]= 1

6[0 + 4(1

2) + 2(2) + 4( 18

4 ) + 8]

= 16

[0 + 2 + 4 + 18 + 8]

= 16

[32]= 53333


int 2

0 2x2dx = 22+1

x2+120

= 23

x320

= 23

(2)3 minus 23

(0)3


= 163

Example 3


int π0 sin(x)dx

Solution

x = πminus04

= π4


= iπ4int π


[f (0) + 2f (π4

) + 2f (2π4

) + 2f (3π4

) + f (π)]

= π8

[sin(0) + 2sin(π4

) + 2sin(π2

) + 2sin(3π4

) + sin(π)]

= π8 [0 + 2(radic 22 ) + 2(1) + 2(radic 22 ) + 0]

= π8

[2radic

2 + 2]asymp 2003


int π0 sin(x)dx

Solution

x = πminus04

= π4

36





= 112

[111 + 4(0947) + 2(0816) + 4(0771) + 0625] asymp 086



1(x+2)2

dx =int 2

11u2

du =int 2


x+2

21

= minus 12+2

minus (minus 11+2

)

= minus14

+ 13

= 112



Example 4


5x4dx


4+1x4+1 + C = x5 + C










Example 5



Solution


Example 6


(sin(x) + 3x2)dx

Solution


2+1x2+1 + C = minuscos(x) + x3 + C

3



Example 7



Solution


0+1x0+1 + C = sin(x) + 4x + C



int 1x

dx = ln|x| + C

int exdx = ex + C

Example 8


exdx

Solution

int ex = ex + C

Example 9


( 1x

+ x4)dx

Solution

int ( 1

x + x4) = ln|x| + 1

4+1x4+1 + C = 1

5x5 + C

Example 10


x3

2 dx

Solution

int x

3

2 dx = 13

2+1

x3

2+1 + C = 1

5

2

x5

2 + C = 25

x5

2 + C

4



Example 11


1x3

dx

Solution

int 1x3

dxint




+ C

Example 12


( 1x2

+ cos(x))dx

Solution

int ( 1

x2 + cos(x))dx =



minus 1x

+ sin(x) + C

5



42 Area

421 Sigma Notation


i=1

ai = a1 + a2 + a3 + a4 + + an



Example 1

Evaluate6sum

i=1

i

Solution

6sumi=1

i = 1 + 2 + 3 + 4 + 5 + 6 = 21

Example 2

Evaluate4sum

i=0

(2i + 1)

Solution4sum

i=0

(2i + 1) = 2(0) + 1 + 2(1) + 1 + 2(2) + 1 + 2(3) + 1 + 2(4) + 1 = 1 + 3 + 5 + 7 + 9 = 25


Example 3

Evaluate4sum

j=0

j2

Solution4sum

j=0

j2 = 12 + 22 + 32 + 42 = 1 + 4 + 9 + 16 = 30

6



Example 4

Evaluate6sum

k=3

k(k minus 2)

Solution6sum

k=3


Example 5

Evaluate5sum

j=3

1

j

Solution

5sum j=3

1

j =

1

3 +

1

4 +

1

5 =

47

60

Example 6

Evaluate4

sum j=1

[(i

minus1)2 + (i + 1)3]

Solution4sum

j=1


1)2 +(4+1)3 = 02 + 23 + 12 + 33 + 22 + 43 + 32 + 53 = 0 + 8 + 1 + 27 + 4 + 64 + 9 + 125 = 237

Example 7


1+1 + 5

1+2 +

51+3

+ + 51+12

Solution


j=1

[ 5

1 + i]

7



Example 8


)2] +

[1 minus (24

)2] + 1 minus [(34

)2] + [1 minus (44

)2]


j=1

[1 + (i

4)2]


1usum

j=1

c = cn

2usum

j=1

i = n(n+1)

2

3usum

j=1

i2 = n(n+1)(2n+1)

6

4usum

j=1

i3 = n

2(n+1)2

4


Example 9


15sum

j=1

(2i minus 3)

Solution15sum

j=1

(2i minus 3) =15sum

j=1

2i minus15sum

j=1

3 = 2 middot15sum

j=1

i minus15sum

j=1

3 = 2 middot 15(16)

2 minus 15(3) = 240 minus 45 = 195

8



Example 10


15

sum j=1

i(i2 + 1)

Solution10sum

j=1

i(i2 + 1) =10sum

j=1

i3 + i

= n

2(n+1)2

4 + n(n+1)

2 (where n=10)

= 102(10+1)2

4 + 10(10+1)

2


2= 25

middot121 + 5

middot11

= 3025 + 55 = 3080





limhrarrinfin




n

9



Example 10



Solution



= 14



= iminus14


= i

4


S (n) =4sum

j=1

f (mi)x

=4

sumi=1

f (iminus 1

4 ) middot (

1

4)

=4sum

i=1

radic iminus 1

4 middot 1

4

=4sum

i=1

radic iminus 1

2 middot 1

4

=4sum

i=1

radic iminus 1

8

=radic 1minus18 +

radic 2minus18 +

radic 3minus18 +

radic 4minus18

=

radic 0

8 +

radic 1

8 +

radic 2

8 +

radic 3

8 = 0 + 125 + 177 + 217 asymp 519

10




S (n) =4sum

j=1

f (M i)x

=4sum

i=1

f (i

4) middot (

1

4)

=4sum

i=1

radic i

4middot 1

4

=

4

sumi=1

radic i

2 middot1

4

=4sum

i=1

radic i

8

=radic 18

+radic 28

+radic 38

+radic 48

= 125 + 177 + 217 + 25 asymp 769

Example 11



n3[n(n+1)(2n+1)

6 ]

Solution

limhrarrinfin


64

n3[n(n + 1)(2n + 1)

6 ]

= limhrarrinfin

64

n3[(n2 + n))(2n + 1)

6 ]

= limhrarrinfin

64

n3[(2n3 + n2 + 2n2 + n

6 ]

= limhrarrinfin

64[2n3 + 3n2 + n

6n3 ]

= limhrarrinfin

128n3

+ 192n2

+ 64n6n3

11



= limhrarrinfin

128n3

6n3 +

192n2

6n3 +

64n

6n3

= limhrarrinfin

64

3 +

32

n +

32

3n2

= 643

+ 0 + 0 = 643

12



43 Riemann Sum




lexi

le1


nsumi=1

f (ci)xi




xi = i2


= i2


n2


n2

= i2minusi

2minus2i+1)n2

= minus2i+1n2


limhrarrinfin

nsumi=1


nsumi=1

f ( i2

n2) middot (

2iminus 1

n2 )

= limhrarrinfin

nsumi=1

radic i2

n2 middot (

2iminus 1

n2 )

= limhrarrinfin

nsumi=1

i

nmiddot 2iminus 1

n2

= limhrarrinfin

nsumi=1

2i2

minusi

n3

13



= 1

n3 limhrarrinfin

nsumi=1

(2i2 minus i)

= 1

n3 limhrarrinfin

[2(n(n + 1)(2n + 1)

6 ) minus n(n + 1)

2 ]

= 1


6 minus n2 + n

2 ]

= 1

n3 limhrarrinfin

((2n2 + 2n)(2n + 1)

6 minus n2 + n

2 ]

= 1

n3 limhrarrinfin

(4n3 + 6n2 + 2n

6 minus n2 + n

2 ]

= 1

n3 limhrarrinfin


6

= 1

n3 limhrarrinfin

4n3 + 3n2 minus n

6n3

= limhrarrinfin

[4n3 + 3n2 minus n

6n3

]

= limhrarrinfin

4n3

6n3 +

3n2

6n3 +

n

6n3

= limhrarrinfin

2

3 +

1

2n +

1

6n2 =

2

3 + 0 + 0 =

2

3




i=1




nsumi=1



nsumi=1

f (ci)xi =int

b

a f (x)dx

14



Example 2


Solution

int 3minus2

xdx

Solution


n = 3minus(minus2)

n


nrarrinfin

nsum

i=1

f (ci)xi

= limnrarrinfin

nsum

i=1

1048616minus 2 +

5i

n

1048617middot 5

n

= limnrarrinfin

nsum

i=1

1048616minus10

n +

25i

n2

1048617

= limnrarrinfin

nsum

i=1

1048616minus10

n +

25

n2 middot n(n + 1)

2

1048617

= limnrarrinfin

nsumi=1

1048616minus10n

+ 25n2

+ 25n2n2

1048617

= limnrarrinfin

nsum

i=1

1048616minus10

n +

25n2

2n2 +

25n

2n21048617

= limnrarrinfin

nsum

i=1

1048616minus10

n +

25

2 +

25

2n

1048617 = minus0 + 125 + 0 = 125

Example 3


Solution

15



int 31 3x2dx

Solution

a = 1 b = 3 x = 3minus1n

= 2n

ci = 1 + i middot 2i = 1 + 2i

n


nrarrinfin

nsum

i=1

f (ci)xi

= limnrarrinfin

nsum

i=1

f (1 + 2i

n ) middot

2

n

= limnrarrinfin

nsum

i=1

3(1 + 2i

n )2 middot

2

n

= limnrarrinfin

nsum

i=1

3(4i2

n2 +

4i

n + 1) middot

2

n

= limnrarrinfin

nsum

i=1

3( 8i2

n3 + 8i

n2 + 2

n)

= limnrarrinfin

nsum

i=1

(24i2

n3 +

24i

n2 +

6

n)

= limnrarrinfin

nsum

i=1

104861624(n(n + 1)(2n + 1))

6n3 +

24n(n + 1)

2n2 +

6n

n

1048617

= limnrarrinfin

nsum

i=1

104861624(2n3 + 3n2 + n)

6n3 +

24n2 + 24n

2n2 +

6n

n

1048617

= limnrarrinfin

n

sumi=1

104861648n3

6n3 + 72n2

6n3 + 24n

6n3 + 24n2

2n2 + 24n

n2 + 61048617

16



=1048616

8 + 12

n +

4

n2 + 12 +

12

n + 6

1048617 = 26

Example 4


Solution

a = 4 b = 10 x = 10minus4n

= 6n


= 4 + 6in

f (x) = 6


nrarrinfin

nsumi=1

f (ci)xi

= limnrarrinfin

nsumi=1

f 983080

4 + 6i

n

983081middot 6

n

= limnrarrinfin

nsumi=1

(6) middot 6

n

= limn

rarrinfin(6n)

middot

6

n = 36

17



Example 5


a = 1 b = 2 x = 2minus1n

= 1n


= 1 + i

n

f (x) = x2 + 1


int 21 (x2 + 1)dx = lim

nrarrinfin

nsumi=1

f (ci)xi

= limnrarrinfin

n

sumi=1 f

9830801 +

i

n983081 middot

1

n

= limnrarrinfin

nsumi=1

9830809830801 +

i

n

9830812+ 1

983081middot 1

n

= limnrarrinfin

nsumi=1

9830809830801 +

i

n

9830819830801 +

i

n

983081+ 1

983081middot 1

n

= limnrarrinfin

nsumi=1

98308010486161 +

2i

n +

i2

n2

983081+ 1

983081middot 1

n

= limnrarrinfin

n

sumi=1

9830802 + 2i

n

+ i2

n2983081 middot

1

n

= limnrarrinfin

nsumi=1

9830802

n +

2i

n2 +

i2

n3

983081

= limnrarrinfin

983080 2

n middot (n) +

2

n2 middot n(n + 1)

2 +

1

n3 middot n(n + 1)(2n + 1)

6

983081

= limnrarrinfin

9830802 +

2n2 + 2n

2n2 +

(n2 + n)(2n + 1)

6n3

983081

= limnrarrinfin

9830802 +

2n2 + 2n

2n2 +

2n3 + 3n2 + n

6n3

983081


2 + 2n2

2n2

+ 2n

2n2

+ 2n3

6n3

+ 3n2

6n3

+ n

6n3983081

= limnrarrinfin

9830802 + 1 +

1

n +

1

3 +

1

2n +

1

6n2

983081 = 2 + 1 + 0 +

1

3 + 0 + 0 =

10

3

18



Example 6


a = 0 b = 3

x = 3minus0

n = 3

n


= 3n

f (x) = 5x


int 3


n

sumi=1

f (ci)

xi

= limnrarrinfin

nsumi=1

f 9830803i

n

983081middot 3

n

= limnrarrinfin

nsumi=1

59830803i

n

983081middot 3

n

= limnrarrinfin

nsumi=1

45i

n2

= limnrarrinfin

nsum

i=1

45

n2 middot n(n + 1)

2

= limnrarrinfin

nsumi=1

98308045n(n + 1)

2n2

983081

= limnrarrinfin

nsumi=1

98308045n2 + 45n

2n2

983081

= limnrarrinfin

nsumi=1

98308045n2

2n2 +

45n

2n2

983081

= limnrarrinfin

nsumi=1

98308045

2 +

45

2n

983081 =

45

2 + 0 =

45

2

19






int b




Example 1


Solution

int 20

4xdx = 41+1

x220

= 2x2|20 = 2(2)2 minus 2(0)2 = 8 minus 0 = 8

Example 2


Solution

int 21

x3 + x2dx =1048616

13+1

x3+1 + 12+1

x2+1)21

=1048616 14

x4 + 13

x3)21

=104861614

24 + 13

23) minus (14

14 + 13

13)

= 4 + 83 minus 1048616

14

+ 13

)

= 203 minus 1

12

= 7912

Example 3


2

dx

Solution

int 20 6x2dx = 6

2+1x2+1

21


= 16 minus 0= 16

20



Example 4


exdx

Solution

int 20



Example 5


x+radic x

2 dx

= [1

2 middot x2

2 + 1

2 middot 1

2 middot x

3

2

]10

= [x2

4 + 1

3 middot x

3

2 ]10

= [12

4 + 1

3 middot 1

3

2 ] minus [02

4 + 1

3 middot 0

3

2 ]14 minus 1

3 minus 0 = 7

12

Solution

int 10

x+radic x

2 dx =

int 10

x2

+radic x

2 dx =


Example 6


Solution

Note

f (x) = 2

xminus 3

xge

3

2



int 32


int 23

2

(2x minus 3)dx

=int 3

2


3

2

(2x minus 3)dx

= (minusx2 + 3x)|3

2

0 + (x2 minus 3x)|232

= [((32

)2 + 3 middot 32



)]= 10

4

21








1bminusa

int b

a f (x)dx


Example 7


Solution



= 14


= 1

4 middot (1

3x3

+ x)2minus2

= 14 middot [1

3(2)3 + 2] minus 1

3 middot [1

3(minus2)3 minus 2]

= 14 middot 14

3 minus 1

3 middot minus14

3

= 76

+ 76

= 146

= 73

Example 8


2

]

Solution

AV = 1π

2minus0int π

2

0 cos(x)dx

= 2π

int π2

0 cos(x)dx

= 2π middot 1048616 sin(x))|

π

2

0


2) minus sin(0)]


= 2π middot 1

= 2

π

22



Example 9


y = x2 + 2 y = 0 x = 0 x = 2

Solution


=1048616 x3

3 + 2x)

2

0

= [23

3 + 2(2)] minus [03

3 + 2(0)]= [8

3 + 4] minus [0 + 0]

= 203 minus 0

= 203

23



Example 10


y = ex y = 0 x = 0 x = 2

Solution

Area=int 20 (ex)dx

= [ex]|20

= e2 minus e0

= e2 minus 1

Example 11


0 (1 + cos(x))dx

Solution


1048616 x + sin(x))|

π

0


24







Example 1


Solution

f (x) = (x2 + 4x)6




Example 2

Evaluateint

(2x + 3)(x2 + 3x)6dx

Solution


int (2x + 3)(x2 + 3x)6dx =

int u6 middot du


6+1u6+1 + C = 1

7u7 + C = 1

7(x2 + 3x)7 + C

Example 3

Evaluateint

3(1 + 3x)3dx

Solution


int 3(1 + 3x)3dx =

int u3 middot du

25




3+1u3+1 + C = 1

4u4 + C = 1

4(1 + 3x)4 + C

Example 4

Evaluateint

x2(2 + x3)4dx

Solution



int 13

u4 middot du


int 13


5u5 + C = 1

15u5 + C = 1

15(2 + x3)5 + C

26



Example 5

Evaluateint

2tradic

t2 + 3dt

Solution


radic t2 + 3dt =


int u

1

2 middot du


1


3

2 + C = 23

(t2 + 3)3

2 + C = 23

(radic

t2 + 3)3 + C

Example 6

Evaluateint

πcos(πx)dx

Solution


int cos(u)du



Example 7

Evaluateint

xradic 1minus4x2 dx

Solution


= xdx

int x

radic 1minus4x2

dx = int minus1

8 middot du



1


1

2 + C = minus14

u1

2 + C = minus14

(1 minus 4x2)1

2 + C = minus14


27



Example 8

Evaluateint

2xe2x2

dx

Solution


= 2xdx

int 2xe2x

2

dx =int

12

eudu


eudu = 12

eu + C = 12

e2x2

+ C

Example 9

Evaluateint

2xsin(2x2)dx

Solution


du = 2xdx

int 2xsin2x2dx =

int 12

sin(u)du


int 12

sin(u)du =minus

1

2cos(u) + C =

minus1

2cos(2x2) + C

28







2int 1u

du = ln|x| + C

Example 1

Evaluateint

13x

dx

Solution


3du = dxint 13x

dx =int

13 middot 1

u du


int 13 middot 1

u du = 1

3ln|u| + C = 1

3ln|3x| + C

Example 2

Evaluateint

15x+3

dx

Solution


du = dx

int 13x

dx =int

15 middot 1

u du


int 15 middot 1

u du = 1

5ln|u| + C = 1

5ln|5x + 3| + C

29



Example 3

Evaluateint

x2

3minusx3 dx

Solution


du = x2dx

int x

2

3minusx3 dx =int

minus13 middot 1

u du


int minus1

3 middot 1

u du = minus1

3ln|u| + C = minus1

3ln|3 minus x3| + C

Example 4

Evaluateint

sinθ

cosθ dθ

Solution


cosθ dθ =

int minus 1

u du


int minus 1


Example 5

Evaluateint

cos(t)3+sin(t)

dt

Solution

Let u = 3 + sin(t)

rArr du = cos(t)dt

int cos(t)3+sin(t)

dθ =int

1u

du


int 1u

du = ln|u| + C = ln|3 + sin(t)| + C

30





Theorem 48



1int

duradic a2minusu2

= arcsin u

a +C

2int

du

a2+u2 = 1a

arctan u

a +C

3int

du

uradic u2minusa2

= 1a

arcsec |u|a

+C



Example 1

Evaluateint

41+9x2 middotdx

Solution


3du = dx


int 41+9x2 middotdx

=int

41+(3x)2

middotdx

= 13

int 4du1+u2

= 43

int du

1+u2

= 4

3arctan(u) + C

= 43

arctan(3x) + C

Example 2

Evaluateint

dx

4+(x+1)2

Solution



31



int dx

4+(x+1)2 =int

du

22+u2

= 12

arctan(u2

) + C

= 12

arctan(x+12

) + C

Example 3

Evaluateint

7radic 4minusx2

middotdx

Solution



int 7

radic 4minusx2

= int 7du

radic 22

minusu2


) + C

= 7arcsin(x2

) + C

Example 4

Evaluateint

dx


Solution



int dx


=int

du

uradic u2minus9

= 13

arcsec(u3

) + C

= 13

arcsec( |x|3

) + C

32





Example 8

Evaluateint


Solution





rArr minus12

du = (x minus 3)dx

Now substitute


=int


8radic 9+(xminus3)2

= minus12

int dvradic


9+(u)2

= 12

int vminus 1


= minus12

(2v1

2 ) + 8arcsin(u3

) + C



) + C


34






int b

a f (x)dx is


f (x)dx = bminusa


Example 1


int 20

2x2dx n = 4

Solution

x = 2minus04

= 24

= 12


= i2


2(4)[f (0) + 2f (1

2) + 2f (1) + 2f (3

2) + f (2)]

= 28

[2(0)2 + 2(2(12

)2) + 2(2(1)2) + 2(2( 32

)2) + 2(22)]

= 1

4

[0 + 2(1

2

) + 2(2) + 2( 18

4

) + 8]= 1

4[0 + 1 + 4 + 9 + 8]

= 14

[22]= 55



a f (x)dx is


int b


Example 2


int 2

0 2x2dx n = 4

Solution

35



x = 2minus04

= 24

= 12


= i2

int 20

2x2dx = 2minus03(4)

[f (0) + 4f (12

) + 2f (1) + 4f (32

) + f (2)]

= 212

[2(0)2 + 4(2(12

)2) + 2(2(1)2) + 4(2(32

)2) + 2(22)]= 1

6[0 + 4(1

2) + 2(2) + 4( 18

4 ) + 8]

= 16

[0 + 2 + 4 + 18 + 8]

= 16

[32]= 53333


int 2

0 2x2dx = 22+1

x2+120

= 23

x320

= 23

(2)3 minus 23

(0)3


= 163

Example 3


int π0 sin(x)dx

Solution

x = πminus04

= π4


= iπ4int π


[f (0) + 2f (π4

) + 2f (2π4

) + 2f (3π4

) + f (π)]

= π8

[sin(0) + 2sin(π4

) + 2sin(π2

) + 2sin(3π4

) + sin(π)]

= π8 [0 + 2(radic 22 ) + 2(1) + 2(radic 22 ) + 0]

= π8

[2radic

2 + 2]asymp 2003


int π0 sin(x)dx

Solution

x = πminus04

= π4

36





= 112

[111 + 4(0947) + 2(0816) + 4(0771) + 0625] asymp 086



1(x+2)2

dx =int 2

11u2

du =int 2


x+2

21

= minus 12+2

minus (minus 11+2

)

= minus14

+ 13

= 112



Example 7



Solution


0+1x0+1 + C = sin(x) + 4x + C



int 1x

dx = ln|x| + C

int exdx = ex + C

Example 8


exdx

Solution

int ex = ex + C

Example 9


( 1x

+ x4)dx

Solution

int ( 1

x + x4) = ln|x| + 1

4+1x4+1 + C = 1

5x5 + C

Example 10


x3

2 dx

Solution

int x

3

2 dx = 13

2+1

x3

2+1 + C = 1

5

2

x5

2 + C = 25

x5

2 + C

4



Example 11


1x3

dx

Solution

int 1x3

dxint




+ C

Example 12


( 1x2

+ cos(x))dx

Solution

int ( 1

x2 + cos(x))dx =



minus 1x

+ sin(x) + C

5



42 Area

421 Sigma Notation


i=1

ai = a1 + a2 + a3 + a4 + + an



Example 1

Evaluate6sum

i=1

i

Solution

6sumi=1

i = 1 + 2 + 3 + 4 + 5 + 6 = 21

Example 2

Evaluate4sum

i=0

(2i + 1)

Solution4sum

i=0

(2i + 1) = 2(0) + 1 + 2(1) + 1 + 2(2) + 1 + 2(3) + 1 + 2(4) + 1 = 1 + 3 + 5 + 7 + 9 = 25


Example 3

Evaluate4sum

j=0

j2

Solution4sum

j=0

j2 = 12 + 22 + 32 + 42 = 1 + 4 + 9 + 16 = 30

6



Example 4

Evaluate6sum

k=3

k(k minus 2)

Solution6sum

k=3


Example 5

Evaluate5sum

j=3

1

j

Solution

5sum j=3

1

j =

1

3 +

1

4 +

1

5 =

47

60

Example 6

Evaluate4

sum j=1

[(i

minus1)2 + (i + 1)3]

Solution4sum

j=1


1)2 +(4+1)3 = 02 + 23 + 12 + 33 + 22 + 43 + 32 + 53 = 0 + 8 + 1 + 27 + 4 + 64 + 9 + 125 = 237

Example 7


1+1 + 5

1+2 +

51+3

+ + 51+12

Solution


j=1

[ 5

1 + i]

7



Example 8


)2] +

[1 minus (24

)2] + 1 minus [(34

)2] + [1 minus (44

)2]


j=1

[1 + (i

4)2]


1usum

j=1

c = cn

2usum

j=1

i = n(n+1)

2

3usum

j=1

i2 = n(n+1)(2n+1)

6

4usum

j=1

i3 = n

2(n+1)2

4


Example 9


15sum

j=1

(2i minus 3)

Solution15sum

j=1

(2i minus 3) =15sum

j=1

2i minus15sum

j=1

3 = 2 middot15sum

j=1

i minus15sum

j=1

3 = 2 middot 15(16)

2 minus 15(3) = 240 minus 45 = 195

8



Example 10


15

sum j=1

i(i2 + 1)

Solution10sum

j=1

i(i2 + 1) =10sum

j=1

i3 + i

= n

2(n+1)2

4 + n(n+1)

2 (where n=10)

= 102(10+1)2

4 + 10(10+1)

2


2= 25

middot121 + 5

middot11

= 3025 + 55 = 3080





limhrarrinfin




n

9



Example 10



Solution



= 14



= iminus14


= i

4


S (n) =4sum

j=1

f (mi)x

=4

sumi=1

f (iminus 1

4 ) middot (

1

4)

=4sum

i=1

radic iminus 1

4 middot 1

4

=4sum

i=1

radic iminus 1

2 middot 1

4

=4sum

i=1

radic iminus 1

8

=radic 1minus18 +

radic 2minus18 +

radic 3minus18 +

radic 4minus18

=

radic 0

8 +

radic 1

8 +

radic 2

8 +

radic 3

8 = 0 + 125 + 177 + 217 asymp 519

10




S (n) =4sum

j=1

f (M i)x

=4sum

i=1

f (i

4) middot (

1

4)

=4sum

i=1

radic i

4middot 1

4

=

4

sumi=1

radic i

2 middot1

4

=4sum

i=1

radic i

8

=radic 18

+radic 28

+radic 38

+radic 48

= 125 + 177 + 217 + 25 asymp 769

Example 11



n3[n(n+1)(2n+1)

6 ]

Solution

limhrarrinfin


64

n3[n(n + 1)(2n + 1)

6 ]

= limhrarrinfin

64

n3[(n2 + n))(2n + 1)

6 ]

= limhrarrinfin

64

n3[(2n3 + n2 + 2n2 + n

6 ]

= limhrarrinfin

64[2n3 + 3n2 + n

6n3 ]

= limhrarrinfin

128n3

+ 192n2

+ 64n6n3

11



= limhrarrinfin

128n3

6n3 +

192n2

6n3 +

64n

6n3

= limhrarrinfin

64

3 +

32

n +

32

3n2

= 643

+ 0 + 0 = 643

12



43 Riemann Sum




lexi

le1


nsumi=1

f (ci)xi




xi = i2


= i2


n2


n2

= i2minusi

2minus2i+1)n2

= minus2i+1n2


limhrarrinfin

nsumi=1


nsumi=1

f ( i2

n2) middot (

2iminus 1

n2 )

= limhrarrinfin

nsumi=1

radic i2

n2 middot (

2iminus 1

n2 )

= limhrarrinfin

nsumi=1

i

nmiddot 2iminus 1

n2

= limhrarrinfin

nsumi=1

2i2

minusi

n3

13



= 1

n3 limhrarrinfin

nsumi=1

(2i2 minus i)

= 1

n3 limhrarrinfin

[2(n(n + 1)(2n + 1)

6 ) minus n(n + 1)

2 ]

= 1


6 minus n2 + n

2 ]

= 1

n3 limhrarrinfin

((2n2 + 2n)(2n + 1)

6 minus n2 + n

2 ]

= 1

n3 limhrarrinfin

(4n3 + 6n2 + 2n

6 minus n2 + n

2 ]

= 1

n3 limhrarrinfin


6

= 1

n3 limhrarrinfin

4n3 + 3n2 minus n

6n3

= limhrarrinfin

[4n3 + 3n2 minus n

6n3

]

= limhrarrinfin

4n3

6n3 +

3n2

6n3 +

n

6n3

= limhrarrinfin

2

3 +

1

2n +

1

6n2 =

2

3 + 0 + 0 =

2

3




i=1




nsumi=1



nsumi=1

f (ci)xi =int

b

a f (x)dx

14



Example 2


Solution

int 3minus2

xdx

Solution


n = 3minus(minus2)

n


nrarrinfin

nsum

i=1

f (ci)xi

= limnrarrinfin

nsum

i=1

1048616minus 2 +

5i

n

1048617middot 5

n

= limnrarrinfin

nsum

i=1

1048616minus10

n +

25i

n2

1048617

= limnrarrinfin

nsum

i=1

1048616minus10

n +

25

n2 middot n(n + 1)

2

1048617

= limnrarrinfin

nsumi=1

1048616minus10n

+ 25n2

+ 25n2n2

1048617

= limnrarrinfin

nsum

i=1

1048616minus10

n +

25n2

2n2 +

25n

2n21048617

= limnrarrinfin

nsum

i=1

1048616minus10

n +

25

2 +

25

2n

1048617 = minus0 + 125 + 0 = 125

Example 3


Solution

15



int 31 3x2dx

Solution

a = 1 b = 3 x = 3minus1n

= 2n

ci = 1 + i middot 2i = 1 + 2i

n


nrarrinfin

nsum

i=1

f (ci)xi

= limnrarrinfin

nsum

i=1

f (1 + 2i

n ) middot

2

n

= limnrarrinfin

nsum

i=1

3(1 + 2i

n )2 middot

2

n

= limnrarrinfin

nsum

i=1

3(4i2

n2 +

4i

n + 1) middot

2

n

= limnrarrinfin

nsum

i=1

3( 8i2

n3 + 8i

n2 + 2

n)

= limnrarrinfin

nsum

i=1

(24i2

n3 +

24i

n2 +

6

n)

= limnrarrinfin

nsum

i=1

104861624(n(n + 1)(2n + 1))

6n3 +

24n(n + 1)

2n2 +

6n

n

1048617

= limnrarrinfin

nsum

i=1

104861624(2n3 + 3n2 + n)

6n3 +

24n2 + 24n

2n2 +

6n

n

1048617

= limnrarrinfin

n

sumi=1

104861648n3

6n3 + 72n2

6n3 + 24n

6n3 + 24n2

2n2 + 24n

n2 + 61048617

16



=1048616

8 + 12

n +

4

n2 + 12 +

12

n + 6

1048617 = 26

Example 4


Solution

a = 4 b = 10 x = 10minus4n

= 6n


= 4 + 6in

f (x) = 6


nrarrinfin

nsumi=1

f (ci)xi

= limnrarrinfin

nsumi=1

f 983080

4 + 6i

n

983081middot 6

n

= limnrarrinfin

nsumi=1

(6) middot 6

n

= limn

rarrinfin(6n)

middot

6

n = 36

17



Example 5


a = 1 b = 2 x = 2minus1n

= 1n


= 1 + i

n

f (x) = x2 + 1


int 21 (x2 + 1)dx = lim

nrarrinfin

nsumi=1

f (ci)xi

= limnrarrinfin

n

sumi=1 f

9830801 +

i

n983081 middot

1

n

= limnrarrinfin

nsumi=1

9830809830801 +

i

n

9830812+ 1

983081middot 1

n

= limnrarrinfin

nsumi=1

9830809830801 +

i

n

9830819830801 +

i

n

983081+ 1

983081middot 1

n

= limnrarrinfin

nsumi=1

98308010486161 +

2i

n +

i2

n2

983081+ 1

983081middot 1

n

= limnrarrinfin

n

sumi=1

9830802 + 2i

n

+ i2

n2983081 middot

1

n

= limnrarrinfin

nsumi=1

9830802

n +

2i

n2 +

i2

n3

983081

= limnrarrinfin

983080 2

n middot (n) +

2

n2 middot n(n + 1)

2 +

1

n3 middot n(n + 1)(2n + 1)

6

983081

= limnrarrinfin

9830802 +

2n2 + 2n

2n2 +

(n2 + n)(2n + 1)

6n3

983081

= limnrarrinfin

9830802 +

2n2 + 2n

2n2 +

2n3 + 3n2 + n

6n3

983081


2 + 2n2

2n2

+ 2n

2n2

+ 2n3

6n3

+ 3n2

6n3

+ n

6n3983081

= limnrarrinfin

9830802 + 1 +

1

n +

1

3 +

1

2n +

1

6n2

983081 = 2 + 1 + 0 +

1

3 + 0 + 0 =

10

3

18



Example 6


a = 0 b = 3

x = 3minus0

n = 3

n


= 3n

f (x) = 5x


int 3


n

sumi=1

f (ci)

xi

= limnrarrinfin

nsumi=1

f 9830803i

n

983081middot 3

n

= limnrarrinfin

nsumi=1

59830803i

n

983081middot 3

n

= limnrarrinfin

nsumi=1

45i

n2

= limnrarrinfin

nsum

i=1

45

n2 middot n(n + 1)

2

= limnrarrinfin

nsumi=1

98308045n(n + 1)

2n2

983081

= limnrarrinfin

nsumi=1

98308045n2 + 45n

2n2

983081

= limnrarrinfin

nsumi=1

98308045n2

2n2 +

45n

2n2

983081

= limnrarrinfin

nsumi=1

98308045

2 +

45

2n

983081 =

45

2 + 0 =

45

2

19






int b




Example 1


Solution

int 20

4xdx = 41+1

x220

= 2x2|20 = 2(2)2 minus 2(0)2 = 8 minus 0 = 8

Example 2


Solution

int 21

x3 + x2dx =1048616

13+1

x3+1 + 12+1

x2+1)21

=1048616 14

x4 + 13

x3)21

=104861614

24 + 13

23) minus (14

14 + 13

13)

= 4 + 83 minus 1048616

14

+ 13

)

= 203 minus 1

12

= 7912

Example 3


2

dx

Solution

int 20 6x2dx = 6

2+1x2+1

21


= 16 minus 0= 16

20



Example 4


exdx

Solution

int 20



Example 5


x+radic x

2 dx

= [1

2 middot x2

2 + 1

2 middot 1

2 middot x

3

2

]10

= [x2

4 + 1

3 middot x

3

2 ]10

= [12

4 + 1

3 middot 1

3

2 ] minus [02

4 + 1

3 middot 0

3

2 ]14 minus 1

3 minus 0 = 7

12

Solution

int 10

x+radic x

2 dx =

int 10

x2

+radic x

2 dx =


Example 6


Solution

Note

f (x) = 2

xminus 3

xge

3

2



int 32


int 23

2

(2x minus 3)dx

=int 3

2


3

2

(2x minus 3)dx

= (minusx2 + 3x)|3

2

0 + (x2 minus 3x)|232

= [((32

)2 + 3 middot 32



)]= 10

4

21








1bminusa

int b

a f (x)dx


Example 7


Solution



= 14


= 1

4 middot (1

3x3

+ x)2minus2

= 14 middot [1

3(2)3 + 2] minus 1

3 middot [1

3(minus2)3 minus 2]

= 14 middot 14

3 minus 1

3 middot minus14

3

= 76

+ 76

= 146

= 73

Example 8


2

]

Solution

AV = 1π

2minus0int π

2

0 cos(x)dx

= 2π

int π2

0 cos(x)dx

= 2π middot 1048616 sin(x))|

π

2

0


2) minus sin(0)]


= 2π middot 1

= 2

π

22



Example 9


y = x2 + 2 y = 0 x = 0 x = 2

Solution


=1048616 x3

3 + 2x)

2

0

= [23

3 + 2(2)] minus [03

3 + 2(0)]= [8

3 + 4] minus [0 + 0]

= 203 minus 0

= 203

23



Example 10


y = ex y = 0 x = 0 x = 2

Solution

Area=int 20 (ex)dx

= [ex]|20

= e2 minus e0

= e2 minus 1

Example 11


0 (1 + cos(x))dx

Solution


1048616 x + sin(x))|

π

0


24







Example 1


Solution

f (x) = (x2 + 4x)6




Example 2

Evaluateint

(2x + 3)(x2 + 3x)6dx

Solution


int (2x + 3)(x2 + 3x)6dx =

int u6 middot du


6+1u6+1 + C = 1

7u7 + C = 1

7(x2 + 3x)7 + C

Example 3

Evaluateint

3(1 + 3x)3dx

Solution


int 3(1 + 3x)3dx =

int u3 middot du

25




3+1u3+1 + C = 1

4u4 + C = 1

4(1 + 3x)4 + C

Example 4

Evaluateint

x2(2 + x3)4dx

Solution



int 13

u4 middot du


int 13


5u5 + C = 1

15u5 + C = 1

15(2 + x3)5 + C

26



Example 5

Evaluateint

2tradic

t2 + 3dt

Solution


radic t2 + 3dt =


int u

1

2 middot du


1


3

2 + C = 23

(t2 + 3)3

2 + C = 23

(radic

t2 + 3)3 + C

Example 6

Evaluateint

πcos(πx)dx

Solution


int cos(u)du



Example 7

Evaluateint

xradic 1minus4x2 dx

Solution


= xdx

int x

radic 1minus4x2

dx = int minus1

8 middot du



1


1

2 + C = minus14

u1

2 + C = minus14

(1 minus 4x2)1

2 + C = minus14


27



Example 8

Evaluateint

2xe2x2

dx

Solution


= 2xdx

int 2xe2x

2

dx =int

12

eudu


eudu = 12

eu + C = 12

e2x2

+ C

Example 9

Evaluateint

2xsin(2x2)dx

Solution


du = 2xdx

int 2xsin2x2dx =

int 12

sin(u)du


int 12

sin(u)du =minus

1

2cos(u) + C =

minus1

2cos(2x2) + C

28







2int 1u

du = ln|x| + C

Example 1

Evaluateint

13x

dx

Solution


3du = dxint 13x

dx =int

13 middot 1

u du


int 13 middot 1

u du = 1

3ln|u| + C = 1

3ln|3x| + C

Example 2

Evaluateint

15x+3

dx

Solution


du = dx

int 13x

dx =int

15 middot 1

u du


int 15 middot 1

u du = 1

5ln|u| + C = 1

5ln|5x + 3| + C

29



Example 3

Evaluateint

x2

3minusx3 dx

Solution


du = x2dx

int x

2

3minusx3 dx =int

minus13 middot 1

u du


int minus1

3 middot 1

u du = minus1

3ln|u| + C = minus1

3ln|3 minus x3| + C

Example 4

Evaluateint

sinθ

cosθ dθ

Solution


cosθ dθ =

int minus 1

u du


int minus 1


Example 5

Evaluateint

cos(t)3+sin(t)

dt

Solution

Let u = 3 + sin(t)

rArr du = cos(t)dt

int cos(t)3+sin(t)

dθ =int

1u

du


int 1u

du = ln|u| + C = ln|3 + sin(t)| + C

30





Theorem 48



1int

duradic a2minusu2

= arcsin u

a +C

2int

du

a2+u2 = 1a

arctan u

a +C

3int

du

uradic u2minusa2

= 1a

arcsec |u|a

+C



Example 1

Evaluateint

41+9x2 middotdx

Solution


3du = dx


int 41+9x2 middotdx

=int

41+(3x)2

middotdx

= 13

int 4du1+u2

= 43

int du

1+u2

= 4

3arctan(u) + C

= 43

arctan(3x) + C

Example 2

Evaluateint

dx

4+(x+1)2

Solution



31



int dx

4+(x+1)2 =int

du

22+u2

= 12

arctan(u2

) + C

= 12

arctan(x+12

) + C

Example 3

Evaluateint

7radic 4minusx2

middotdx

Solution



int 7

radic 4minusx2

= int 7du

radic 22

minusu2


) + C

= 7arcsin(x2

) + C

Example 4

Evaluateint

dx


Solution



int dx


=int

du

uradic u2minus9

= 13

arcsec(u3

) + C

= 13

arcsec( |x|3

) + C

32





Example 8

Evaluateint


Solution





rArr minus12

du = (x minus 3)dx

Now substitute


=int


8radic 9+(xminus3)2

= minus12

int dvradic


9+(u)2

= 12

int vminus 1


= minus12

(2v1

2 ) + 8arcsin(u3

) + C



) + C


34






int b

a f (x)dx is


f (x)dx = bminusa


Example 1


int 20

2x2dx n = 4

Solution

x = 2minus04

= 24

= 12


= i2


2(4)[f (0) + 2f (1

2) + 2f (1) + 2f (3

2) + f (2)]

= 28

[2(0)2 + 2(2(12

)2) + 2(2(1)2) + 2(2( 32

)2) + 2(22)]

= 1

4

[0 + 2(1

2

) + 2(2) + 2( 18

4

) + 8]= 1

4[0 + 1 + 4 + 9 + 8]

= 14

[22]= 55



a f (x)dx is


int b


Example 2


int 2

0 2x2dx n = 4

Solution

35



x = 2minus04

= 24

= 12


= i2

int 20

2x2dx = 2minus03(4)

[f (0) + 4f (12

) + 2f (1) + 4f (32

) + f (2)]

= 212

[2(0)2 + 4(2(12

)2) + 2(2(1)2) + 4(2(32

)2) + 2(22)]= 1

6[0 + 4(1

2) + 2(2) + 4( 18

4 ) + 8]

= 16

[0 + 2 + 4 + 18 + 8]

= 16

[32]= 53333


int 2

0 2x2dx = 22+1

x2+120

= 23

x320

= 23

(2)3 minus 23

(0)3


= 163

Example 3


int π0 sin(x)dx

Solution

x = πminus04

= π4


= iπ4int π


[f (0) + 2f (π4

) + 2f (2π4

) + 2f (3π4

) + f (π)]

= π8

[sin(0) + 2sin(π4

) + 2sin(π2

) + 2sin(3π4

) + sin(π)]

= π8 [0 + 2(radic 22 ) + 2(1) + 2(radic 22 ) + 0]

= π8

[2radic

2 + 2]asymp 2003


int π0 sin(x)dx

Solution

x = πminus04

= π4

36





= 112

[111 + 4(0947) + 2(0816) + 4(0771) + 0625] asymp 086



1(x+2)2

dx =int 2

11u2

du =int 2


x+2

21

= minus 12+2

minus (minus 11+2

)

= minus14

+ 13

= 112



Example 11


1x3

dx

Solution

int 1x3

dxint




+ C

Example 12


( 1x2

+ cos(x))dx

Solution

int ( 1

x2 + cos(x))dx =



minus 1x

+ sin(x) + C

5



42 Area

421 Sigma Notation


i=1

ai = a1 + a2 + a3 + a4 + + an



Example 1

Evaluate6sum

i=1

i

Solution

6sumi=1

i = 1 + 2 + 3 + 4 + 5 + 6 = 21

Example 2

Evaluate4sum

i=0

(2i + 1)

Solution4sum

i=0

(2i + 1) = 2(0) + 1 + 2(1) + 1 + 2(2) + 1 + 2(3) + 1 + 2(4) + 1 = 1 + 3 + 5 + 7 + 9 = 25


Example 3

Evaluate4sum

j=0

j2

Solution4sum

j=0

j2 = 12 + 22 + 32 + 42 = 1 + 4 + 9 + 16 = 30

6



Example 4

Evaluate6sum

k=3

k(k minus 2)

Solution6sum

k=3


Example 5

Evaluate5sum

j=3

1

j

Solution

5sum j=3

1

j =

1

3 +

1

4 +

1

5 =

47

60

Example 6

Evaluate4

sum j=1

[(i

minus1)2 + (i + 1)3]

Solution4sum

j=1


1)2 +(4+1)3 = 02 + 23 + 12 + 33 + 22 + 43 + 32 + 53 = 0 + 8 + 1 + 27 + 4 + 64 + 9 + 125 = 237

Example 7


1+1 + 5

1+2 +

51+3

+ + 51+12

Solution


j=1

[ 5

1 + i]

7



Example 8


)2] +

[1 minus (24

)2] + 1 minus [(34

)2] + [1 minus (44

)2]


j=1

[1 + (i

4)2]


1usum

j=1

c = cn

2usum

j=1

i = n(n+1)

2

3usum

j=1

i2 = n(n+1)(2n+1)

6

4usum

j=1

i3 = n

2(n+1)2

4


Example 9


15sum

j=1

(2i minus 3)

Solution15sum

j=1

(2i minus 3) =15sum

j=1

2i minus15sum

j=1

3 = 2 middot15sum

j=1

i minus15sum

j=1

3 = 2 middot 15(16)

2 minus 15(3) = 240 minus 45 = 195

8



Example 10


15

sum j=1

i(i2 + 1)

Solution10sum

j=1

i(i2 + 1) =10sum

j=1

i3 + i

= n

2(n+1)2

4 + n(n+1)

2 (where n=10)

= 102(10+1)2

4 + 10(10+1)

2


2= 25

middot121 + 5

middot11

= 3025 + 55 = 3080





limhrarrinfin




n

9



Example 10



Solution



= 14



= iminus14


= i

4


S (n) =4sum

j=1

f (mi)x

=4

sumi=1

f (iminus 1

4 ) middot (

1

4)

=4sum

i=1

radic iminus 1

4 middot 1

4

=4sum

i=1

radic iminus 1

2 middot 1

4

=4sum

i=1

radic iminus 1

8

=radic 1minus18 +

radic 2minus18 +

radic 3minus18 +

radic 4minus18

=

radic 0

8 +

radic 1

8 +

radic 2

8 +

radic 3

8 = 0 + 125 + 177 + 217 asymp 519

10




S (n) =4sum

j=1

f (M i)x

=4sum

i=1

f (i

4) middot (

1

4)

=4sum

i=1

radic i

4middot 1

4

=

4

sumi=1

radic i

2 middot1

4

=4sum

i=1

radic i

8

=radic 18

+radic 28

+radic 38

+radic 48

= 125 + 177 + 217 + 25 asymp 769

Example 11



n3[n(n+1)(2n+1)

6 ]

Solution

limhrarrinfin


64

n3[n(n + 1)(2n + 1)

6 ]

= limhrarrinfin

64

n3[(n2 + n))(2n + 1)

6 ]

= limhrarrinfin

64

n3[(2n3 + n2 + 2n2 + n

6 ]

= limhrarrinfin

64[2n3 + 3n2 + n

6n3 ]

= limhrarrinfin

128n3

+ 192n2

+ 64n6n3

11



= limhrarrinfin

128n3

6n3 +

192n2

6n3 +

64n

6n3

= limhrarrinfin

64

3 +

32

n +

32

3n2

= 643

+ 0 + 0 = 643

12



43 Riemann Sum




lexi

le1


nsumi=1

f (ci)xi




xi = i2


= i2


n2


n2

= i2minusi

2minus2i+1)n2

= minus2i+1n2


limhrarrinfin

nsumi=1


nsumi=1

f ( i2

n2) middot (

2iminus 1

n2 )

= limhrarrinfin

nsumi=1

radic i2

n2 middot (

2iminus 1

n2 )

= limhrarrinfin

nsumi=1

i

nmiddot 2iminus 1

n2

= limhrarrinfin

nsumi=1

2i2

minusi

n3

13



= 1

n3 limhrarrinfin

nsumi=1

(2i2 minus i)

= 1

n3 limhrarrinfin

[2(n(n + 1)(2n + 1)

6 ) minus n(n + 1)

2 ]

= 1


6 minus n2 + n

2 ]

= 1

n3 limhrarrinfin

((2n2 + 2n)(2n + 1)

6 minus n2 + n

2 ]

= 1

n3 limhrarrinfin

(4n3 + 6n2 + 2n

6 minus n2 + n

2 ]

= 1

n3 limhrarrinfin


6

= 1

n3 limhrarrinfin

4n3 + 3n2 minus n

6n3

= limhrarrinfin

[4n3 + 3n2 minus n

6n3

]

= limhrarrinfin

4n3

6n3 +

3n2

6n3 +

n

6n3

= limhrarrinfin

2

3 +

1

2n +

1

6n2 =

2

3 + 0 + 0 =

2

3




i=1




nsumi=1



nsumi=1

f (ci)xi =int

b

a f (x)dx

14



Example 2


Solution

int 3minus2

xdx

Solution


n = 3minus(minus2)

n


nrarrinfin

nsum

i=1

f (ci)xi

= limnrarrinfin

nsum

i=1

1048616minus 2 +

5i

n

1048617middot 5

n

= limnrarrinfin

nsum

i=1

1048616minus10

n +

25i

n2

1048617

= limnrarrinfin

nsum

i=1

1048616minus10

n +

25

n2 middot n(n + 1)

2

1048617

= limnrarrinfin

nsumi=1

1048616minus10n

+ 25n2

+ 25n2n2

1048617

= limnrarrinfin

nsum

i=1

1048616minus10

n +

25n2

2n2 +

25n

2n21048617

= limnrarrinfin

nsum

i=1

1048616minus10

n +

25

2 +

25

2n

1048617 = minus0 + 125 + 0 = 125

Example 3


Solution

15



int 31 3x2dx

Solution

a = 1 b = 3 x = 3minus1n

= 2n

ci = 1 + i middot 2i = 1 + 2i

n


nrarrinfin

nsum

i=1

f (ci)xi

= limnrarrinfin

nsum

i=1

f (1 + 2i

n ) middot

2

n

= limnrarrinfin

nsum

i=1

3(1 + 2i

n )2 middot

2

n

= limnrarrinfin

nsum

i=1

3(4i2

n2 +

4i

n + 1) middot

2

n

= limnrarrinfin

nsum

i=1

3( 8i2

n3 + 8i

n2 + 2

n)

= limnrarrinfin

nsum

i=1

(24i2

n3 +

24i

n2 +

6

n)

= limnrarrinfin

nsum

i=1

104861624(n(n + 1)(2n + 1))

6n3 +

24n(n + 1)

2n2 +

6n

n

1048617

= limnrarrinfin

nsum

i=1

104861624(2n3 + 3n2 + n)

6n3 +

24n2 + 24n

2n2 +

6n

n

1048617

= limnrarrinfin

n

sumi=1

104861648n3

6n3 + 72n2

6n3 + 24n

6n3 + 24n2

2n2 + 24n

n2 + 61048617

16



=1048616

8 + 12

n +

4

n2 + 12 +

12

n + 6

1048617 = 26

Example 4


Solution

a = 4 b = 10 x = 10minus4n

= 6n


= 4 + 6in

f (x) = 6


nrarrinfin

nsumi=1

f (ci)xi

= limnrarrinfin

nsumi=1

f 983080

4 + 6i

n

983081middot 6

n

= limnrarrinfin

nsumi=1

(6) middot 6

n

= limn

rarrinfin(6n)

middot

6

n = 36

17



Example 5


a = 1 b = 2 x = 2minus1n

= 1n


= 1 + i

n

f (x) = x2 + 1


int 21 (x2 + 1)dx = lim

nrarrinfin

nsumi=1

f (ci)xi

= limnrarrinfin

n

sumi=1 f

9830801 +

i

n983081 middot

1

n

= limnrarrinfin

nsumi=1

9830809830801 +

i

n

9830812+ 1

983081middot 1

n

= limnrarrinfin

nsumi=1

9830809830801 +

i

n

9830819830801 +

i

n

983081+ 1

983081middot 1

n

= limnrarrinfin

nsumi=1

98308010486161 +

2i

n +

i2

n2

983081+ 1

983081middot 1

n

= limnrarrinfin

n

sumi=1

9830802 + 2i

n

+ i2

n2983081 middot

1

n

= limnrarrinfin

nsumi=1

9830802

n +

2i

n2 +

i2

n3

983081

= limnrarrinfin

983080 2

n middot (n) +

2

n2 middot n(n + 1)

2 +

1

n3 middot n(n + 1)(2n + 1)

6

983081

= limnrarrinfin

9830802 +

2n2 + 2n

2n2 +

(n2 + n)(2n + 1)

6n3

983081

= limnrarrinfin

9830802 +

2n2 + 2n

2n2 +

2n3 + 3n2 + n

6n3

983081


2 + 2n2

2n2

+ 2n

2n2

+ 2n3

6n3

+ 3n2

6n3

+ n

6n3983081

= limnrarrinfin

9830802 + 1 +

1

n +

1

3 +

1

2n +

1

6n2

983081 = 2 + 1 + 0 +

1

3 + 0 + 0 =

10

3

18



Example 6


a = 0 b = 3

x = 3minus0

n = 3

n


= 3n

f (x) = 5x


int 3


n

sumi=1

f (ci)

xi

= limnrarrinfin

nsumi=1

f 9830803i

n

983081middot 3

n

= limnrarrinfin

nsumi=1

59830803i

n

983081middot 3

n

= limnrarrinfin

nsumi=1

45i

n2

= limnrarrinfin

nsum

i=1

45

n2 middot n(n + 1)

2

= limnrarrinfin

nsumi=1

98308045n(n + 1)

2n2

983081

= limnrarrinfin

nsumi=1

98308045n2 + 45n

2n2

983081

= limnrarrinfin

nsumi=1

98308045n2

2n2 +

45n

2n2

983081

= limnrarrinfin

nsumi=1

98308045

2 +

45

2n

983081 =

45

2 + 0 =

45

2

19






int b




Example 1


Solution

int 20

4xdx = 41+1

x220

= 2x2|20 = 2(2)2 minus 2(0)2 = 8 minus 0 = 8

Example 2


Solution

int 21

x3 + x2dx =1048616

13+1

x3+1 + 12+1

x2+1)21

=1048616 14

x4 + 13

x3)21

=104861614

24 + 13

23) minus (14

14 + 13

13)

= 4 + 83 minus 1048616

14

+ 13

)

= 203 minus 1

12

= 7912

Example 3


2

dx

Solution

int 20 6x2dx = 6

2+1x2+1

21


= 16 minus 0= 16

20



Example 4


exdx

Solution

int 20



Example 5


x+radic x

2 dx

= [1

2 middot x2

2 + 1

2 middot 1

2 middot x

3

2

]10

= [x2

4 + 1

3 middot x

3

2 ]10

= [12

4 + 1

3 middot 1

3

2 ] minus [02

4 + 1

3 middot 0

3

2 ]14 minus 1

3 minus 0 = 7

12

Solution

int 10

x+radic x

2 dx =

int 10

x2

+radic x

2 dx =


Example 6


Solution

Note

f (x) = 2

xminus 3

xge

3

2



int 32


int 23

2

(2x minus 3)dx

=int 3

2


3

2

(2x minus 3)dx

= (minusx2 + 3x)|3

2

0 + (x2 minus 3x)|232

= [((32

)2 + 3 middot 32



)]= 10

4

21








1bminusa

int b

a f (x)dx


Example 7


Solution



= 14


= 1

4 middot (1

3x3

+ x)2minus2

= 14 middot [1

3(2)3 + 2] minus 1

3 middot [1

3(minus2)3 minus 2]

= 14 middot 14

3 minus 1

3 middot minus14

3

= 76

+ 76

= 146

= 73

Example 8


2

]

Solution

AV = 1π

2minus0int π

2

0 cos(x)dx

= 2π

int π2

0 cos(x)dx

= 2π middot 1048616 sin(x))|

π

2

0


2) minus sin(0)]


= 2π middot 1

= 2

π

22



Example 9


y = x2 + 2 y = 0 x = 0 x = 2

Solution


=1048616 x3

3 + 2x)

2

0

= [23

3 + 2(2)] minus [03

3 + 2(0)]= [8

3 + 4] minus [0 + 0]

= 203 minus 0

= 203

23



Example 10


y = ex y = 0 x = 0 x = 2

Solution

Area=int 20 (ex)dx

= [ex]|20

= e2 minus e0

= e2 minus 1

Example 11


0 (1 + cos(x))dx

Solution


1048616 x + sin(x))|

π

0


24







Example 1


Solution

f (x) = (x2 + 4x)6




Example 2

Evaluateint

(2x + 3)(x2 + 3x)6dx

Solution


int (2x + 3)(x2 + 3x)6dx =

int u6 middot du


6+1u6+1 + C = 1

7u7 + C = 1

7(x2 + 3x)7 + C

Example 3

Evaluateint

3(1 + 3x)3dx

Solution


int 3(1 + 3x)3dx =

int u3 middot du

25




3+1u3+1 + C = 1

4u4 + C = 1

4(1 + 3x)4 + C

Example 4

Evaluateint

x2(2 + x3)4dx

Solution



int 13

u4 middot du


int 13


5u5 + C = 1

15u5 + C = 1

15(2 + x3)5 + C

26



Example 5

Evaluateint

2tradic

t2 + 3dt

Solution


radic t2 + 3dt =


int u

1

2 middot du


1


3

2 + C = 23

(t2 + 3)3

2 + C = 23

(radic

t2 + 3)3 + C

Example 6

Evaluateint

πcos(πx)dx

Solution


int cos(u)du



Example 7

Evaluateint

xradic 1minus4x2 dx

Solution


= xdx

int x

radic 1minus4x2

dx = int minus1

8 middot du



1


1

2 + C = minus14

u1

2 + C = minus14

(1 minus 4x2)1

2 + C = minus14


27



Example 8

Evaluateint

2xe2x2

dx

Solution


= 2xdx

int 2xe2x

2

dx =int

12

eudu


eudu = 12

eu + C = 12

e2x2

+ C

Example 9

Evaluateint

2xsin(2x2)dx

Solution


du = 2xdx

int 2xsin2x2dx =

int 12

sin(u)du


int 12

sin(u)du =minus

1

2cos(u) + C =

minus1

2cos(2x2) + C

28







2int 1u

du = ln|x| + C

Example 1

Evaluateint

13x

dx

Solution


3du = dxint 13x

dx =int

13 middot 1

u du


int 13 middot 1

u du = 1

3ln|u| + C = 1

3ln|3x| + C

Example 2

Evaluateint

15x+3

dx

Solution


du = dx

int 13x

dx =int

15 middot 1

u du


int 15 middot 1

u du = 1

5ln|u| + C = 1

5ln|5x + 3| + C

29



Example 3

Evaluateint

x2

3minusx3 dx

Solution


du = x2dx

int x

2

3minusx3 dx =int

minus13 middot 1

u du


int minus1

3 middot 1

u du = minus1

3ln|u| + C = minus1

3ln|3 minus x3| + C

Example 4

Evaluateint

sinθ

cosθ dθ

Solution


cosθ dθ =

int minus 1

u du


int minus 1


Example 5

Evaluateint

cos(t)3+sin(t)

dt

Solution

Let u = 3 + sin(t)

rArr du = cos(t)dt

int cos(t)3+sin(t)

dθ =int

1u

du


int 1u

du = ln|u| + C = ln|3 + sin(t)| + C

30





Theorem 48



1int

duradic a2minusu2

= arcsin u

a +C

2int

du

a2+u2 = 1a

arctan u

a +C

3int

du

uradic u2minusa2

= 1a

arcsec |u|a

+C



Example 1

Evaluateint

41+9x2 middotdx

Solution


3du = dx


int 41+9x2 middotdx

=int

41+(3x)2

middotdx

= 13

int 4du1+u2

= 43

int du

1+u2

= 4

3arctan(u) + C

= 43

arctan(3x) + C

Example 2

Evaluateint

dx

4+(x+1)2

Solution



31



int dx

4+(x+1)2 =int

du

22+u2

= 12

arctan(u2

) + C

= 12

arctan(x+12

) + C

Example 3

Evaluateint

7radic 4minusx2

middotdx

Solution



int 7

radic 4minusx2

= int 7du

radic 22

minusu2


) + C

= 7arcsin(x2

) + C

Example 4

Evaluateint

dx


Solution



int dx


=int

du

uradic u2minus9

= 13

arcsec(u3

) + C

= 13

arcsec( |x|3

) + C

32





Example 8

Evaluateint


Solution





rArr minus12

du = (x minus 3)dx

Now substitute


=int


8radic 9+(xminus3)2

= minus12

int dvradic


9+(u)2

= 12

int vminus 1


= minus12

(2v1

2 ) + 8arcsin(u3

) + C



) + C


34






int b

a f (x)dx is


f (x)dx = bminusa


Example 1


int 20

2x2dx n = 4

Solution

x = 2minus04

= 24

= 12


= i2


2(4)[f (0) + 2f (1

2) + 2f (1) + 2f (3

2) + f (2)]

= 28

[2(0)2 + 2(2(12

)2) + 2(2(1)2) + 2(2( 32

)2) + 2(22)]

= 1

4

[0 + 2(1

2

) + 2(2) + 2( 18

4

) + 8]= 1

4[0 + 1 + 4 + 9 + 8]

= 14

[22]= 55



a f (x)dx is


int b


Example 2


int 2

0 2x2dx n = 4

Solution

35



x = 2minus04

= 24

= 12


= i2

int 20

2x2dx = 2minus03(4)

[f (0) + 4f (12

) + 2f (1) + 4f (32

) + f (2)]

= 212

[2(0)2 + 4(2(12

)2) + 2(2(1)2) + 4(2(32

)2) + 2(22)]= 1

6[0 + 4(1

2) + 2(2) + 4( 18

4 ) + 8]

= 16

[0 + 2 + 4 + 18 + 8]

= 16

[32]= 53333


int 2

0 2x2dx = 22+1

x2+120

= 23

x320

= 23

(2)3 minus 23

(0)3


= 163

Example 3


int π0 sin(x)dx

Solution

x = πminus04

= π4


= iπ4int π


[f (0) + 2f (π4

) + 2f (2π4

) + 2f (3π4

) + f (π)]

= π8

[sin(0) + 2sin(π4

) + 2sin(π2

) + 2sin(3π4

) + sin(π)]

= π8 [0 + 2(radic 22 ) + 2(1) + 2(radic 22 ) + 0]

= π8

[2radic

2 + 2]asymp 2003


int π0 sin(x)dx

Solution

x = πminus04

= π4

36





= 112

[111 + 4(0947) + 2(0816) + 4(0771) + 0625] asymp 086



1(x+2)2

dx =int 2

11u2

du =int 2


x+2

21

= minus 12+2

minus (minus 11+2

)

= minus14

+ 13

= 112



42 Area

421 Sigma Notation


i=1

ai = a1 + a2 + a3 + a4 + + an



Example 1

Evaluate6sum

i=1

i

Solution

6sumi=1

i = 1 + 2 + 3 + 4 + 5 + 6 = 21

Example 2

Evaluate4sum

i=0

(2i + 1)

Solution4sum

i=0

(2i + 1) = 2(0) + 1 + 2(1) + 1 + 2(2) + 1 + 2(3) + 1 + 2(4) + 1 = 1 + 3 + 5 + 7 + 9 = 25


Example 3

Evaluate4sum

j=0

j2

Solution4sum

j=0

j2 = 12 + 22 + 32 + 42 = 1 + 4 + 9 + 16 = 30

6



Example 4

Evaluate6sum

k=3

k(k minus 2)

Solution6sum

k=3


Example 5

Evaluate5sum

j=3

1

j

Solution

5sum j=3

1

j =

1

3 +

1

4 +

1

5 =

47

60

Example 6

Evaluate4

sum j=1

[(i

minus1)2 + (i + 1)3]

Solution4sum

j=1


1)2 +(4+1)3 = 02 + 23 + 12 + 33 + 22 + 43 + 32 + 53 = 0 + 8 + 1 + 27 + 4 + 64 + 9 + 125 = 237

Example 7


1+1 + 5

1+2 +

51+3

+ + 51+12

Solution


j=1

[ 5

1 + i]

7



Example 8


)2] +

[1 minus (24

)2] + 1 minus [(34

)2] + [1 minus (44

)2]


j=1

[1 + (i

4)2]


1usum

j=1

c = cn

2usum

j=1

i = n(n+1)

2

3usum

j=1

i2 = n(n+1)(2n+1)

6

4usum

j=1

i3 = n

2(n+1)2

4


Example 9


15sum

j=1

(2i minus 3)

Solution15sum

j=1

(2i minus 3) =15sum

j=1

2i minus15sum

j=1

3 = 2 middot15sum

j=1

i minus15sum

j=1

3 = 2 middot 15(16)

2 minus 15(3) = 240 minus 45 = 195

8



Example 10


15

sum j=1

i(i2 + 1)

Solution10sum

j=1

i(i2 + 1) =10sum

j=1

i3 + i

= n

2(n+1)2

4 + n(n+1)

2 (where n=10)

= 102(10+1)2

4 + 10(10+1)

2


2= 25

middot121 + 5

middot11

= 3025 + 55 = 3080





limhrarrinfin




n

9



Example 10



Solution



= 14



= iminus14


= i

4


S (n) =4sum

j=1

f (mi)x

=4

sumi=1

f (iminus 1

4 ) middot (

1

4)

=4sum

i=1

radic iminus 1

4 middot 1

4

=4sum

i=1

radic iminus 1

2 middot 1

4

=4sum

i=1

radic iminus 1

8

=radic 1minus18 +

radic 2minus18 +

radic 3minus18 +

radic 4minus18

=

radic 0

8 +

radic 1

8 +

radic 2

8 +

radic 3

8 = 0 + 125 + 177 + 217 asymp 519

10




S (n) =4sum

j=1

f (M i)x

=4sum

i=1

f (i

4) middot (

1

4)

=4sum

i=1

radic i

4middot 1

4

=

4

sumi=1

radic i

2 middot1

4

=4sum

i=1

radic i

8

=radic 18

+radic 28

+radic 38

+radic 48

= 125 + 177 + 217 + 25 asymp 769

Example 11



n3[n(n+1)(2n+1)

6 ]

Solution

limhrarrinfin


64

n3[n(n + 1)(2n + 1)

6 ]

= limhrarrinfin

64

n3[(n2 + n))(2n + 1)

6 ]

= limhrarrinfin

64

n3[(2n3 + n2 + 2n2 + n

6 ]

= limhrarrinfin

64[2n3 + 3n2 + n

6n3 ]

= limhrarrinfin

128n3

+ 192n2

+ 64n6n3

11



= limhrarrinfin

128n3

6n3 +

192n2

6n3 +

64n

6n3

= limhrarrinfin

64

3 +

32

n +

32

3n2

= 643

+ 0 + 0 = 643

12



43 Riemann Sum




lexi

le1


nsumi=1

f (ci)xi




xi = i2


= i2


n2


n2

= i2minusi

2minus2i+1)n2

= minus2i+1n2


limhrarrinfin

nsumi=1


nsumi=1

f ( i2

n2) middot (

2iminus 1

n2 )

= limhrarrinfin

nsumi=1

radic i2

n2 middot (

2iminus 1

n2 )

= limhrarrinfin

nsumi=1

i

nmiddot 2iminus 1

n2

= limhrarrinfin

nsumi=1

2i2

minusi

n3

13



= 1

n3 limhrarrinfin

nsumi=1

(2i2 minus i)

= 1

n3 limhrarrinfin

[2(n(n + 1)(2n + 1)

6 ) minus n(n + 1)

2 ]

= 1


6 minus n2 + n

2 ]

= 1

n3 limhrarrinfin

((2n2 + 2n)(2n + 1)

6 minus n2 + n

2 ]

= 1

n3 limhrarrinfin

(4n3 + 6n2 + 2n

6 minus n2 + n

2 ]

= 1

n3 limhrarrinfin


6

= 1

n3 limhrarrinfin

4n3 + 3n2 minus n

6n3

= limhrarrinfin

[4n3 + 3n2 minus n

6n3

]

= limhrarrinfin

4n3

6n3 +

3n2

6n3 +

n

6n3

= limhrarrinfin

2

3 +

1

2n +

1

6n2 =

2

3 + 0 + 0 =

2

3




i=1




nsumi=1



nsumi=1

f (ci)xi =int

b

a f (x)dx

14



Example 2


Solution

int 3minus2

xdx

Solution


n = 3minus(minus2)

n


nrarrinfin

nsum

i=1

f (ci)xi

= limnrarrinfin

nsum

i=1

1048616minus 2 +

5i

n

1048617middot 5

n

= limnrarrinfin

nsum

i=1

1048616minus10

n +

25i

n2

1048617

= limnrarrinfin

nsum

i=1

1048616minus10

n +

25

n2 middot n(n + 1)

2

1048617

= limnrarrinfin

nsumi=1

1048616minus10n

+ 25n2

+ 25n2n2

1048617

= limnrarrinfin

nsum

i=1

1048616minus10

n +

25n2

2n2 +

25n

2n21048617

= limnrarrinfin

nsum

i=1

1048616minus10

n +

25

2 +

25

2n

1048617 = minus0 + 125 + 0 = 125

Example 3


Solution

15



int 31 3x2dx

Solution

a = 1 b = 3 x = 3minus1n

= 2n

ci = 1 + i middot 2i = 1 + 2i

n


nrarrinfin

nsum

i=1

f (ci)xi

= limnrarrinfin

nsum

i=1

f (1 + 2i

n ) middot

2

n

= limnrarrinfin

nsum

i=1

3(1 + 2i

n )2 middot

2

n

= limnrarrinfin

nsum

i=1

3(4i2

n2 +

4i

n + 1) middot

2

n

= limnrarrinfin

nsum

i=1

3( 8i2

n3 + 8i

n2 + 2

n)

= limnrarrinfin

nsum

i=1

(24i2

n3 +

24i

n2 +

6

n)

= limnrarrinfin

nsum

i=1

104861624(n(n + 1)(2n + 1))

6n3 +

24n(n + 1)

2n2 +

6n

n

1048617

= limnrarrinfin

nsum

i=1

104861624(2n3 + 3n2 + n)

6n3 +

24n2 + 24n

2n2 +

6n

n

1048617

= limnrarrinfin

n

sumi=1

104861648n3

6n3 + 72n2

6n3 + 24n

6n3 + 24n2

2n2 + 24n

n2 + 61048617

16



=1048616

8 + 12

n +

4

n2 + 12 +

12

n + 6

1048617 = 26

Example 4


Solution

a = 4 b = 10 x = 10minus4n

= 6n


= 4 + 6in

f (x) = 6


nrarrinfin

nsumi=1

f (ci)xi

= limnrarrinfin

nsumi=1

f 983080

4 + 6i

n

983081middot 6

n

= limnrarrinfin

nsumi=1

(6) middot 6

n

= limn

rarrinfin(6n)

middot

6

n = 36

17



Example 5


a = 1 b = 2 x = 2minus1n

= 1n


= 1 + i

n

f (x) = x2 + 1


int 21 (x2 + 1)dx = lim

nrarrinfin

nsumi=1

f (ci)xi

= limnrarrinfin

n

sumi=1 f

9830801 +

i

n983081 middot

1

n

= limnrarrinfin

nsumi=1

9830809830801 +

i

n

9830812+ 1

983081middot 1

n

= limnrarrinfin

nsumi=1

9830809830801 +

i

n

9830819830801 +

i

n

983081+ 1

983081middot 1

n

= limnrarrinfin

nsumi=1

98308010486161 +

2i

n +

i2

n2

983081+ 1

983081middot 1

n

= limnrarrinfin

n

sumi=1

9830802 + 2i

n

+ i2

n2983081 middot

1

n

= limnrarrinfin

nsumi=1

9830802

n +

2i

n2 +

i2

n3

983081

= limnrarrinfin

983080 2

n middot (n) +

2

n2 middot n(n + 1)

2 +

1

n3 middot n(n + 1)(2n + 1)

6

983081

= limnrarrinfin

9830802 +

2n2 + 2n

2n2 +

(n2 + n)(2n + 1)

6n3

983081

= limnrarrinfin

9830802 +

2n2 + 2n

2n2 +

2n3 + 3n2 + n

6n3

983081


2 + 2n2

2n2

+ 2n

2n2

+ 2n3

6n3

+ 3n2

6n3

+ n

6n3983081

= limnrarrinfin

9830802 + 1 +

1

n +

1

3 +

1

2n +

1

6n2

983081 = 2 + 1 + 0 +

1

3 + 0 + 0 =

10

3

18



Example 6


a = 0 b = 3

x = 3minus0

n = 3

n


= 3n

f (x) = 5x


int 3


n

sumi=1

f (ci)

xi

= limnrarrinfin

nsumi=1

f 9830803i

n

983081middot 3

n

= limnrarrinfin

nsumi=1

59830803i

n

983081middot 3

n

= limnrarrinfin

nsumi=1

45i

n2

= limnrarrinfin

nsum

i=1

45

n2 middot n(n + 1)

2

= limnrarrinfin

nsumi=1

98308045n(n + 1)

2n2

983081

= limnrarrinfin

nsumi=1

98308045n2 + 45n

2n2

983081

= limnrarrinfin

nsumi=1

98308045n2

2n2 +

45n

2n2

983081

= limnrarrinfin

nsumi=1

98308045

2 +

45

2n

983081 =

45

2 + 0 =

45

2

19






int b




Example 1


Solution

int 20

4xdx = 41+1

x220

= 2x2|20 = 2(2)2 minus 2(0)2 = 8 minus 0 = 8

Example 2


Solution

int 21

x3 + x2dx =1048616

13+1

x3+1 + 12+1

x2+1)21

=1048616 14

x4 + 13

x3)21

=104861614

24 + 13

23) minus (14

14 + 13

13)

= 4 + 83 minus 1048616

14

+ 13

)

= 203 minus 1

12

= 7912

Example 3


2

dx

Solution

int 20 6x2dx = 6

2+1x2+1

21


= 16 minus 0= 16

20



Example 4


exdx

Solution

int 20



Example 5


x+radic x

2 dx

= [1

2 middot x2

2 + 1

2 middot 1

2 middot x

3

2

]10

= [x2

4 + 1

3 middot x

3

2 ]10

= [12

4 + 1

3 middot 1

3

2 ] minus [02

4 + 1

3 middot 0

3

2 ]14 minus 1

3 minus 0 = 7

12

Solution

int 10

x+radic x

2 dx =

int 10

x2

+radic x

2 dx =


Example 6


Solution

Note

f (x) = 2

xminus 3

xge

3

2



int 32


int 23

2

(2x minus 3)dx

=int 3

2


3

2

(2x minus 3)dx

= (minusx2 + 3x)|3

2

0 + (x2 minus 3x)|232

= [((32

)2 + 3 middot 32



)]= 10

4

21








1bminusa

int b

a f (x)dx


Example 7


Solution



= 14


= 1

4 middot (1

3x3

+ x)2minus2

= 14 middot [1

3(2)3 + 2] minus 1

3 middot [1

3(minus2)3 minus 2]

= 14 middot 14

3 minus 1

3 middot minus14

3

= 76

+ 76

= 146

= 73

Example 8


2

]

Solution

AV = 1π

2minus0int π

2

0 cos(x)dx

= 2π

int π2

0 cos(x)dx

= 2π middot 1048616 sin(x))|

π

2

0


2) minus sin(0)]


= 2π middot 1

= 2

π

22



Example 9


y = x2 + 2 y = 0 x = 0 x = 2

Solution


=1048616 x3

3 + 2x)

2

0

= [23

3 + 2(2)] minus [03

3 + 2(0)]= [8

3 + 4] minus [0 + 0]

= 203 minus 0

= 203

23



Example 10


y = ex y = 0 x = 0 x = 2

Solution

Area=int 20 (ex)dx

= [ex]|20

= e2 minus e0

= e2 minus 1

Example 11


0 (1 + cos(x))dx

Solution


1048616 x + sin(x))|

π

0


24







Example 1


Solution

f (x) = (x2 + 4x)6




Example 2

Evaluateint

(2x + 3)(x2 + 3x)6dx

Solution


int (2x + 3)(x2 + 3x)6dx =

int u6 middot du


6+1u6+1 + C = 1

7u7 + C = 1

7(x2 + 3x)7 + C

Example 3

Evaluateint

3(1 + 3x)3dx

Solution


int 3(1 + 3x)3dx =

int u3 middot du

25




3+1u3+1 + C = 1

4u4 + C = 1

4(1 + 3x)4 + C

Example 4

Evaluateint

x2(2 + x3)4dx

Solution



int 13

u4 middot du


int 13


5u5 + C = 1

15u5 + C = 1

15(2 + x3)5 + C

26



Example 5

Evaluateint

2tradic

t2 + 3dt

Solution


radic t2 + 3dt =


int u

1

2 middot du


1


3

2 + C = 23

(t2 + 3)3

2 + C = 23

(radic

t2 + 3)3 + C

Example 6

Evaluateint

πcos(πx)dx

Solution


int cos(u)du



Example 7

Evaluateint

xradic 1minus4x2 dx

Solution


= xdx

int x

radic 1minus4x2

dx = int minus1

8 middot du



1


1

2 + C = minus14

u1

2 + C = minus14

(1 minus 4x2)1

2 + C = minus14


27



Example 8

Evaluateint

2xe2x2

dx

Solution


= 2xdx

int 2xe2x

2

dx =int

12

eudu


eudu = 12

eu + C = 12

e2x2

+ C

Example 9

Evaluateint

2xsin(2x2)dx

Solution


du = 2xdx

int 2xsin2x2dx =

int 12

sin(u)du


int 12

sin(u)du =minus

1

2cos(u) + C =

minus1

2cos(2x2) + C

28







2int 1u

du = ln|x| + C

Example 1

Evaluateint

13x

dx

Solution


3du = dxint 13x

dx =int

13 middot 1

u du


int 13 middot 1

u du = 1

3ln|u| + C = 1

3ln|3x| + C

Example 2

Evaluateint

15x+3

dx

Solution


du = dx

int 13x

dx =int

15 middot 1

u du


int 15 middot 1

u du = 1

5ln|u| + C = 1

5ln|5x + 3| + C

29



Example 3

Evaluateint

x2

3minusx3 dx

Solution


du = x2dx

int x

2

3minusx3 dx =int

minus13 middot 1

u du


int minus1

3 middot 1

u du = minus1

3ln|u| + C = minus1

3ln|3 minus x3| + C

Example 4

Evaluateint

sinθ

cosθ dθ

Solution


cosθ dθ =

int minus 1

u du


int minus 1


Example 5

Evaluateint

cos(t)3+sin(t)

dt

Solution

Let u = 3 + sin(t)

rArr du = cos(t)dt

int cos(t)3+sin(t)

dθ =int

1u

du


int 1u

du = ln|u| + C = ln|3 + sin(t)| + C

30





Theorem 48



1int

duradic a2minusu2

= arcsin u

a +C

2int

du

a2+u2 = 1a

arctan u

a +C

3int

du

uradic u2minusa2

= 1a

arcsec |u|a

+C



Example 1

Evaluateint

41+9x2 middotdx

Solution


3du = dx


int 41+9x2 middotdx

=int

41+(3x)2

middotdx

= 13

int 4du1+u2

= 43

int du

1+u2

= 4

3arctan(u) + C

= 43

arctan(3x) + C

Example 2

Evaluateint

dx

4+(x+1)2

Solution



31



int dx

4+(x+1)2 =int

du

22+u2

= 12

arctan(u2

) + C

= 12

arctan(x+12

) + C

Example 3

Evaluateint

7radic 4minusx2

middotdx

Solution



int 7

radic 4minusx2

= int 7du

radic 22

minusu2


) + C

= 7arcsin(x2

) + C

Example 4

Evaluateint

dx


Solution



int dx


=int

du

uradic u2minus9

= 13

arcsec(u3

) + C

= 13

arcsec( |x|3

) + C

32





Example 8

Evaluateint


Solution





rArr minus12

du = (x minus 3)dx

Now substitute


=int


8radic 9+(xminus3)2

= minus12

int dvradic


9+(u)2

= 12

int vminus 1


= minus12

(2v1

2 ) + 8arcsin(u3

) + C



) + C


34






int b

a f (x)dx is


f (x)dx = bminusa


Example 1


int 20

2x2dx n = 4

Solution

x = 2minus04

= 24

= 12


= i2


2(4)[f (0) + 2f (1

2) + 2f (1) + 2f (3

2) + f (2)]

= 28

[2(0)2 + 2(2(12

)2) + 2(2(1)2) + 2(2( 32

)2) + 2(22)]

= 1

4

[0 + 2(1

2

) + 2(2) + 2( 18

4

) + 8]= 1

4[0 + 1 + 4 + 9 + 8]

= 14

[22]= 55



a f (x)dx is


int b


Example 2


int 2

0 2x2dx n = 4

Solution

35



x = 2minus04

= 24

= 12


= i2

int 20

2x2dx = 2minus03(4)

[f (0) + 4f (12

) + 2f (1) + 4f (32

) + f (2)]

= 212

[2(0)2 + 4(2(12

)2) + 2(2(1)2) + 4(2(32

)2) + 2(22)]= 1

6[0 + 4(1

2) + 2(2) + 4( 18

4 ) + 8]

= 16

[0 + 2 + 4 + 18 + 8]

= 16

[32]= 53333


int 2

0 2x2dx = 22+1

x2+120

= 23

x320

= 23

(2)3 minus 23

(0)3


= 163

Example 3


int π0 sin(x)dx

Solution

x = πminus04

= π4


= iπ4int π


[f (0) + 2f (π4

) + 2f (2π4

) + 2f (3π4

) + f (π)]

= π8

[sin(0) + 2sin(π4

) + 2sin(π2

) + 2sin(3π4

) + sin(π)]

= π8 [0 + 2(radic 22 ) + 2(1) + 2(radic 22 ) + 0]

= π8

[2radic

2 + 2]asymp 2003


int π0 sin(x)dx

Solution

x = πminus04

= π4

36





= 112

[111 + 4(0947) + 2(0816) + 4(0771) + 0625] asymp 086



1(x+2)2

dx =int 2

11u2

du =int 2


x+2

21

= minus 12+2

minus (minus 11+2

)

= minus14

+ 13

= 112



Example 4

Evaluate6sum

k=3

k(k minus 2)

Solution6sum

k=3


Example 5

Evaluate5sum

j=3

1

j

Solution

5sum j=3

1

j =

1

3 +

1

4 +

1

5 =

47

60

Example 6

Evaluate4

sum j=1

[(i

minus1)2 + (i + 1)3]

Solution4sum

j=1


1)2 +(4+1)3 = 02 + 23 + 12 + 33 + 22 + 43 + 32 + 53 = 0 + 8 + 1 + 27 + 4 + 64 + 9 + 125 = 237

Example 7


1+1 + 5

1+2 +

51+3

+ + 51+12

Solution


j=1

[ 5

1 + i]

7



Example 8


)2] +

[1 minus (24

)2] + 1 minus [(34

)2] + [1 minus (44

)2]


j=1

[1 + (i

4)2]


1usum

j=1

c = cn

2usum

j=1

i = n(n+1)

2

3usum

j=1

i2 = n(n+1)(2n+1)

6

4usum

j=1

i3 = n

2(n+1)2

4


Example 9


15sum

j=1

(2i minus 3)

Solution15sum

j=1

(2i minus 3) =15sum

j=1

2i minus15sum

j=1

3 = 2 middot15sum

j=1

i minus15sum

j=1

3 = 2 middot 15(16)

2 minus 15(3) = 240 minus 45 = 195

8



Example 10


15

sum j=1

i(i2 + 1)

Solution10sum

j=1

i(i2 + 1) =10sum

j=1

i3 + i

= n

2(n+1)2

4 + n(n+1)

2 (where n=10)

= 102(10+1)2

4 + 10(10+1)

2


2= 25

middot121 + 5

middot11

= 3025 + 55 = 3080





limhrarrinfin




n

9



Example 10



Solution



= 14



= iminus14


= i

4


S (n) =4sum

j=1

f (mi)x

=4

sumi=1

f (iminus 1

4 ) middot (

1

4)

=4sum

i=1

radic iminus 1

4 middot 1

4

=4sum

i=1

radic iminus 1

2 middot 1

4

=4sum

i=1

radic iminus 1

8

=radic 1minus18 +

radic 2minus18 +

radic 3minus18 +

radic 4minus18

=

radic 0

8 +

radic 1

8 +

radic 2

8 +

radic 3

8 = 0 + 125 + 177 + 217 asymp 519

10




S (n) =4sum

j=1

f (M i)x

=4sum

i=1

f (i

4) middot (

1

4)

=4sum

i=1

radic i

4middot 1

4

=

4

sumi=1

radic i

2 middot1

4

=4sum

i=1

radic i

8

=radic 18

+radic 28

+radic 38

+radic 48

= 125 + 177 + 217 + 25 asymp 769

Example 11



n3[n(n+1)(2n+1)

6 ]

Solution

limhrarrinfin


64

n3[n(n + 1)(2n + 1)

6 ]

= limhrarrinfin

64

n3[(n2 + n))(2n + 1)

6 ]

= limhrarrinfin

64

n3[(2n3 + n2 + 2n2 + n

6 ]

= limhrarrinfin

64[2n3 + 3n2 + n

6n3 ]

= limhrarrinfin

128n3

+ 192n2

+ 64n6n3

11



= limhrarrinfin

128n3

6n3 +

192n2

6n3 +

64n

6n3

= limhrarrinfin

64

3 +

32

n +

32

3n2

= 643

+ 0 + 0 = 643

12



43 Riemann Sum




lexi

le1


nsumi=1

f (ci)xi




xi = i2


= i2


n2


n2

= i2minusi

2minus2i+1)n2

= minus2i+1n2


limhrarrinfin

nsumi=1


nsumi=1

f ( i2

n2) middot (

2iminus 1

n2 )

= limhrarrinfin

nsumi=1

radic i2

n2 middot (

2iminus 1

n2 )

= limhrarrinfin

nsumi=1

i

nmiddot 2iminus 1

n2

= limhrarrinfin

nsumi=1

2i2

minusi

n3

13



= 1

n3 limhrarrinfin

nsumi=1

(2i2 minus i)

= 1

n3 limhrarrinfin

[2(n(n + 1)(2n + 1)

6 ) minus n(n + 1)

2 ]

= 1


6 minus n2 + n

2 ]

= 1

n3 limhrarrinfin

((2n2 + 2n)(2n + 1)

6 minus n2 + n

2 ]

= 1

n3 limhrarrinfin

(4n3 + 6n2 + 2n

6 minus n2 + n

2 ]

= 1

n3 limhrarrinfin


6

= 1

n3 limhrarrinfin

4n3 + 3n2 minus n

6n3

= limhrarrinfin

[4n3 + 3n2 minus n

6n3

]

= limhrarrinfin

4n3

6n3 +

3n2

6n3 +

n

6n3

= limhrarrinfin

2

3 +

1

2n +

1

6n2 =

2

3 + 0 + 0 =

2

3




i=1




nsumi=1



nsumi=1

f (ci)xi =int

b

a f (x)dx

14



Example 2


Solution

int 3minus2

xdx

Solution


n = 3minus(minus2)

n


nrarrinfin

nsum

i=1

f (ci)xi

= limnrarrinfin

nsum

i=1

1048616minus 2 +

5i

n

1048617middot 5

n

= limnrarrinfin

nsum

i=1

1048616minus10

n +

25i

n2

1048617

= limnrarrinfin

nsum

i=1

1048616minus10

n +

25

n2 middot n(n + 1)

2

1048617

= limnrarrinfin

nsumi=1

1048616minus10n

+ 25n2

+ 25n2n2

1048617

= limnrarrinfin

nsum

i=1

1048616minus10

n +

25n2

2n2 +

25n

2n21048617

= limnrarrinfin

nsum

i=1

1048616minus10

n +

25

2 +

25

2n

1048617 = minus0 + 125 + 0 = 125

Example 3


Solution

15



int 31 3x2dx

Solution

a = 1 b = 3 x = 3minus1n

= 2n

ci = 1 + i middot 2i = 1 + 2i

n


nrarrinfin

nsum

i=1

f (ci)xi

= limnrarrinfin

nsum

i=1

f (1 + 2i

n ) middot

2

n

= limnrarrinfin

nsum

i=1

3(1 + 2i

n )2 middot

2

n

= limnrarrinfin

nsum

i=1

3(4i2

n2 +

4i

n + 1) middot

2

n

= limnrarrinfin

nsum

i=1

3( 8i2

n3 + 8i

n2 + 2

n)

= limnrarrinfin

nsum

i=1

(24i2

n3 +

24i

n2 +

6

n)

= limnrarrinfin

nsum

i=1

104861624(n(n + 1)(2n + 1))

6n3 +

24n(n + 1)

2n2 +

6n

n

1048617

= limnrarrinfin

nsum

i=1

104861624(2n3 + 3n2 + n)

6n3 +

24n2 + 24n

2n2 +

6n

n

1048617

= limnrarrinfin

n

sumi=1

104861648n3

6n3 + 72n2

6n3 + 24n

6n3 + 24n2

2n2 + 24n

n2 + 61048617

16



=1048616

8 + 12

n +

4

n2 + 12 +

12

n + 6

1048617 = 26

Example 4


Solution

a = 4 b = 10 x = 10minus4n

= 6n


= 4 + 6in

f (x) = 6


nrarrinfin

nsumi=1

f (ci)xi

= limnrarrinfin

nsumi=1

f 983080

4 + 6i

n

983081middot 6

n

= limnrarrinfin

nsumi=1

(6) middot 6

n

= limn

rarrinfin(6n)

middot

6

n = 36

17



Example 5


a = 1 b = 2 x = 2minus1n

= 1n


= 1 + i

n

f (x) = x2 + 1


int 21 (x2 + 1)dx = lim

nrarrinfin

nsumi=1

f (ci)xi

= limnrarrinfin

n

sumi=1 f

9830801 +

i

n983081 middot

1

n

= limnrarrinfin

nsumi=1

9830809830801 +

i

n

9830812+ 1

983081middot 1

n

= limnrarrinfin

nsumi=1

9830809830801 +

i

n

9830819830801 +

i

n

983081+ 1

983081middot 1

n

= limnrarrinfin

nsumi=1

98308010486161 +

2i

n +

i2

n2

983081+ 1

983081middot 1

n

= limnrarrinfin

n

sumi=1

9830802 + 2i

n

+ i2

n2983081 middot

1

n

= limnrarrinfin

nsumi=1

9830802

n +

2i

n2 +

i2

n3

983081

= limnrarrinfin

983080 2

n middot (n) +

2

n2 middot n(n + 1)

2 +

1

n3 middot n(n + 1)(2n + 1)

6

983081

= limnrarrinfin

9830802 +

2n2 + 2n

2n2 +

(n2 + n)(2n + 1)

6n3

983081

= limnrarrinfin

9830802 +

2n2 + 2n

2n2 +

2n3 + 3n2 + n

6n3

983081


2 + 2n2

2n2

+ 2n

2n2

+ 2n3

6n3

+ 3n2

6n3

+ n

6n3983081

= limnrarrinfin

9830802 + 1 +

1

n +

1

3 +

1

2n +

1

6n2

983081 = 2 + 1 + 0 +

1

3 + 0 + 0 =

10

3

18



Example 6


a = 0 b = 3

x = 3minus0

n = 3

n


= 3n

f (x) = 5x


int 3


n

sumi=1

f (ci)

xi

= limnrarrinfin

nsumi=1

f 9830803i

n

983081middot 3

n

= limnrarrinfin

nsumi=1

59830803i

n

983081middot 3

n

= limnrarrinfin

nsumi=1

45i

n2

= limnrarrinfin

nsum

i=1

45

n2 middot n(n + 1)

2

= limnrarrinfin

nsumi=1

98308045n(n + 1)

2n2

983081

= limnrarrinfin

nsumi=1

98308045n2 + 45n

2n2

983081

= limnrarrinfin

nsumi=1

98308045n2

2n2 +

45n

2n2

983081

= limnrarrinfin

nsumi=1

98308045

2 +

45

2n

983081 =

45

2 + 0 =

45

2

19






int b




Example 1


Solution

int 20

4xdx = 41+1

x220

= 2x2|20 = 2(2)2 minus 2(0)2 = 8 minus 0 = 8

Example 2


Solution

int 21

x3 + x2dx =1048616

13+1

x3+1 + 12+1

x2+1)21

=1048616 14

x4 + 13

x3)21

=104861614

24 + 13

23) minus (14

14 + 13

13)

= 4 + 83 minus 1048616

14

+ 13

)

= 203 minus 1

12

= 7912

Example 3


2

dx

Solution

int 20 6x2dx = 6

2+1x2+1

21


= 16 minus 0= 16

20



Example 4


exdx

Solution

int 20



Example 5


x+radic x

2 dx

= [1

2 middot x2

2 + 1

2 middot 1

2 middot x

3

2

]10

= [x2

4 + 1

3 middot x

3

2 ]10

= [12

4 + 1

3 middot 1

3

2 ] minus [02

4 + 1

3 middot 0

3

2 ]14 minus 1

3 minus 0 = 7

12

Solution

int 10

x+radic x

2 dx =

int 10

x2

+radic x

2 dx =


Example 6


Solution

Note

f (x) = 2

xminus 3

xge

3

2



int 32


int 23

2

(2x minus 3)dx

=int 3

2


3

2

(2x minus 3)dx

= (minusx2 + 3x)|3

2

0 + (x2 minus 3x)|232

= [((32

)2 + 3 middot 32



)]= 10

4

21








1bminusa

int b

a f (x)dx


Example 7


Solution



= 14


= 1

4 middot (1

3x3

+ x)2minus2

= 14 middot [1

3(2)3 + 2] minus 1

3 middot [1

3(minus2)3 minus 2]

= 14 middot 14

3 minus 1

3 middot minus14

3

= 76

+ 76

= 146

= 73

Example 8


2

]

Solution

AV = 1π

2minus0int π

2

0 cos(x)dx

= 2π

int π2

0 cos(x)dx

= 2π middot 1048616 sin(x))|

π

2

0


2) minus sin(0)]


= 2π middot 1

= 2

π

22



Example 9


y = x2 + 2 y = 0 x = 0 x = 2

Solution


=1048616 x3

3 + 2x)

2

0

= [23

3 + 2(2)] minus [03

3 + 2(0)]= [8

3 + 4] minus [0 + 0]

= 203 minus 0

= 203

23



Example 10


y = ex y = 0 x = 0 x = 2

Solution

Area=int 20 (ex)dx

= [ex]|20

= e2 minus e0

= e2 minus 1

Example 11


0 (1 + cos(x))dx

Solution


1048616 x + sin(x))|

π

0


24







Example 1


Solution

f (x) = (x2 + 4x)6




Example 2

Evaluateint

(2x + 3)(x2 + 3x)6dx

Solution


int (2x + 3)(x2 + 3x)6dx =

int u6 middot du


6+1u6+1 + C = 1

7u7 + C = 1

7(x2 + 3x)7 + C

Example 3

Evaluateint

3(1 + 3x)3dx

Solution


int 3(1 + 3x)3dx =

int u3 middot du

25




3+1u3+1 + C = 1

4u4 + C = 1

4(1 + 3x)4 + C

Example 4

Evaluateint

x2(2 + x3)4dx

Solution



int 13

u4 middot du


int 13


5u5 + C = 1

15u5 + C = 1

15(2 + x3)5 + C

26



Example 5

Evaluateint

2tradic

t2 + 3dt

Solution


radic t2 + 3dt =


int u

1

2 middot du


1


3

2 + C = 23

(t2 + 3)3

2 + C = 23

(radic

t2 + 3)3 + C

Example 6

Evaluateint

πcos(πx)dx

Solution


int cos(u)du



Example 7

Evaluateint

xradic 1minus4x2 dx

Solution


= xdx

int x

radic 1minus4x2

dx = int minus1

8 middot du



1


1

2 + C = minus14

u1

2 + C = minus14

(1 minus 4x2)1

2 + C = minus14


27



Example 8

Evaluateint

2xe2x2

dx

Solution


= 2xdx

int 2xe2x

2

dx =int

12

eudu


eudu = 12

eu + C = 12

e2x2

+ C

Example 9

Evaluateint

2xsin(2x2)dx

Solution


du = 2xdx

int 2xsin2x2dx =

int 12

sin(u)du


int 12

sin(u)du =minus

1

2cos(u) + C =

minus1

2cos(2x2) + C

28







2int 1u

du = ln|x| + C

Example 1

Evaluateint

13x

dx

Solution


3du = dxint 13x

dx =int

13 middot 1

u du


int 13 middot 1

u du = 1

3ln|u| + C = 1

3ln|3x| + C

Example 2

Evaluateint

15x+3

dx

Solution


du = dx

int 13x

dx =int

15 middot 1

u du


int 15 middot 1

u du = 1

5ln|u| + C = 1

5ln|5x + 3| + C

29



Example 3

Evaluateint

x2

3minusx3 dx

Solution


du = x2dx

int x

2

3minusx3 dx =int

minus13 middot 1

u du


int minus1

3 middot 1

u du = minus1

3ln|u| + C = minus1

3ln|3 minus x3| + C

Example 4

Evaluateint

sinθ

cosθ dθ

Solution


cosθ dθ =

int minus 1

u du


int minus 1


Example 5

Evaluateint

cos(t)3+sin(t)

dt

Solution

Let u = 3 + sin(t)

rArr du = cos(t)dt

int cos(t)3+sin(t)

dθ =int

1u

du


int 1u

du = ln|u| + C = ln|3 + sin(t)| + C

30





Theorem 48



1int

duradic a2minusu2

= arcsin u

a +C

2int

du

a2+u2 = 1a

arctan u

a +C

3int

du

uradic u2minusa2

= 1a

arcsec |u|a

+C



Example 1

Evaluateint

41+9x2 middotdx

Solution


3du = dx


int 41+9x2 middotdx

=int

41+(3x)2

middotdx

= 13

int 4du1+u2

= 43

int du

1+u2

= 4

3arctan(u) + C

= 43

arctan(3x) + C

Example 2

Evaluateint

dx

4+(x+1)2

Solution



31



int dx

4+(x+1)2 =int

du

22+u2

= 12

arctan(u2

) + C

= 12

arctan(x+12

) + C

Example 3

Evaluateint

7radic 4minusx2

middotdx

Solution



int 7

radic 4minusx2

= int 7du

radic 22

minusu2


) + C

= 7arcsin(x2

) + C

Example 4

Evaluateint

dx


Solution



int dx


=int

du

uradic u2minus9

= 13

arcsec(u3

) + C

= 13

arcsec( |x|3

) + C

32





Example 8

Evaluateint


Solution





rArr minus12

du = (x minus 3)dx

Now substitute


=int


8radic 9+(xminus3)2

= minus12

int dvradic


9+(u)2

= 12

int vminus 1


= minus12

(2v1

2 ) + 8arcsin(u3

) + C



) + C


34






int b

a f (x)dx is


f (x)dx = bminusa


Example 1


int 20

2x2dx n = 4

Solution

x = 2minus04

= 24

= 12


= i2


2(4)[f (0) + 2f (1

2) + 2f (1) + 2f (3

2) + f (2)]

= 28

[2(0)2 + 2(2(12

)2) + 2(2(1)2) + 2(2( 32

)2) + 2(22)]

= 1

4

[0 + 2(1

2

) + 2(2) + 2( 18

4

) + 8]= 1

4[0 + 1 + 4 + 9 + 8]

= 14

[22]= 55



a f (x)dx is


int b


Example 2


int 2

0 2x2dx n = 4

Solution

35



x = 2minus04

= 24

= 12


= i2

int 20

2x2dx = 2minus03(4)

[f (0) + 4f (12

) + 2f (1) + 4f (32

) + f (2)]

= 212

[2(0)2 + 4(2(12

)2) + 2(2(1)2) + 4(2(32

)2) + 2(22)]= 1

6[0 + 4(1

2) + 2(2) + 4( 18

4 ) + 8]

= 16

[0 + 2 + 4 + 18 + 8]

= 16

[32]= 53333


int 2

0 2x2dx = 22+1

x2+120

= 23

x320

= 23

(2)3 minus 23

(0)3


= 163

Example 3


int π0 sin(x)dx

Solution

x = πminus04

= π4


= iπ4int π


[f (0) + 2f (π4

) + 2f (2π4

) + 2f (3π4

) + f (π)]

= π8

[sin(0) + 2sin(π4

) + 2sin(π2

) + 2sin(3π4

) + sin(π)]

= π8 [0 + 2(radic 22 ) + 2(1) + 2(radic 22 ) + 0]

= π8

[2radic

2 + 2]asymp 2003


int π0 sin(x)dx

Solution

x = πminus04

= π4

36





= 112

[111 + 4(0947) + 2(0816) + 4(0771) + 0625] asymp 086



1(x+2)2

dx =int 2

11u2

du =int 2


x+2

21

= minus 12+2

minus (minus 11+2

)

= minus14

+ 13

= 112



Example 8


)2] +

[1 minus (24

)2] + 1 minus [(34

)2] + [1 minus (44

)2]


j=1

[1 + (i

4)2]


1usum

j=1

c = cn

2usum

j=1

i = n(n+1)

2

3usum

j=1

i2 = n(n+1)(2n+1)

6

4usum

j=1

i3 = n

2(n+1)2

4


Example 9


15sum

j=1

(2i minus 3)

Solution15sum

j=1

(2i minus 3) =15sum

j=1

2i minus15sum

j=1

3 = 2 middot15sum

j=1

i minus15sum

j=1

3 = 2 middot 15(16)

2 minus 15(3) = 240 minus 45 = 195

8



Example 10


15

sum j=1

i(i2 + 1)

Solution10sum

j=1

i(i2 + 1) =10sum

j=1

i3 + i

= n

2(n+1)2

4 + n(n+1)

2 (where n=10)

= 102(10+1)2

4 + 10(10+1)

2


2= 25

middot121 + 5

middot11

= 3025 + 55 = 3080





limhrarrinfin




n

9



Example 10



Solution



= 14



= iminus14


= i

4


S (n) =4sum

j=1

f (mi)x

=4

sumi=1

f (iminus 1

4 ) middot (

1

4)

=4sum

i=1

radic iminus 1

4 middot 1

4

=4sum

i=1

radic iminus 1

2 middot 1

4

=4sum

i=1

radic iminus 1

8

=radic 1minus18 +

radic 2minus18 +

radic 3minus18 +

radic 4minus18

=

radic 0

8 +

radic 1

8 +

radic 2

8 +

radic 3

8 = 0 + 125 + 177 + 217 asymp 519

10




S (n) =4sum

j=1

f (M i)x

=4sum

i=1

f (i

4) middot (

1

4)

=4sum

i=1

radic i

4middot 1

4

=

4

sumi=1

radic i

2 middot1

4

=4sum

i=1

radic i

8

=radic 18

+radic 28

+radic 38

+radic 48

= 125 + 177 + 217 + 25 asymp 769

Example 11



n3[n(n+1)(2n+1)

6 ]

Solution

limhrarrinfin


64

n3[n(n + 1)(2n + 1)

6 ]

= limhrarrinfin

64

n3[(n2 + n))(2n + 1)

6 ]

= limhrarrinfin

64

n3[(2n3 + n2 + 2n2 + n

6 ]

= limhrarrinfin

64[2n3 + 3n2 + n

6n3 ]

= limhrarrinfin

128n3

+ 192n2

+ 64n6n3

11



= limhrarrinfin

128n3

6n3 +

192n2

6n3 +

64n

6n3

= limhrarrinfin

64

3 +

32

n +

32

3n2

= 643

+ 0 + 0 = 643

12



43 Riemann Sum




lexi

le1


nsumi=1

f (ci)xi




xi = i2


= i2


n2


n2

= i2minusi

2minus2i+1)n2

= minus2i+1n2


limhrarrinfin

nsumi=1


nsumi=1

f ( i2

n2) middot (

2iminus 1

n2 )

= limhrarrinfin

nsumi=1

radic i2

n2 middot (

2iminus 1

n2 )

= limhrarrinfin

nsumi=1

i

nmiddot 2iminus 1

n2

= limhrarrinfin

nsumi=1

2i2

minusi

n3

13



= 1

n3 limhrarrinfin

nsumi=1

(2i2 minus i)

= 1

n3 limhrarrinfin

[2(n(n + 1)(2n + 1)

6 ) minus n(n + 1)

2 ]

= 1


6 minus n2 + n

2 ]

= 1

n3 limhrarrinfin

((2n2 + 2n)(2n + 1)

6 minus n2 + n

2 ]

= 1

n3 limhrarrinfin

(4n3 + 6n2 + 2n

6 minus n2 + n

2 ]

= 1

n3 limhrarrinfin


6

= 1

n3 limhrarrinfin

4n3 + 3n2 minus n

6n3

= limhrarrinfin

[4n3 + 3n2 minus n

6n3

]

= limhrarrinfin

4n3

6n3 +

3n2

6n3 +

n

6n3

= limhrarrinfin

2

3 +

1

2n +

1

6n2 =

2

3 + 0 + 0 =

2

3




i=1




nsumi=1



nsumi=1

f (ci)xi =int

b

a f (x)dx

14



Example 2


Solution

int 3minus2

xdx

Solution


n = 3minus(minus2)

n


nrarrinfin

nsum

i=1

f (ci)xi

= limnrarrinfin

nsum

i=1

1048616minus 2 +

5i

n

1048617middot 5

n

= limnrarrinfin

nsum

i=1

1048616minus10

n +

25i

n2

1048617

= limnrarrinfin

nsum

i=1

1048616minus10

n +

25

n2 middot n(n + 1)

2

1048617

= limnrarrinfin

nsumi=1

1048616minus10n

+ 25n2

+ 25n2n2

1048617

= limnrarrinfin

nsum

i=1

1048616minus10

n +

25n2

2n2 +

25n

2n21048617

= limnrarrinfin

nsum

i=1

1048616minus10

n +

25

2 +

25

2n

1048617 = minus0 + 125 + 0 = 125

Example 3


Solution

15



int 31 3x2dx

Solution

a = 1 b = 3 x = 3minus1n

= 2n

ci = 1 + i middot 2i = 1 + 2i

n


nrarrinfin

nsum

i=1

f (ci)xi

= limnrarrinfin

nsum

i=1

f (1 + 2i

n ) middot

2

n

= limnrarrinfin

nsum

i=1

3(1 + 2i

n )2 middot

2

n

= limnrarrinfin

nsum

i=1

3(4i2

n2 +

4i

n + 1) middot

2

n

= limnrarrinfin

nsum

i=1

3( 8i2

n3 + 8i

n2 + 2

n)

= limnrarrinfin

nsum

i=1

(24i2

n3 +

24i

n2 +

6

n)

= limnrarrinfin

nsum

i=1

104861624(n(n + 1)(2n + 1))

6n3 +

24n(n + 1)

2n2 +

6n

n

1048617

= limnrarrinfin

nsum

i=1

104861624(2n3 + 3n2 + n)

6n3 +

24n2 + 24n

2n2 +

6n

n

1048617

= limnrarrinfin

n

sumi=1

104861648n3

6n3 + 72n2

6n3 + 24n

6n3 + 24n2

2n2 + 24n

n2 + 61048617

16



=1048616

8 + 12

n +

4

n2 + 12 +

12

n + 6

1048617 = 26

Example 4


Solution

a = 4 b = 10 x = 10minus4n

= 6n


= 4 + 6in

f (x) = 6


nrarrinfin

nsumi=1

f (ci)xi

= limnrarrinfin

nsumi=1

f 983080

4 + 6i

n

983081middot 6

n

= limnrarrinfin

nsumi=1

(6) middot 6

n

= limn

rarrinfin(6n)

middot

6

n = 36

17



Example 5


a = 1 b = 2 x = 2minus1n

= 1n


= 1 + i

n

f (x) = x2 + 1


int 21 (x2 + 1)dx = lim

nrarrinfin

nsumi=1

f (ci)xi

= limnrarrinfin

n

sumi=1 f

9830801 +

i

n983081 middot

1

n

= limnrarrinfin

nsumi=1

9830809830801 +

i

n

9830812+ 1

983081middot 1

n

= limnrarrinfin

nsumi=1

9830809830801 +

i

n

9830819830801 +

i

n

983081+ 1

983081middot 1

n

= limnrarrinfin

nsumi=1

98308010486161 +

2i

n +

i2

n2

983081+ 1

983081middot 1

n

= limnrarrinfin

n

sumi=1

9830802 + 2i

n

+ i2

n2983081 middot

1

n

= limnrarrinfin

nsumi=1

9830802

n +

2i

n2 +

i2

n3

983081

= limnrarrinfin

983080 2

n middot (n) +

2

n2 middot n(n + 1)

2 +

1

n3 middot n(n + 1)(2n + 1)

6

983081

= limnrarrinfin

9830802 +

2n2 + 2n

2n2 +

(n2 + n)(2n + 1)

6n3

983081

= limnrarrinfin

9830802 +

2n2 + 2n

2n2 +

2n3 + 3n2 + n

6n3

983081


2 + 2n2

2n2

+ 2n

2n2

+ 2n3

6n3

+ 3n2

6n3

+ n

6n3983081

= limnrarrinfin

9830802 + 1 +

1

n +

1

3 +

1

2n +

1

6n2

983081 = 2 + 1 + 0 +

1

3 + 0 + 0 =

10

3

18



Example 6


a = 0 b = 3

x = 3minus0

n = 3

n


= 3n

f (x) = 5x


int 3


n

sumi=1

f (ci)

xi

= limnrarrinfin

nsumi=1

f 9830803i

n

983081middot 3

n

= limnrarrinfin

nsumi=1

59830803i

n

983081middot 3

n

= limnrarrinfin

nsumi=1

45i

n2

= limnrarrinfin

nsum

i=1

45

n2 middot n(n + 1)

2

= limnrarrinfin

nsumi=1

98308045n(n + 1)

2n2

983081

= limnrarrinfin

nsumi=1

98308045n2 + 45n

2n2

983081

= limnrarrinfin

nsumi=1

98308045n2

2n2 +

45n

2n2

983081

= limnrarrinfin

nsumi=1

98308045

2 +

45

2n

983081 =

45

2 + 0 =

45

2

19






int b




Example 1


Solution

int 20

4xdx = 41+1

x220

= 2x2|20 = 2(2)2 minus 2(0)2 = 8 minus 0 = 8

Example 2


Solution

int 21

x3 + x2dx =1048616

13+1

x3+1 + 12+1

x2+1)21

=1048616 14

x4 + 13

x3)21

=104861614

24 + 13

23) minus (14

14 + 13

13)

= 4 + 83 minus 1048616

14

+ 13

)

= 203 minus 1

12

= 7912

Example 3


2

dx

Solution

int 20 6x2dx = 6

2+1x2+1

21


= 16 minus 0= 16

20



Example 4


exdx

Solution

int 20



Example 5


x+radic x

2 dx

= [1

2 middot x2

2 + 1

2 middot 1

2 middot x

3

2

]10

= [x2

4 + 1

3 middot x

3

2 ]10

= [12

4 + 1

3 middot 1

3

2 ] minus [02

4 + 1

3 middot 0

3

2 ]14 minus 1

3 minus 0 = 7

12

Solution

int 10

x+radic x

2 dx =

int 10

x2

+radic x

2 dx =


Example 6


Solution

Note

f (x) = 2

xminus 3

xge

3

2



int 32


int 23

2

(2x minus 3)dx

=int 3

2


3

2

(2x minus 3)dx

= (minusx2 + 3x)|3

2

0 + (x2 minus 3x)|232

= [((32

)2 + 3 middot 32



)]= 10

4

21








1bminusa

int b

a f (x)dx


Example 7


Solution



= 14


= 1

4 middot (1

3x3

+ x)2minus2

= 14 middot [1

3(2)3 + 2] minus 1

3 middot [1

3(minus2)3 minus 2]

= 14 middot 14

3 minus 1

3 middot minus14

3

= 76

+ 76

= 146

= 73

Example 8


2

]

Solution

AV = 1π

2minus0int π

2

0 cos(x)dx

= 2π

int π2

0 cos(x)dx

= 2π middot 1048616 sin(x))|

π

2

0


2) minus sin(0)]


= 2π middot 1

= 2

π

22



Example 9


y = x2 + 2 y = 0 x = 0 x = 2

Solution


=1048616 x3

3 + 2x)

2

0

= [23

3 + 2(2)] minus [03

3 + 2(0)]= [8

3 + 4] minus [0 + 0]

= 203 minus 0

= 203

23



Example 10


y = ex y = 0 x = 0 x = 2

Solution

Area=int 20 (ex)dx

= [ex]|20

= e2 minus e0

= e2 minus 1

Example 11


0 (1 + cos(x))dx

Solution


1048616 x + sin(x))|

π

0


24







Example 1


Solution

f (x) = (x2 + 4x)6




Example 2

Evaluateint

(2x + 3)(x2 + 3x)6dx

Solution


int (2x + 3)(x2 + 3x)6dx =

int u6 middot du


6+1u6+1 + C = 1

7u7 + C = 1

7(x2 + 3x)7 + C

Example 3

Evaluateint

3(1 + 3x)3dx

Solution


int 3(1 + 3x)3dx =

int u3 middot du

25




3+1u3+1 + C = 1

4u4 + C = 1

4(1 + 3x)4 + C

Example 4

Evaluateint

x2(2 + x3)4dx

Solution



int 13

u4 middot du


int 13


5u5 + C = 1

15u5 + C = 1

15(2 + x3)5 + C

26



Example 5

Evaluateint

2tradic

t2 + 3dt

Solution


radic t2 + 3dt =


int u

1

2 middot du


1


3

2 + C = 23

(t2 + 3)3

2 + C = 23

(radic

t2 + 3)3 + C

Example 6

Evaluateint

πcos(πx)dx

Solution


int cos(u)du



Example 7

Evaluateint

xradic 1minus4x2 dx

Solution


= xdx

int x

radic 1minus4x2

dx = int minus1

8 middot du



1


1

2 + C = minus14

u1

2 + C = minus14

(1 minus 4x2)1

2 + C = minus14


27



Example 8

Evaluateint

2xe2x2

dx

Solution


= 2xdx

int 2xe2x

2

dx =int

12

eudu


eudu = 12

eu + C = 12

e2x2

+ C

Example 9

Evaluateint

2xsin(2x2)dx

Solution


du = 2xdx

int 2xsin2x2dx =

int 12

sin(u)du


int 12

sin(u)du =minus

1

2cos(u) + C =

minus1

2cos(2x2) + C

28







2int 1u

du = ln|x| + C

Example 1

Evaluateint

13x

dx

Solution


3du = dxint 13x

dx =int

13 middot 1

u du


int 13 middot 1

u du = 1

3ln|u| + C = 1

3ln|3x| + C

Example 2

Evaluateint

15x+3

dx

Solution


du = dx

int 13x

dx =int

15 middot 1

u du


int 15 middot 1

u du = 1

5ln|u| + C = 1

5ln|5x + 3| + C

29



Example 3

Evaluateint

x2

3minusx3 dx

Solution


du = x2dx

int x

2

3minusx3 dx =int

minus13 middot 1

u du


int minus1

3 middot 1

u du = minus1

3ln|u| + C = minus1

3ln|3 minus x3| + C

Example 4

Evaluateint

sinθ

cosθ dθ

Solution


cosθ dθ =

int minus 1

u du


int minus 1


Example 5

Evaluateint

cos(t)3+sin(t)

dt

Solution

Let u = 3 + sin(t)

rArr du = cos(t)dt

int cos(t)3+sin(t)

dθ =int

1u

du


int 1u

du = ln|u| + C = ln|3 + sin(t)| + C

30





Theorem 48



1int

duradic a2minusu2

= arcsin u

a +C

2int

du

a2+u2 = 1a

arctan u

a +C

3int

du

uradic u2minusa2

= 1a

arcsec |u|a

+C



Example 1

Evaluateint

41+9x2 middotdx

Solution


3du = dx


int 41+9x2 middotdx

=int

41+(3x)2

middotdx

= 13

int 4du1+u2

= 43

int du

1+u2

= 4

3arctan(u) + C

= 43

arctan(3x) + C

Example 2

Evaluateint

dx

4+(x+1)2

Solution



31



int dx

4+(x+1)2 =int

du

22+u2

= 12

arctan(u2

) + C

= 12

arctan(x+12

) + C

Example 3

Evaluateint

7radic 4minusx2

middotdx

Solution



int 7

radic 4minusx2

= int 7du

radic 22

minusu2


) + C

= 7arcsin(x2

) + C

Example 4

Evaluateint

dx


Solution



int dx


=int

du

uradic u2minus9

= 13

arcsec(u3

) + C

= 13

arcsec( |x|3

) + C

32





Example 8

Evaluateint


Solution





rArr minus12

du = (x minus 3)dx

Now substitute


=int


8radic 9+(xminus3)2

= minus12

int dvradic


9+(u)2

= 12

int vminus 1


= minus12

(2v1

2 ) + 8arcsin(u3

) + C



) + C


34






int b

a f (x)dx is


f (x)dx = bminusa


Example 1


int 20

2x2dx n = 4

Solution

x = 2minus04

= 24

= 12


= i2


2(4)[f (0) + 2f (1

2) + 2f (1) + 2f (3

2) + f (2)]

= 28

[2(0)2 + 2(2(12

)2) + 2(2(1)2) + 2(2( 32

)2) + 2(22)]

= 1

4

[0 + 2(1

2

) + 2(2) + 2( 18

4

) + 8]= 1

4[0 + 1 + 4 + 9 + 8]

= 14

[22]= 55



a f (x)dx is


int b


Example 2


int 2

0 2x2dx n = 4

Solution

35



x = 2minus04

= 24

= 12


= i2

int 20

2x2dx = 2minus03(4)

[f (0) + 4f (12

) + 2f (1) + 4f (32

) + f (2)]

= 212

[2(0)2 + 4(2(12

)2) + 2(2(1)2) + 4(2(32

)2) + 2(22)]= 1

6[0 + 4(1

2) + 2(2) + 4( 18

4 ) + 8]

= 16

[0 + 2 + 4 + 18 + 8]

= 16

[32]= 53333


int 2

0 2x2dx = 22+1

x2+120

= 23

x320

= 23

(2)3 minus 23

(0)3


= 163

Example 3


int π0 sin(x)dx

Solution

x = πminus04

= π4


= iπ4int π


[f (0) + 2f (π4

) + 2f (2π4

) + 2f (3π4

) + f (π)]

= π8

[sin(0) + 2sin(π4

) + 2sin(π2

) + 2sin(3π4

) + sin(π)]

= π8 [0 + 2(radic 22 ) + 2(1) + 2(radic 22 ) + 0]

= π8

[2radic

2 + 2]asymp 2003


int π0 sin(x)dx

Solution

x = πminus04

= π4

36





= 112

[111 + 4(0947) + 2(0816) + 4(0771) + 0625] asymp 086



1(x+2)2

dx =int 2

11u2

du =int 2


x+2

21

= minus 12+2

minus (minus 11+2

)

= minus14

+ 13

= 112



Example 10


15

sum j=1

i(i2 + 1)

Solution10sum

j=1

i(i2 + 1) =10sum

j=1

i3 + i

= n

2(n+1)2

4 + n(n+1)

2 (where n=10)

= 102(10+1)2

4 + 10(10+1)

2


2= 25

middot121 + 5

middot11

= 3025 + 55 = 3080





limhrarrinfin




n

9



Example 10



Solution



= 14



= iminus14


= i

4


S (n) =4sum

j=1

f (mi)x

=4

sumi=1

f (iminus 1

4 ) middot (

1

4)

=4sum

i=1

radic iminus 1

4 middot 1

4

=4sum

i=1

radic iminus 1

2 middot 1

4

=4sum

i=1

radic iminus 1

8

=radic 1minus18 +

radic 2minus18 +

radic 3minus18 +

radic 4minus18

=

radic 0

8 +

radic 1

8 +

radic 2

8 +

radic 3

8 = 0 + 125 + 177 + 217 asymp 519

10




S (n) =4sum

j=1

f (M i)x

=4sum

i=1

f (i

4) middot (

1

4)

=4sum

i=1

radic i

4middot 1

4

=

4

sumi=1

radic i

2 middot1

4

=4sum

i=1

radic i

8

=radic 18

+radic 28

+radic 38

+radic 48

= 125 + 177 + 217 + 25 asymp 769

Example 11



n3[n(n+1)(2n+1)

6 ]

Solution

limhrarrinfin


64

n3[n(n + 1)(2n + 1)

6 ]

= limhrarrinfin

64

n3[(n2 + n))(2n + 1)

6 ]

= limhrarrinfin

64

n3[(2n3 + n2 + 2n2 + n

6 ]

= limhrarrinfin

64[2n3 + 3n2 + n

6n3 ]

= limhrarrinfin

128n3

+ 192n2

+ 64n6n3

11



= limhrarrinfin

128n3

6n3 +

192n2

6n3 +

64n

6n3

= limhrarrinfin

64

3 +

32

n +

32

3n2

= 643

+ 0 + 0 = 643

12



43 Riemann Sum




lexi

le1


nsumi=1

f (ci)xi




xi = i2


= i2


n2


n2

= i2minusi

2minus2i+1)n2

= minus2i+1n2


limhrarrinfin

nsumi=1


nsumi=1

f ( i2

n2) middot (

2iminus 1

n2 )

= limhrarrinfin

nsumi=1

radic i2

n2 middot (

2iminus 1

n2 )

= limhrarrinfin

nsumi=1

i

nmiddot 2iminus 1

n2

= limhrarrinfin

nsumi=1

2i2

minusi

n3

13



= 1

n3 limhrarrinfin

nsumi=1

(2i2 minus i)

= 1

n3 limhrarrinfin

[2(n(n + 1)(2n + 1)

6 ) minus n(n + 1)

2 ]

= 1


6 minus n2 + n

2 ]

= 1

n3 limhrarrinfin

((2n2 + 2n)(2n + 1)

6 minus n2 + n

2 ]

= 1

n3 limhrarrinfin

(4n3 + 6n2 + 2n

6 minus n2 + n

2 ]

= 1

n3 limhrarrinfin


6

= 1

n3 limhrarrinfin

4n3 + 3n2 minus n

6n3

= limhrarrinfin

[4n3 + 3n2 minus n

6n3

]

= limhrarrinfin

4n3

6n3 +

3n2

6n3 +

n

6n3

= limhrarrinfin

2

3 +

1

2n +

1

6n2 =

2

3 + 0 + 0 =

2

3




i=1




nsumi=1



nsumi=1

f (ci)xi =int

b

a f (x)dx

14



Example 2


Solution

int 3minus2

xdx

Solution


n = 3minus(minus2)

n


nrarrinfin

nsum

i=1

f (ci)xi

= limnrarrinfin

nsum

i=1

1048616minus 2 +

5i

n

1048617middot 5

n

= limnrarrinfin

nsum

i=1

1048616minus10

n +

25i

n2

1048617

= limnrarrinfin

nsum

i=1

1048616minus10

n +

25

n2 middot n(n + 1)

2

1048617

= limnrarrinfin

nsumi=1

1048616minus10n

+ 25n2

+ 25n2n2

1048617

= limnrarrinfin

nsum

i=1

1048616minus10

n +

25n2

2n2 +

25n

2n21048617

= limnrarrinfin

nsum

i=1

1048616minus10

n +

25

2 +

25

2n

1048617 = minus0 + 125 + 0 = 125

Example 3


Solution

15



int 31 3x2dx

Solution

a = 1 b = 3 x = 3minus1n

= 2n

ci = 1 + i middot 2i = 1 + 2i

n


nrarrinfin

nsum

i=1

f (ci)xi

= limnrarrinfin

nsum

i=1

f (1 + 2i

n ) middot

2

n

= limnrarrinfin

nsum

i=1

3(1 + 2i

n )2 middot

2

n

= limnrarrinfin

nsum

i=1

3(4i2

n2 +

4i

n + 1) middot

2

n

= limnrarrinfin

nsum

i=1

3( 8i2

n3 + 8i

n2 + 2

n)

= limnrarrinfin

nsum

i=1

(24i2

n3 +

24i

n2 +

6

n)

= limnrarrinfin

nsum

i=1

104861624(n(n + 1)(2n + 1))

6n3 +

24n(n + 1)

2n2 +

6n

n

1048617

= limnrarrinfin

nsum

i=1

104861624(2n3 + 3n2 + n)

6n3 +

24n2 + 24n

2n2 +

6n

n

1048617

= limnrarrinfin

n

sumi=1

104861648n3

6n3 + 72n2

6n3 + 24n

6n3 + 24n2

2n2 + 24n

n2 + 61048617

16



=1048616

8 + 12

n +

4

n2 + 12 +

12

n + 6

1048617 = 26

Example 4


Solution

a = 4 b = 10 x = 10minus4n

= 6n


= 4 + 6in

f (x) = 6


nrarrinfin

nsumi=1

f (ci)xi

= limnrarrinfin

nsumi=1

f 983080

4 + 6i

n

983081middot 6

n

= limnrarrinfin

nsumi=1

(6) middot 6

n

= limn

rarrinfin(6n)

middot

6

n = 36

17



Example 5


a = 1 b = 2 x = 2minus1n

= 1n


= 1 + i

n

f (x) = x2 + 1


int 21 (x2 + 1)dx = lim

nrarrinfin

nsumi=1

f (ci)xi

= limnrarrinfin

n

sumi=1 f

9830801 +

i

n983081 middot

1

n

= limnrarrinfin

nsumi=1

9830809830801 +

i

n

9830812+ 1

983081middot 1

n

= limnrarrinfin

nsumi=1

9830809830801 +

i

n

9830819830801 +

i

n

983081+ 1

983081middot 1

n

= limnrarrinfin

nsumi=1

98308010486161 +

2i

n +

i2

n2

983081+ 1

983081middot 1

n

= limnrarrinfin

n

sumi=1

9830802 + 2i

n

+ i2

n2983081 middot

1

n

= limnrarrinfin

nsumi=1

9830802

n +

2i

n2 +

i2

n3

983081

= limnrarrinfin

983080 2

n middot (n) +

2

n2 middot n(n + 1)

2 +

1

n3 middot n(n + 1)(2n + 1)

6

983081

= limnrarrinfin

9830802 +

2n2 + 2n

2n2 +

(n2 + n)(2n + 1)

6n3

983081

= limnrarrinfin

9830802 +

2n2 + 2n

2n2 +

2n3 + 3n2 + n

6n3

983081


2 + 2n2

2n2

+ 2n

2n2

+ 2n3

6n3

+ 3n2

6n3

+ n

6n3983081

= limnrarrinfin

9830802 + 1 +

1

n +

1

3 +

1

2n +

1

6n2

983081 = 2 + 1 + 0 +

1

3 + 0 + 0 =

10

3

18



Example 6


a = 0 b = 3

x = 3minus0

n = 3

n


= 3n

f (x) = 5x


int 3


n

sumi=1

f (ci)

xi

= limnrarrinfin

nsumi=1

f 9830803i

n

983081middot 3

n

= limnrarrinfin

nsumi=1

59830803i

n

983081middot 3

n

= limnrarrinfin

nsumi=1

45i

n2

= limnrarrinfin

nsum

i=1

45

n2 middot n(n + 1)

2

= limnrarrinfin

nsumi=1

98308045n(n + 1)

2n2

983081

= limnrarrinfin

nsumi=1

98308045n2 + 45n

2n2

983081

= limnrarrinfin

nsumi=1

98308045n2

2n2 +

45n

2n2

983081

= limnrarrinfin

nsumi=1

98308045

2 +

45

2n

983081 =

45

2 + 0 =

45

2

19






int b




Example 1


Solution

int 20

4xdx = 41+1

x220

= 2x2|20 = 2(2)2 minus 2(0)2 = 8 minus 0 = 8

Example 2


Solution

int 21

x3 + x2dx =1048616

13+1

x3+1 + 12+1

x2+1)21

=1048616 14

x4 + 13

x3)21

=104861614

24 + 13

23) minus (14

14 + 13

13)

= 4 + 83 minus 1048616

14

+ 13

)

= 203 minus 1

12

= 7912

Example 3


2

dx

Solution

int 20 6x2dx = 6

2+1x2+1

21


= 16 minus 0= 16

20



Example 4


exdx

Solution

int 20



Example 5


x+radic x

2 dx

= [1

2 middot x2

2 + 1

2 middot 1

2 middot x

3

2

]10

= [x2

4 + 1

3 middot x

3

2 ]10

= [12

4 + 1

3 middot 1

3

2 ] minus [02

4 + 1

3 middot 0

3

2 ]14 minus 1

3 minus 0 = 7

12

Solution

int 10

x+radic x

2 dx =

int 10

x2

+radic x

2 dx =


Example 6


Solution

Note

f (x) = 2

xminus 3

xge

3

2



int 32


int 23

2

(2x minus 3)dx

=int 3

2


3

2

(2x minus 3)dx

= (minusx2 + 3x)|3

2

0 + (x2 minus 3x)|232

= [((32

)2 + 3 middot 32



)]= 10

4

21








1bminusa

int b

a f (x)dx


Example 7


Solution



= 14


= 1

4 middot (1

3x3

+ x)2minus2

= 14 middot [1

3(2)3 + 2] minus 1

3 middot [1

3(minus2)3 minus 2]

= 14 middot 14

3 minus 1

3 middot minus14

3

= 76

+ 76

= 146

= 73

Example 8


2

]

Solution

AV = 1π

2minus0int π

2

0 cos(x)dx

= 2π

int π2

0 cos(x)dx

= 2π middot 1048616 sin(x))|

π

2

0


2) minus sin(0)]


= 2π middot 1

= 2

π

22



Example 9


y = x2 + 2 y = 0 x = 0 x = 2

Solution


=1048616 x3

3 + 2x)

2

0

= [23

3 + 2(2)] minus [03

3 + 2(0)]= [8

3 + 4] minus [0 + 0]

= 203 minus 0

= 203

23



Example 10


y = ex y = 0 x = 0 x = 2

Solution

Area=int 20 (ex)dx

= [ex]|20

= e2 minus e0

= e2 minus 1

Example 11


0 (1 + cos(x))dx

Solution


1048616 x + sin(x))|

π

0


24







Example 1


Solution

f (x) = (x2 + 4x)6




Example 2

Evaluateint

(2x + 3)(x2 + 3x)6dx

Solution


int (2x + 3)(x2 + 3x)6dx =

int u6 middot du


6+1u6+1 + C = 1

7u7 + C = 1

7(x2 + 3x)7 + C

Example 3

Evaluateint

3(1 + 3x)3dx

Solution


int 3(1 + 3x)3dx =

int u3 middot du

25




3+1u3+1 + C = 1

4u4 + C = 1

4(1 + 3x)4 + C

Example 4

Evaluateint

x2(2 + x3)4dx

Solution



int 13

u4 middot du


int 13


5u5 + C = 1

15u5 + C = 1

15(2 + x3)5 + C

26



Example 5

Evaluateint

2tradic

t2 + 3dt

Solution


radic t2 + 3dt =


int u

1

2 middot du


1


3

2 + C = 23

(t2 + 3)3

2 + C = 23

(radic

t2 + 3)3 + C

Example 6

Evaluateint

πcos(πx)dx

Solution


int cos(u)du



Example 7

Evaluateint

xradic 1minus4x2 dx

Solution


= xdx

int x

radic 1minus4x2

dx = int minus1

8 middot du



1


1

2 + C = minus14

u1

2 + C = minus14

(1 minus 4x2)1

2 + C = minus14


27



Example 8

Evaluateint

2xe2x2

dx

Solution


= 2xdx

int 2xe2x

2

dx =int

12

eudu


eudu = 12

eu + C = 12

e2x2

+ C

Example 9

Evaluateint

2xsin(2x2)dx

Solution


du = 2xdx

int 2xsin2x2dx =

int 12

sin(u)du


int 12

sin(u)du =minus

1

2cos(u) + C =

minus1

2cos(2x2) + C

28







2int 1u

du = ln|x| + C

Example 1

Evaluateint

13x

dx

Solution


3du = dxint 13x

dx =int

13 middot 1

u du


int 13 middot 1

u du = 1

3ln|u| + C = 1

3ln|3x| + C

Example 2

Evaluateint

15x+3

dx

Solution


du = dx

int 13x

dx =int

15 middot 1

u du


int 15 middot 1

u du = 1

5ln|u| + C = 1

5ln|5x + 3| + C

29



Example 3

Evaluateint

x2

3minusx3 dx

Solution


du = x2dx

int x

2

3minusx3 dx =int

minus13 middot 1

u du


int minus1

3 middot 1

u du = minus1

3ln|u| + C = minus1

3ln|3 minus x3| + C

Example 4

Evaluateint

sinθ

cosθ dθ

Solution


cosθ dθ =

int minus 1

u du


int minus 1


Example 5

Evaluateint

cos(t)3+sin(t)

dt

Solution

Let u = 3 + sin(t)

rArr du = cos(t)dt

int cos(t)3+sin(t)

dθ =int

1u

du


int 1u

du = ln|u| + C = ln|3 + sin(t)| + C

30





Theorem 48



1int

duradic a2minusu2

= arcsin u

a +C

2int

du

a2+u2 = 1a

arctan u

a +C

3int

du

uradic u2minusa2

= 1a

arcsec |u|a

+C



Example 1

Evaluateint

41+9x2 middotdx

Solution


3du = dx


int 41+9x2 middotdx

=int

41+(3x)2

middotdx

= 13

int 4du1+u2

= 43

int du

1+u2

= 4

3arctan(u) + C

= 43

arctan(3x) + C

Example 2

Evaluateint

dx

4+(x+1)2

Solution



31



int dx

4+(x+1)2 =int

du

22+u2

= 12

arctan(u2

) + C

= 12

arctan(x+12

) + C

Example 3

Evaluateint

7radic 4minusx2

middotdx

Solution



int 7

radic 4minusx2

= int 7du

radic 22

minusu2


) + C

= 7arcsin(x2

) + C

Example 4

Evaluateint

dx


Solution



int dx


=int

du

uradic u2minus9

= 13

arcsec(u3

) + C

= 13

arcsec( |x|3

) + C

32





Example 8

Evaluateint


Solution





rArr minus12

du = (x minus 3)dx

Now substitute


=int


8radic 9+(xminus3)2

= minus12

int dvradic


9+(u)2

= 12

int vminus 1


= minus12

(2v1

2 ) + 8arcsin(u3

) + C



) + C


34






int b

a f (x)dx is


f (x)dx = bminusa


Example 1


int 20

2x2dx n = 4

Solution

x = 2minus04

= 24

= 12


= i2


2(4)[f (0) + 2f (1

2) + 2f (1) + 2f (3

2) + f (2)]

= 28

[2(0)2 + 2(2(12

)2) + 2(2(1)2) + 2(2( 32

)2) + 2(22)]

= 1

4

[0 + 2(1

2

) + 2(2) + 2( 18

4

) + 8]= 1

4[0 + 1 + 4 + 9 + 8]

= 14

[22]= 55



a f (x)dx is


int b


Example 2


int 2

0 2x2dx n = 4

Solution

35



x = 2minus04

= 24

= 12


= i2

int 20

2x2dx = 2minus03(4)

[f (0) + 4f (12

) + 2f (1) + 4f (32

) + f (2)]

= 212

[2(0)2 + 4(2(12

)2) + 2(2(1)2) + 4(2(32

)2) + 2(22)]= 1

6[0 + 4(1

2) + 2(2) + 4( 18

4 ) + 8]

= 16

[0 + 2 + 4 + 18 + 8]

= 16

[32]= 53333


int 2

0 2x2dx = 22+1

x2+120

= 23

x320

= 23

(2)3 minus 23

(0)3


= 163

Example 3


int π0 sin(x)dx

Solution

x = πminus04

= π4


= iπ4int π


[f (0) + 2f (π4

) + 2f (2π4

) + 2f (3π4

) + f (π)]

= π8

[sin(0) + 2sin(π4

) + 2sin(π2

) + 2sin(3π4

) + sin(π)]

= π8 [0 + 2(radic 22 ) + 2(1) + 2(radic 22 ) + 0]

= π8

[2radic

2 + 2]asymp 2003


int π0 sin(x)dx

Solution

x = πminus04

= π4

36





= 112

[111 + 4(0947) + 2(0816) + 4(0771) + 0625] asymp 086



1(x+2)2

dx =int 2

11u2

du =int 2


x+2

21

= minus 12+2

minus (minus 11+2

)

= minus14

+ 13

= 112



Example 10



Solution



= 14



= iminus14


= i

4


S (n) =4sum

j=1

f (mi)x

=4

sumi=1

f (iminus 1

4 ) middot (

1

4)

=4sum

i=1

radic iminus 1

4 middot 1

4

=4sum

i=1

radic iminus 1

2 middot 1

4

=4sum

i=1

radic iminus 1

8

=radic 1minus18 +

radic 2minus18 +

radic 3minus18 +

radic 4minus18

=

radic 0

8 +

radic 1

8 +

radic 2

8 +

radic 3

8 = 0 + 125 + 177 + 217 asymp 519

10




S (n) =4sum

j=1

f (M i)x

=4sum

i=1

f (i

4) middot (

1

4)

=4sum

i=1

radic i

4middot 1

4

=

4

sumi=1

radic i

2 middot1

4

=4sum

i=1

radic i

8

=radic 18

+radic 28

+radic 38

+radic 48

= 125 + 177 + 217 + 25 asymp 769

Example 11



n3[n(n+1)(2n+1)

6 ]

Solution

limhrarrinfin


64

n3[n(n + 1)(2n + 1)

6 ]

= limhrarrinfin

64

n3[(n2 + n))(2n + 1)

6 ]

= limhrarrinfin

64

n3[(2n3 + n2 + 2n2 + n

6 ]

= limhrarrinfin

64[2n3 + 3n2 + n

6n3 ]

= limhrarrinfin

128n3

+ 192n2

+ 64n6n3

11



= limhrarrinfin

128n3

6n3 +

192n2

6n3 +

64n

6n3

= limhrarrinfin

64

3 +

32

n +

32

3n2

= 643

+ 0 + 0 = 643

12



43 Riemann Sum




lexi

le1


nsumi=1

f (ci)xi




xi = i2


= i2


n2


n2

= i2minusi

2minus2i+1)n2

= minus2i+1n2


limhrarrinfin

nsumi=1


nsumi=1

f ( i2

n2) middot (

2iminus 1

n2 )

= limhrarrinfin

nsumi=1

radic i2

n2 middot (

2iminus 1

n2 )

= limhrarrinfin

nsumi=1

i

nmiddot 2iminus 1

n2

= limhrarrinfin

nsumi=1

2i2

minusi

n3

13



= 1

n3 limhrarrinfin

nsumi=1

(2i2 minus i)

= 1

n3 limhrarrinfin

[2(n(n + 1)(2n + 1)

6 ) minus n(n + 1)

2 ]

= 1


6 minus n2 + n

2 ]

= 1

n3 limhrarrinfin

((2n2 + 2n)(2n + 1)

6 minus n2 + n

2 ]

= 1

n3 limhrarrinfin

(4n3 + 6n2 + 2n

6 minus n2 + n

2 ]

= 1

n3 limhrarrinfin


6

= 1

n3 limhrarrinfin

4n3 + 3n2 minus n

6n3

= limhrarrinfin

[4n3 + 3n2 minus n

6n3

]

= limhrarrinfin

4n3

6n3 +

3n2

6n3 +

n

6n3

= limhrarrinfin

2

3 +

1

2n +

1

6n2 =

2

3 + 0 + 0 =

2

3




i=1




nsumi=1



nsumi=1

f (ci)xi =int

b

a f (x)dx

14



Example 2


Solution

int 3minus2

xdx

Solution


n = 3minus(minus2)

n


nrarrinfin

nsum

i=1

f (ci)xi

= limnrarrinfin

nsum

i=1

1048616minus 2 +

5i

n

1048617middot 5

n

= limnrarrinfin

nsum

i=1

1048616minus10

n +

25i

n2

1048617

= limnrarrinfin

nsum

i=1

1048616minus10

n +

25

n2 middot n(n + 1)

2

1048617

= limnrarrinfin

nsumi=1

1048616minus10n

+ 25n2

+ 25n2n2

1048617

= limnrarrinfin

nsum

i=1

1048616minus10

n +

25n2

2n2 +

25n

2n21048617

= limnrarrinfin

nsum

i=1

1048616minus10

n +

25

2 +

25

2n

1048617 = minus0 + 125 + 0 = 125

Example 3


Solution

15



int 31 3x2dx

Solution

a = 1 b = 3 x = 3minus1n

= 2n

ci = 1 + i middot 2i = 1 + 2i

n


nrarrinfin

nsum

i=1

f (ci)xi

= limnrarrinfin

nsum

i=1

f (1 + 2i

n ) middot

2

n

= limnrarrinfin

nsum

i=1

3(1 + 2i

n )2 middot

2

n

= limnrarrinfin

nsum

i=1

3(4i2

n2 +

4i

n + 1) middot

2

n

= limnrarrinfin

nsum

i=1

3( 8i2

n3 + 8i

n2 + 2

n)

= limnrarrinfin

nsum

i=1

(24i2

n3 +

24i

n2 +

6

n)

= limnrarrinfin

nsum

i=1

104861624(n(n + 1)(2n + 1))

6n3 +

24n(n + 1)

2n2 +

6n

n

1048617

= limnrarrinfin

nsum

i=1

104861624(2n3 + 3n2 + n)

6n3 +

24n2 + 24n

2n2 +

6n

n

1048617

= limnrarrinfin

n

sumi=1

104861648n3

6n3 + 72n2

6n3 + 24n

6n3 + 24n2

2n2 + 24n

n2 + 61048617

16



=1048616

8 + 12

n +

4

n2 + 12 +

12

n + 6

1048617 = 26

Example 4


Solution

a = 4 b = 10 x = 10minus4n

= 6n


= 4 + 6in

f (x) = 6


nrarrinfin

nsumi=1

f (ci)xi

= limnrarrinfin

nsumi=1

f 983080

4 + 6i

n

983081middot 6

n

= limnrarrinfin

nsumi=1

(6) middot 6

n

= limn

rarrinfin(6n)

middot

6

n = 36

17



Example 5


a = 1 b = 2 x = 2minus1n

= 1n


= 1 + i

n

f (x) = x2 + 1


int 21 (x2 + 1)dx = lim

nrarrinfin

nsumi=1

f (ci)xi

= limnrarrinfin

n

sumi=1 f

9830801 +

i

n983081 middot

1

n

= limnrarrinfin

nsumi=1

9830809830801 +

i

n

9830812+ 1

983081middot 1

n

= limnrarrinfin

nsumi=1

9830809830801 +

i

n

9830819830801 +

i

n

983081+ 1

983081middot 1

n

= limnrarrinfin

nsumi=1

98308010486161 +

2i

n +

i2

n2

983081+ 1

983081middot 1

n

= limnrarrinfin

n

sumi=1

9830802 + 2i

n

+ i2

n2983081 middot

1

n

= limnrarrinfin

nsumi=1

9830802

n +

2i

n2 +

i2

n3

983081

= limnrarrinfin

983080 2

n middot (n) +

2

n2 middot n(n + 1)

2 +

1

n3 middot n(n + 1)(2n + 1)

6

983081

= limnrarrinfin

9830802 +

2n2 + 2n

2n2 +

(n2 + n)(2n + 1)

6n3

983081

= limnrarrinfin

9830802 +

2n2 + 2n

2n2 +

2n3 + 3n2 + n

6n3

983081


2 + 2n2

2n2

+ 2n

2n2

+ 2n3

6n3

+ 3n2

6n3

+ n

6n3983081

= limnrarrinfin

9830802 + 1 +

1

n +

1

3 +

1

2n +

1

6n2

983081 = 2 + 1 + 0 +

1

3 + 0 + 0 =

10

3

18



Example 6


a = 0 b = 3

x = 3minus0

n = 3

n


= 3n

f (x) = 5x


int 3


n

sumi=1

f (ci)

xi

= limnrarrinfin

nsumi=1

f 9830803i

n

983081middot 3

n

= limnrarrinfin

nsumi=1

59830803i

n

983081middot 3

n

= limnrarrinfin

nsumi=1

45i

n2

= limnrarrinfin

nsum

i=1

45

n2 middot n(n + 1)

2

= limnrarrinfin

nsumi=1

98308045n(n + 1)

2n2

983081

= limnrarrinfin

nsumi=1

98308045n2 + 45n

2n2

983081

= limnrarrinfin

nsumi=1

98308045n2

2n2 +

45n

2n2

983081

= limnrarrinfin

nsumi=1

98308045

2 +

45

2n

983081 =

45

2 + 0 =

45

2

19






int b




Example 1


Solution

int 20

4xdx = 41+1

x220

= 2x2|20 = 2(2)2 minus 2(0)2 = 8 minus 0 = 8

Example 2


Solution

int 21

x3 + x2dx =1048616

13+1

x3+1 + 12+1

x2+1)21

=1048616 14

x4 + 13

x3)21

=104861614

24 + 13

23) minus (14

14 + 13

13)

= 4 + 83 minus 1048616

14

+ 13

)

= 203 minus 1

12

= 7912

Example 3


2

dx

Solution

int 20 6x2dx = 6

2+1x2+1

21


= 16 minus 0= 16

20



Example 4


exdx

Solution

int 20



Example 5


x+radic x

2 dx

= [1

2 middot x2

2 + 1

2 middot 1

2 middot x

3

2

]10

= [x2

4 + 1

3 middot x

3

2 ]10

= [12

4 + 1

3 middot 1

3

2 ] minus [02

4 + 1

3 middot 0

3

2 ]14 minus 1

3 minus 0 = 7

12

Solution

int 10

x+radic x

2 dx =

int 10

x2

+radic x

2 dx =


Example 6


Solution

Note

f (x) = 2

xminus 3

xge

3

2



int 32


int 23

2

(2x minus 3)dx

=int 3

2


3

2

(2x minus 3)dx

= (minusx2 + 3x)|3

2

0 + (x2 minus 3x)|232

= [((32

)2 + 3 middot 32



)]= 10

4

21








1bminusa

int b

a f (x)dx


Example 7


Solution



= 14


= 1

4 middot (1

3x3

+ x)2minus2

= 14 middot [1

3(2)3 + 2] minus 1

3 middot [1

3(minus2)3 minus 2]

= 14 middot 14

3 minus 1

3 middot minus14

3

= 76

+ 76

= 146

= 73

Example 8


2

]

Solution

AV = 1π

2minus0int π

2

0 cos(x)dx

= 2π

int π2

0 cos(x)dx

= 2π middot 1048616 sin(x))|

π

2

0


2) minus sin(0)]


= 2π middot 1

= 2

π

22



Example 9


y = x2 + 2 y = 0 x = 0 x = 2

Solution


=1048616 x3

3 + 2x)

2

0

= [23

3 + 2(2)] minus [03

3 + 2(0)]= [8

3 + 4] minus [0 + 0]

= 203 minus 0

= 203

23



Example 10


y = ex y = 0 x = 0 x = 2

Solution

Area=int 20 (ex)dx

= [ex]|20

= e2 minus e0

= e2 minus 1

Example 11


0 (1 + cos(x))dx

Solution


1048616 x + sin(x))|

π

0


24







Example 1


Solution

f (x) = (x2 + 4x)6




Example 2

Evaluateint

(2x + 3)(x2 + 3x)6dx

Solution


int (2x + 3)(x2 + 3x)6dx =

int u6 middot du


6+1u6+1 + C = 1

7u7 + C = 1

7(x2 + 3x)7 + C

Example 3

Evaluateint

3(1 + 3x)3dx

Solution


int 3(1 + 3x)3dx =

int u3 middot du

25




3+1u3+1 + C = 1

4u4 + C = 1

4(1 + 3x)4 + C

Example 4

Evaluateint

x2(2 + x3)4dx

Solution



int 13

u4 middot du


int 13


5u5 + C = 1

15u5 + C = 1

15(2 + x3)5 + C

26



Example 5

Evaluateint

2tradic

t2 + 3dt

Solution


radic t2 + 3dt =


int u

1

2 middot du


1


3

2 + C = 23

(t2 + 3)3

2 + C = 23

(radic

t2 + 3)3 + C

Example 6

Evaluateint

πcos(πx)dx

Solution


int cos(u)du



Example 7

Evaluateint

xradic 1minus4x2 dx

Solution


= xdx

int x

radic 1minus4x2

dx = int minus1

8 middot du



1


1

2 + C = minus14

u1

2 + C = minus14

(1 minus 4x2)1

2 + C = minus14


27



Example 8

Evaluateint

2xe2x2

dx

Solution


= 2xdx

int 2xe2x

2

dx =int

12

eudu


eudu = 12

eu + C = 12

e2x2

+ C

Example 9

Evaluateint

2xsin(2x2)dx

Solution


du = 2xdx

int 2xsin2x2dx =

int 12

sin(u)du


int 12

sin(u)du =minus

1

2cos(u) + C =

minus1

2cos(2x2) + C

28







2int 1u

du = ln|x| + C

Example 1

Evaluateint

13x

dx

Solution


3du = dxint 13x

dx =int

13 middot 1

u du


int 13 middot 1

u du = 1

3ln|u| + C = 1

3ln|3x| + C

Example 2

Evaluateint

15x+3

dx

Solution


du = dx

int 13x

dx =int

15 middot 1

u du


int 15 middot 1

u du = 1

5ln|u| + C = 1

5ln|5x + 3| + C

29



Example 3

Evaluateint

x2

3minusx3 dx

Solution


du = x2dx

int x

2

3minusx3 dx =int

minus13 middot 1

u du


int minus1

3 middot 1

u du = minus1

3ln|u| + C = minus1

3ln|3 minus x3| + C

Example 4

Evaluateint

sinθ

cosθ dθ

Solution


cosθ dθ =

int minus 1

u du


int minus 1


Example 5

Evaluateint

cos(t)3+sin(t)

dt

Solution

Let u = 3 + sin(t)

rArr du = cos(t)dt

int cos(t)3+sin(t)

dθ =int

1u

du


int 1u

du = ln|u| + C = ln|3 + sin(t)| + C

30





Theorem 48



1int

duradic a2minusu2

= arcsin u

a +C

2int

du

a2+u2 = 1a

arctan u

a +C

3int

du

uradic u2minusa2

= 1a

arcsec |u|a

+C



Example 1

Evaluateint

41+9x2 middotdx

Solution


3du = dx


int 41+9x2 middotdx

=int

41+(3x)2

middotdx

= 13

int 4du1+u2

= 43

int du

1+u2

= 4

3arctan(u) + C

= 43

arctan(3x) + C

Example 2

Evaluateint

dx

4+(x+1)2

Solution



31



int dx

4+(x+1)2 =int

du

22+u2

= 12

arctan(u2

) + C

= 12

arctan(x+12

) + C

Example 3

Evaluateint

7radic 4minusx2

middotdx

Solution



int 7

radic 4minusx2

= int 7du

radic 22

minusu2


) + C

= 7arcsin(x2

) + C

Example 4

Evaluateint

dx


Solution



int dx


=int

du

uradic u2minus9

= 13

arcsec(u3

) + C

= 13

arcsec( |x|3

) + C

32





Example 8

Evaluateint


Solution





rArr minus12

du = (x minus 3)dx

Now substitute


=int


8radic 9+(xminus3)2

= minus12

int dvradic


9+(u)2

= 12

int vminus 1


= minus12

(2v1

2 ) + 8arcsin(u3

) + C



) + C


34






int b

a f (x)dx is


f (x)dx = bminusa


Example 1


int 20

2x2dx n = 4

Solution

x = 2minus04

= 24

= 12


= i2


2(4)[f (0) + 2f (1

2) + 2f (1) + 2f (3

2) + f (2)]

= 28

[2(0)2 + 2(2(12

)2) + 2(2(1)2) + 2(2( 32

)2) + 2(22)]

= 1

4

[0 + 2(1

2

) + 2(2) + 2( 18

4

) + 8]= 1

4[0 + 1 + 4 + 9 + 8]

= 14

[22]= 55



a f (x)dx is


int b


Example 2


int 2

0 2x2dx n = 4

Solution

35



x = 2minus04

= 24

= 12


= i2

int 20

2x2dx = 2minus03(4)

[f (0) + 4f (12

) + 2f (1) + 4f (32

) + f (2)]

= 212

[2(0)2 + 4(2(12

)2) + 2(2(1)2) + 4(2(32

)2) + 2(22)]= 1

6[0 + 4(1

2) + 2(2) + 4( 18

4 ) + 8]

= 16

[0 + 2 + 4 + 18 + 8]

= 16

[32]= 53333


int 2

0 2x2dx = 22+1

x2+120

= 23

x320

= 23

(2)3 minus 23

(0)3


= 163

Example 3


int π0 sin(x)dx

Solution

x = πminus04

= π4


= iπ4int π


[f (0) + 2f (π4

) + 2f (2π4

) + 2f (3π4

) + f (π)]

= π8

[sin(0) + 2sin(π4

) + 2sin(π2

) + 2sin(3π4

) + sin(π)]

= π8 [0 + 2(radic 22 ) + 2(1) + 2(radic 22 ) + 0]

= π8

[2radic

2 + 2]asymp 2003


int π0 sin(x)dx

Solution

x = πminus04

= π4

36





= 112

[111 + 4(0947) + 2(0816) + 4(0771) + 0625] asymp 086



1(x+2)2

dx =int 2

11u2

du =int 2


x+2

21

= minus 12+2

minus (minus 11+2

)

= minus14

+ 13

= 112




S (n) =4sum

j=1

f (M i)x

=4sum

i=1

f (i

4) middot (

1

4)

=4sum

i=1

radic i

4middot 1

4

=

4

sumi=1

radic i

2 middot1

4

=4sum

i=1

radic i

8

=radic 18

+radic 28

+radic 38

+radic 48

= 125 + 177 + 217 + 25 asymp 769

Example 11



n3[n(n+1)(2n+1)

6 ]

Solution

limhrarrinfin


64

n3[n(n + 1)(2n + 1)

6 ]

= limhrarrinfin

64

n3[(n2 + n))(2n + 1)

6 ]

= limhrarrinfin

64

n3[(2n3 + n2 + 2n2 + n

6 ]

= limhrarrinfin

64[2n3 + 3n2 + n

6n3 ]

= limhrarrinfin

128n3

+ 192n2

+ 64n6n3

11



= limhrarrinfin

128n3

6n3 +

192n2

6n3 +

64n

6n3

= limhrarrinfin

64

3 +

32

n +

32

3n2

= 643

+ 0 + 0 = 643

12



43 Riemann Sum




lexi

le1


nsumi=1

f (ci)xi




xi = i2


= i2


n2


n2

= i2minusi

2minus2i+1)n2

= minus2i+1n2


limhrarrinfin

nsumi=1


nsumi=1

f ( i2

n2) middot (

2iminus 1

n2 )

= limhrarrinfin

nsumi=1

radic i2

n2 middot (

2iminus 1

n2 )

= limhrarrinfin

nsumi=1

i

nmiddot 2iminus 1

n2

= limhrarrinfin

nsumi=1

2i2

minusi

n3

13



= 1

n3 limhrarrinfin

nsumi=1

(2i2 minus i)

= 1

n3 limhrarrinfin

[2(n(n + 1)(2n + 1)

6 ) minus n(n + 1)

2 ]

= 1


6 minus n2 + n

2 ]

= 1

n3 limhrarrinfin

((2n2 + 2n)(2n + 1)

6 minus n2 + n

2 ]

= 1

n3 limhrarrinfin

(4n3 + 6n2 + 2n

6 minus n2 + n

2 ]

= 1

n3 limhrarrinfin


6

= 1

n3 limhrarrinfin

4n3 + 3n2 minus n

6n3

= limhrarrinfin

[4n3 + 3n2 minus n

6n3

]

= limhrarrinfin

4n3

6n3 +

3n2

6n3 +

n

6n3

= limhrarrinfin

2

3 +

1

2n +

1

6n2 =

2

3 + 0 + 0 =

2

3




i=1




nsumi=1



nsumi=1

f (ci)xi =int

b

a f (x)dx

14



Example 2


Solution

int 3minus2

xdx

Solution


n = 3minus(minus2)

n


nrarrinfin

nsum

i=1

f (ci)xi

= limnrarrinfin

nsum

i=1

1048616minus 2 +

5i

n

1048617middot 5

n

= limnrarrinfin

nsum

i=1

1048616minus10

n +

25i

n2

1048617

= limnrarrinfin

nsum

i=1

1048616minus10

n +

25

n2 middot n(n + 1)

2

1048617

= limnrarrinfin

nsumi=1

1048616minus10n

+ 25n2

+ 25n2n2

1048617

= limnrarrinfin

nsum

i=1

1048616minus10

n +

25n2

2n2 +

25n

2n21048617

= limnrarrinfin

nsum

i=1

1048616minus10

n +

25

2 +

25

2n

1048617 = minus0 + 125 + 0 = 125

Example 3


Solution

15



int 31 3x2dx

Solution

a = 1 b = 3 x = 3minus1n

= 2n

ci = 1 + i middot 2i = 1 + 2i

n


nrarrinfin

nsum

i=1

f (ci)xi

= limnrarrinfin

nsum

i=1

f (1 + 2i

n ) middot

2

n

= limnrarrinfin

nsum

i=1

3(1 + 2i

n )2 middot

2

n

= limnrarrinfin

nsum

i=1

3(4i2

n2 +

4i

n + 1) middot

2

n

= limnrarrinfin

nsum

i=1

3( 8i2

n3 + 8i

n2 + 2

n)

= limnrarrinfin

nsum

i=1

(24i2

n3 +

24i

n2 +

6

n)

= limnrarrinfin

nsum

i=1

104861624(n(n + 1)(2n + 1))

6n3 +

24n(n + 1)

2n2 +

6n

n

1048617

= limnrarrinfin

nsum

i=1

104861624(2n3 + 3n2 + n)

6n3 +

24n2 + 24n

2n2 +

6n

n

1048617

= limnrarrinfin

n

sumi=1

104861648n3

6n3 + 72n2

6n3 + 24n

6n3 + 24n2

2n2 + 24n

n2 + 61048617

16



=1048616

8 + 12

n +

4

n2 + 12 +

12

n + 6

1048617 = 26

Example 4


Solution

a = 4 b = 10 x = 10minus4n

= 6n


= 4 + 6in

f (x) = 6


nrarrinfin

nsumi=1

f (ci)xi

= limnrarrinfin

nsumi=1

f 983080

4 + 6i

n

983081middot 6

n

= limnrarrinfin

nsumi=1

(6) middot 6

n

= limn

rarrinfin(6n)

middot

6

n = 36

17



Example 5


a = 1 b = 2 x = 2minus1n

= 1n


= 1 + i

n

f (x) = x2 + 1


int 21 (x2 + 1)dx = lim

nrarrinfin

nsumi=1

f (ci)xi

= limnrarrinfin

n

sumi=1 f

9830801 +

i

n983081 middot

1

n

= limnrarrinfin

nsumi=1

9830809830801 +

i

n

9830812+ 1

983081middot 1

n

= limnrarrinfin

nsumi=1

9830809830801 +

i

n

9830819830801 +

i

n

983081+ 1

983081middot 1

n

= limnrarrinfin

nsumi=1

98308010486161 +

2i

n +

i2

n2

983081+ 1

983081middot 1

n

= limnrarrinfin

n

sumi=1

9830802 + 2i

n

+ i2

n2983081 middot

1

n

= limnrarrinfin

nsumi=1

9830802

n +

2i

n2 +

i2

n3

983081

= limnrarrinfin

983080 2

n middot (n) +

2

n2 middot n(n + 1)

2 +

1

n3 middot n(n + 1)(2n + 1)

6

983081

= limnrarrinfin

9830802 +

2n2 + 2n

2n2 +

(n2 + n)(2n + 1)

6n3

983081

= limnrarrinfin

9830802 +

2n2 + 2n

2n2 +

2n3 + 3n2 + n

6n3

983081


2 + 2n2

2n2

+ 2n

2n2

+ 2n3

6n3

+ 3n2

6n3

+ n

6n3983081

= limnrarrinfin

9830802 + 1 +

1

n +

1

3 +

1

2n +

1

6n2

983081 = 2 + 1 + 0 +

1

3 + 0 + 0 =

10

3

18



Example 6


a = 0 b = 3

x = 3minus0

n = 3

n


= 3n

f (x) = 5x


int 3


n

sumi=1

f (ci)

xi

= limnrarrinfin

nsumi=1

f 9830803i

n

983081middot 3

n

= limnrarrinfin

nsumi=1

59830803i

n

983081middot 3

n

= limnrarrinfin

nsumi=1

45i

n2

= limnrarrinfin

nsum

i=1

45

n2 middot n(n + 1)

2

= limnrarrinfin

nsumi=1

98308045n(n + 1)

2n2

983081

= limnrarrinfin

nsumi=1

98308045n2 + 45n

2n2

983081

= limnrarrinfin

nsumi=1

98308045n2

2n2 +

45n

2n2

983081

= limnrarrinfin

nsumi=1

98308045

2 +

45

2n

983081 =

45

2 + 0 =

45

2

19






int b




Example 1


Solution

int 20

4xdx = 41+1

x220

= 2x2|20 = 2(2)2 minus 2(0)2 = 8 minus 0 = 8

Example 2


Solution

int 21

x3 + x2dx =1048616

13+1

x3+1 + 12+1

x2+1)21

=1048616 14

x4 + 13

x3)21

=104861614

24 + 13

23) minus (14

14 + 13

13)

= 4 + 83 minus 1048616

14

+ 13

)

= 203 minus 1

12

= 7912

Example 3


2

dx

Solution

int 20 6x2dx = 6

2+1x2+1

21


= 16 minus 0= 16

20



Example 4


exdx

Solution

int 20



Example 5


x+radic x

2 dx

= [1

2 middot x2

2 + 1

2 middot 1

2 middot x

3

2

]10

= [x2

4 + 1

3 middot x

3

2 ]10

= [12

4 + 1

3 middot 1

3

2 ] minus [02

4 + 1

3 middot 0

3

2 ]14 minus 1

3 minus 0 = 7

12

Solution

int 10

x+radic x

2 dx =

int 10

x2

+radic x

2 dx =


Example 6


Solution

Note

f (x) = 2

xminus 3

xge

3

2



int 32


int 23

2

(2x minus 3)dx

=int 3

2


3

2

(2x minus 3)dx

= (minusx2 + 3x)|3

2

0 + (x2 minus 3x)|232

= [((32

)2 + 3 middot 32



)]= 10

4

21








1bminusa

int b

a f (x)dx


Example 7


Solution



= 14


= 1

4 middot (1

3x3

+ x)2minus2

= 14 middot [1

3(2)3 + 2] minus 1

3 middot [1

3(minus2)3 minus 2]

= 14 middot 14

3 minus 1

3 middot minus14

3

= 76

+ 76

= 146

= 73

Example 8


2

]

Solution

AV = 1π

2minus0int π

2

0 cos(x)dx

= 2π

int π2

0 cos(x)dx

= 2π middot 1048616 sin(x))|

π

2

0


2) minus sin(0)]


= 2π middot 1

= 2

π

22



Example 9


y = x2 + 2 y = 0 x = 0 x = 2

Solution


=1048616 x3

3 + 2x)

2

0

= [23

3 + 2(2)] minus [03

3 + 2(0)]= [8

3 + 4] minus [0 + 0]

= 203 minus 0

= 203

23



Example 10


y = ex y = 0 x = 0 x = 2

Solution

Area=int 20 (ex)dx

= [ex]|20

= e2 minus e0

= e2 minus 1

Example 11


0 (1 + cos(x))dx

Solution


1048616 x + sin(x))|

π

0


24







Example 1


Solution

f (x) = (x2 + 4x)6




Example 2

Evaluateint

(2x + 3)(x2 + 3x)6dx

Solution


int (2x + 3)(x2 + 3x)6dx =

int u6 middot du


6+1u6+1 + C = 1

7u7 + C = 1

7(x2 + 3x)7 + C

Example 3

Evaluateint

3(1 + 3x)3dx

Solution


int 3(1 + 3x)3dx =

int u3 middot du

25




3+1u3+1 + C = 1

4u4 + C = 1

4(1 + 3x)4 + C

Example 4

Evaluateint

x2(2 + x3)4dx

Solution



int 13

u4 middot du


int 13


5u5 + C = 1

15u5 + C = 1

15(2 + x3)5 + C

26



Example 5

Evaluateint

2tradic

t2 + 3dt

Solution


radic t2 + 3dt =


int u

1

2 middot du


1


3

2 + C = 23

(t2 + 3)3

2 + C = 23

(radic

t2 + 3)3 + C

Example 6

Evaluateint

πcos(πx)dx

Solution


int cos(u)du



Example 7

Evaluateint

xradic 1minus4x2 dx

Solution


= xdx

int x

radic 1minus4x2

dx = int minus1

8 middot du



1


1

2 + C = minus14

u1

2 + C = minus14

(1 minus 4x2)1

2 + C = minus14


27



Example 8

Evaluateint

2xe2x2

dx

Solution


= 2xdx

int 2xe2x

2

dx =int

12

eudu


eudu = 12

eu + C = 12

e2x2

+ C

Example 9

Evaluateint

2xsin(2x2)dx

Solution


du = 2xdx

int 2xsin2x2dx =

int 12

sin(u)du


int 12

sin(u)du =minus

1

2cos(u) + C =

minus1

2cos(2x2) + C

28







2int 1u

du = ln|x| + C

Example 1

Evaluateint

13x

dx

Solution


3du = dxint 13x

dx =int

13 middot 1

u du


int 13 middot 1

u du = 1

3ln|u| + C = 1

3ln|3x| + C

Example 2

Evaluateint

15x+3

dx

Solution


du = dx

int 13x

dx =int

15 middot 1

u du


int 15 middot 1

u du = 1

5ln|u| + C = 1

5ln|5x + 3| + C

29



Example 3

Evaluateint

x2

3minusx3 dx

Solution


du = x2dx

int x

2

3minusx3 dx =int

minus13 middot 1

u du


int minus1

3 middot 1

u du = minus1

3ln|u| + C = minus1

3ln|3 minus x3| + C

Example 4

Evaluateint

sinθ

cosθ dθ

Solution


cosθ dθ =

int minus 1

u du


int minus 1


Example 5

Evaluateint

cos(t)3+sin(t)

dt

Solution

Let u = 3 + sin(t)

rArr du = cos(t)dt

int cos(t)3+sin(t)

dθ =int

1u

du


int 1u

du = ln|u| + C = ln|3 + sin(t)| + C

30





Theorem 48



1int

duradic a2minusu2

= arcsin u

a +C

2int

du

a2+u2 = 1a

arctan u

a +C

3int

du

uradic u2minusa2

= 1a

arcsec |u|a

+C



Example 1

Evaluateint

41+9x2 middotdx

Solution


3du = dx


int 41+9x2 middotdx

=int

41+(3x)2

middotdx

= 13

int 4du1+u2

= 43

int du

1+u2

= 4

3arctan(u) + C

= 43

arctan(3x) + C

Example 2

Evaluateint

dx

4+(x+1)2

Solution



31



int dx

4+(x+1)2 =int

du

22+u2

= 12

arctan(u2

) + C

= 12

arctan(x+12

) + C

Example 3

Evaluateint

7radic 4minusx2

middotdx

Solution



int 7

radic 4minusx2

= int 7du

radic 22

minusu2


) + C

= 7arcsin(x2

) + C

Example 4

Evaluateint

dx


Solution



int dx


=int

du

uradic u2minus9

= 13

arcsec(u3

) + C

= 13

arcsec( |x|3

) + C

32





Example 8

Evaluateint


Solution





rArr minus12

du = (x minus 3)dx

Now substitute


=int


8radic 9+(xminus3)2

= minus12

int dvradic


9+(u)2

= 12

int vminus 1


= minus12

(2v1

2 ) + 8arcsin(u3

) + C



) + C


34






int b

a f (x)dx is


f (x)dx = bminusa


Example 1


int 20

2x2dx n = 4

Solution

x = 2minus04

= 24

= 12


= i2


2(4)[f (0) + 2f (1

2) + 2f (1) + 2f (3

2) + f (2)]

= 28

[2(0)2 + 2(2(12

)2) + 2(2(1)2) + 2(2( 32

)2) + 2(22)]

= 1

4

[0 + 2(1

2

) + 2(2) + 2( 18

4

) + 8]= 1

4[0 + 1 + 4 + 9 + 8]

= 14

[22]= 55



a f (x)dx is


int b


Example 2


int 2

0 2x2dx n = 4

Solution

35



x = 2minus04

= 24

= 12


= i2

int 20

2x2dx = 2minus03(4)

[f (0) + 4f (12

) + 2f (1) + 4f (32

) + f (2)]

= 212

[2(0)2 + 4(2(12

)2) + 2(2(1)2) + 4(2(32

)2) + 2(22)]= 1

6[0 + 4(1

2) + 2(2) + 4( 18

4 ) + 8]

= 16

[0 + 2 + 4 + 18 + 8]

= 16

[32]= 53333


int 2

0 2x2dx = 22+1

x2+120

= 23

x320

= 23

(2)3 minus 23

(0)3


= 163

Example 3


int π0 sin(x)dx

Solution

x = πminus04

= π4


= iπ4int π


[f (0) + 2f (π4

) + 2f (2π4

) + 2f (3π4

) + f (π)]

= π8

[sin(0) + 2sin(π4

) + 2sin(π2

) + 2sin(3π4

) + sin(π)]

= π8 [0 + 2(radic 22 ) + 2(1) + 2(radic 22 ) + 0]

= π8

[2radic

2 + 2]asymp 2003


int π0 sin(x)dx

Solution

x = πminus04

= π4

36





= 112

[111 + 4(0947) + 2(0816) + 4(0771) + 0625] asymp 086



1(x+2)2

dx =int 2

11u2

du =int 2


x+2

21

= minus 12+2

minus (minus 11+2

)

= minus14

+ 13

= 112



= limhrarrinfin

128n3

6n3 +

192n2

6n3 +

64n

6n3

= limhrarrinfin

64

3 +

32

n +

32

3n2

= 643

+ 0 + 0 = 643

12



43 Riemann Sum




lexi

le1


nsumi=1

f (ci)xi




xi = i2


= i2


n2


n2

= i2minusi

2minus2i+1)n2

= minus2i+1n2


limhrarrinfin

nsumi=1


nsumi=1

f ( i2

n2) middot (

2iminus 1

n2 )

= limhrarrinfin

nsumi=1

radic i2

n2 middot (

2iminus 1

n2 )

= limhrarrinfin

nsumi=1

i

nmiddot 2iminus 1

n2

= limhrarrinfin

nsumi=1

2i2

minusi

n3

13



= 1

n3 limhrarrinfin

nsumi=1

(2i2 minus i)

= 1

n3 limhrarrinfin

[2(n(n + 1)(2n + 1)

6 ) minus n(n + 1)

2 ]

= 1


6 minus n2 + n

2 ]

= 1

n3 limhrarrinfin

((2n2 + 2n)(2n + 1)

6 minus n2 + n

2 ]

= 1

n3 limhrarrinfin

(4n3 + 6n2 + 2n

6 minus n2 + n

2 ]

= 1

n3 limhrarrinfin


6

= 1

n3 limhrarrinfin

4n3 + 3n2 minus n

6n3

= limhrarrinfin

[4n3 + 3n2 minus n

6n3

]

= limhrarrinfin

4n3

6n3 +

3n2

6n3 +

n

6n3

= limhrarrinfin

2

3 +

1

2n +

1

6n2 =

2

3 + 0 + 0 =

2

3




i=1




nsumi=1



nsumi=1

f (ci)xi =int

b

a f (x)dx

14



Example 2


Solution

int 3minus2

xdx

Solution


n = 3minus(minus2)

n


nrarrinfin

nsum

i=1

f (ci)xi

= limnrarrinfin

nsum

i=1

1048616minus 2 +

5i

n

1048617middot 5

n

= limnrarrinfin

nsum

i=1

1048616minus10

n +

25i

n2

1048617

= limnrarrinfin

nsum

i=1

1048616minus10

n +

25

n2 middot n(n + 1)

2

1048617

= limnrarrinfin

nsumi=1

1048616minus10n

+ 25n2

+ 25n2n2

1048617

= limnrarrinfin

nsum

i=1

1048616minus10

n +

25n2

2n2 +

25n

2n21048617

= limnrarrinfin

nsum

i=1

1048616minus10

n +

25

2 +

25

2n

1048617 = minus0 + 125 + 0 = 125

Example 3


Solution

15



int 31 3x2dx

Solution

a = 1 b = 3 x = 3minus1n

= 2n

ci = 1 + i middot 2i = 1 + 2i

n


nrarrinfin

nsum

i=1

f (ci)xi

= limnrarrinfin

nsum

i=1

f (1 + 2i

n ) middot

2

n

= limnrarrinfin

nsum

i=1

3(1 + 2i

n )2 middot

2

n

= limnrarrinfin

nsum

i=1

3(4i2

n2 +

4i

n + 1) middot

2

n

= limnrarrinfin

nsum

i=1

3( 8i2

n3 + 8i

n2 + 2

n)

= limnrarrinfin

nsum

i=1

(24i2

n3 +

24i

n2 +

6

n)

= limnrarrinfin

nsum

i=1

104861624(n(n + 1)(2n + 1))

6n3 +

24n(n + 1)

2n2 +

6n

n

1048617

= limnrarrinfin

nsum

i=1

104861624(2n3 + 3n2 + n)

6n3 +

24n2 + 24n

2n2 +

6n

n

1048617

= limnrarrinfin

n

sumi=1

104861648n3

6n3 + 72n2

6n3 + 24n

6n3 + 24n2

2n2 + 24n

n2 + 61048617

16



=1048616

8 + 12

n +

4

n2 + 12 +

12

n + 6

1048617 = 26

Example 4


Solution

a = 4 b = 10 x = 10minus4n

= 6n


= 4 + 6in

f (x) = 6


nrarrinfin

nsumi=1

f (ci)xi

= limnrarrinfin

nsumi=1

f 983080

4 + 6i

n

983081middot 6

n

= limnrarrinfin

nsumi=1

(6) middot 6

n

= limn

rarrinfin(6n)

middot

6

n = 36

17



Example 5


a = 1 b = 2 x = 2minus1n

= 1n


= 1 + i

n

f (x) = x2 + 1


int 21 (x2 + 1)dx = lim

nrarrinfin

nsumi=1

f (ci)xi

= limnrarrinfin

n

sumi=1 f

9830801 +

i

n983081 middot

1

n

= limnrarrinfin

nsumi=1

9830809830801 +

i

n

9830812+ 1

983081middot 1

n

= limnrarrinfin

nsumi=1

9830809830801 +

i

n

9830819830801 +

i

n

983081+ 1

983081middot 1

n

= limnrarrinfin

nsumi=1

98308010486161 +

2i

n +

i2

n2

983081+ 1

983081middot 1

n

= limnrarrinfin

n

sumi=1

9830802 + 2i

n

+ i2

n2983081 middot

1

n

= limnrarrinfin

nsumi=1

9830802

n +

2i

n2 +

i2

n3

983081

= limnrarrinfin

983080 2

n middot (n) +

2

n2 middot n(n + 1)

2 +

1

n3 middot n(n + 1)(2n + 1)

6

983081

= limnrarrinfin

9830802 +

2n2 + 2n

2n2 +

(n2 + n)(2n + 1)

6n3

983081

= limnrarrinfin

9830802 +

2n2 + 2n

2n2 +

2n3 + 3n2 + n

6n3

983081


2 + 2n2

2n2

+ 2n

2n2

+ 2n3

6n3

+ 3n2

6n3

+ n

6n3983081

= limnrarrinfin

9830802 + 1 +

1

n +

1

3 +

1

2n +

1

6n2

983081 = 2 + 1 + 0 +

1

3 + 0 + 0 =

10

3

18



Example 6


a = 0 b = 3

x = 3minus0

n = 3

n


= 3n

f (x) = 5x


int 3


n

sumi=1

f (ci)

xi

= limnrarrinfin

nsumi=1

f 9830803i

n

983081middot 3

n

= limnrarrinfin

nsumi=1

59830803i

n

983081middot 3

n

= limnrarrinfin

nsumi=1

45i

n2

= limnrarrinfin

nsum

i=1

45

n2 middot n(n + 1)

2

= limnrarrinfin

nsumi=1

98308045n(n + 1)

2n2

983081

= limnrarrinfin

nsumi=1

98308045n2 + 45n

2n2

983081

= limnrarrinfin

nsumi=1

98308045n2

2n2 +

45n

2n2

983081

= limnrarrinfin

nsumi=1

98308045

2 +

45

2n

983081 =

45

2 + 0 =

45

2

19






int b




Example 1


Solution

int 20

4xdx = 41+1

x220

= 2x2|20 = 2(2)2 minus 2(0)2 = 8 minus 0 = 8

Example 2


Solution

int 21

x3 + x2dx =1048616

13+1

x3+1 + 12+1

x2+1)21

=1048616 14

x4 + 13

x3)21

=104861614

24 + 13

23) minus (14

14 + 13

13)

= 4 + 83 minus 1048616

14

+ 13

)

= 203 minus 1

12

= 7912

Example 3


2

dx

Solution

int 20 6x2dx = 6

2+1x2+1

21


= 16 minus 0= 16

20



Example 4


exdx

Solution

int 20



Example 5


x+radic x

2 dx

= [1

2 middot x2

2 + 1

2 middot 1

2 middot x

3

2

]10

= [x2

4 + 1

3 middot x

3

2 ]10

= [12

4 + 1

3 middot 1

3

2 ] minus [02

4 + 1

3 middot 0

3

2 ]14 minus 1

3 minus 0 = 7

12

Solution

int 10

x+radic x

2 dx =

int 10

x2

+radic x

2 dx =


Example 6


Solution

Note

f (x) = 2

xminus 3

xge

3

2



int 32


int 23

2

(2x minus 3)dx

=int 3

2


3

2

(2x minus 3)dx

= (minusx2 + 3x)|3

2

0 + (x2 minus 3x)|232

= [((32

)2 + 3 middot 32



)]= 10

4

21








1bminusa

int b

a f (x)dx


Example 7


Solution



= 14


= 1

4 middot (1

3x3

+ x)2minus2

= 14 middot [1

3(2)3 + 2] minus 1

3 middot [1

3(minus2)3 minus 2]

= 14 middot 14

3 minus 1

3 middot minus14

3

= 76

+ 76

= 146

= 73

Example 8


2

]

Solution

AV = 1π

2minus0int π

2

0 cos(x)dx

= 2π

int π2

0 cos(x)dx

= 2π middot 1048616 sin(x))|

π

2

0


2) minus sin(0)]


= 2π middot 1

= 2

π

22



Example 9


y = x2 + 2 y = 0 x = 0 x = 2

Solution


=1048616 x3

3 + 2x)

2

0

= [23

3 + 2(2)] minus [03

3 + 2(0)]= [8

3 + 4] minus [0 + 0]

= 203 minus 0

= 203

23



Example 10


y = ex y = 0 x = 0 x = 2

Solution

Area=int 20 (ex)dx

= [ex]|20

= e2 minus e0

= e2 minus 1

Example 11


0 (1 + cos(x))dx

Solution


1048616 x + sin(x))|

π

0


24







Example 1


Solution

f (x) = (x2 + 4x)6




Example 2

Evaluateint

(2x + 3)(x2 + 3x)6dx

Solution


int (2x + 3)(x2 + 3x)6dx =

int u6 middot du


6+1u6+1 + C = 1

7u7 + C = 1

7(x2 + 3x)7 + C

Example 3

Evaluateint

3(1 + 3x)3dx

Solution


int 3(1 + 3x)3dx =

int u3 middot du

25




3+1u3+1 + C = 1

4u4 + C = 1

4(1 + 3x)4 + C

Example 4

Evaluateint

x2(2 + x3)4dx

Solution



int 13

u4 middot du


int 13


5u5 + C = 1

15u5 + C = 1

15(2 + x3)5 + C

26



Example 5

Evaluateint

2tradic

t2 + 3dt

Solution


radic t2 + 3dt =


int u

1

2 middot du


1


3

2 + C = 23

(t2 + 3)3

2 + C = 23

(radic

t2 + 3)3 + C

Example 6

Evaluateint

πcos(πx)dx

Solution


int cos(u)du



Example 7

Evaluateint

xradic 1minus4x2 dx

Solution


= xdx

int x

radic 1minus4x2

dx = int minus1

8 middot du



1


1

2 + C = minus14

u1

2 + C = minus14

(1 minus 4x2)1

2 + C = minus14


27



Example 8

Evaluateint

2xe2x2

dx

Solution


= 2xdx

int 2xe2x

2

dx =int

12

eudu


eudu = 12

eu + C = 12

e2x2

+ C

Example 9

Evaluateint

2xsin(2x2)dx

Solution


du = 2xdx

int 2xsin2x2dx =

int 12

sin(u)du


int 12

sin(u)du =minus

1

2cos(u) + C =

minus1

2cos(2x2) + C

28







2int 1u

du = ln|x| + C

Example 1

Evaluateint

13x

dx

Solution


3du = dxint 13x

dx =int

13 middot 1

u du


int 13 middot 1

u du = 1

3ln|u| + C = 1

3ln|3x| + C

Example 2

Evaluateint

15x+3

dx

Solution


du = dx

int 13x

dx =int

15 middot 1

u du


int 15 middot 1

u du = 1

5ln|u| + C = 1

5ln|5x + 3| + C

29



Example 3

Evaluateint

x2

3minusx3 dx

Solution


du = x2dx

int x

2

3minusx3 dx =int

minus13 middot 1

u du


int minus1

3 middot 1

u du = minus1

3ln|u| + C = minus1

3ln|3 minus x3| + C

Example 4

Evaluateint

sinθ

cosθ dθ

Solution


cosθ dθ =

int minus 1

u du


int minus 1


Example 5

Evaluateint

cos(t)3+sin(t)

dt

Solution

Let u = 3 + sin(t)

rArr du = cos(t)dt

int cos(t)3+sin(t)

dθ =int

1u

du


int 1u

du = ln|u| + C = ln|3 + sin(t)| + C

30





Theorem 48



1int

duradic a2minusu2

= arcsin u

a +C

2int

du

a2+u2 = 1a

arctan u

a +C

3int

du

uradic u2minusa2

= 1a

arcsec |u|a

+C



Example 1

Evaluateint

41+9x2 middotdx

Solution


3du = dx


int 41+9x2 middotdx

=int

41+(3x)2

middotdx

= 13

int 4du1+u2

= 43

int du

1+u2

= 4

3arctan(u) + C

= 43

arctan(3x) + C

Example 2

Evaluateint

dx

4+(x+1)2

Solution



31



int dx

4+(x+1)2 =int

du

22+u2

= 12

arctan(u2

) + C

= 12

arctan(x+12

) + C

Example 3

Evaluateint

7radic 4minusx2

middotdx

Solution



int 7

radic 4minusx2

= int 7du

radic 22

minusu2


) + C

= 7arcsin(x2

) + C

Example 4

Evaluateint

dx


Solution



int dx


=int

du

uradic u2minus9

= 13

arcsec(u3

) + C

= 13

arcsec( |x|3

) + C

32





Example 8

Evaluateint


Solution





rArr minus12

du = (x minus 3)dx

Now substitute


=int


8radic 9+(xminus3)2

= minus12

int dvradic


9+(u)2

= 12

int vminus 1


= minus12

(2v1

2 ) + 8arcsin(u3

) + C



) + C


34






int b

a f (x)dx is


f (x)dx = bminusa


Example 1


int 20

2x2dx n = 4

Solution

x = 2minus04

= 24

= 12


= i2


2(4)[f (0) + 2f (1

2) + 2f (1) + 2f (3

2) + f (2)]

= 28

[2(0)2 + 2(2(12

)2) + 2(2(1)2) + 2(2( 32

)2) + 2(22)]

= 1

4

[0 + 2(1

2

) + 2(2) + 2( 18

4

) + 8]= 1

4[0 + 1 + 4 + 9 + 8]

= 14

[22]= 55



a f (x)dx is


int b


Example 2


int 2

0 2x2dx n = 4

Solution

35



x = 2minus04

= 24

= 12


= i2

int 20

2x2dx = 2minus03(4)

[f (0) + 4f (12

) + 2f (1) + 4f (32

) + f (2)]

= 212

[2(0)2 + 4(2(12

)2) + 2(2(1)2) + 4(2(32

)2) + 2(22)]= 1

6[0 + 4(1

2) + 2(2) + 4( 18

4 ) + 8]

= 16

[0 + 2 + 4 + 18 + 8]

= 16

[32]= 53333


int 2

0 2x2dx = 22+1

x2+120

= 23

x320

= 23

(2)3 minus 23

(0)3


= 163

Example 3


int π0 sin(x)dx

Solution

x = πminus04

= π4


= iπ4int π


[f (0) + 2f (π4

) + 2f (2π4

) + 2f (3π4

) + f (π)]

= π8

[sin(0) + 2sin(π4

) + 2sin(π2

) + 2sin(3π4

) + sin(π)]

= π8 [0 + 2(radic 22 ) + 2(1) + 2(radic 22 ) + 0]

= π8

[2radic

2 + 2]asymp 2003


int π0 sin(x)dx

Solution

x = πminus04

= π4

36





= 112

[111 + 4(0947) + 2(0816) + 4(0771) + 0625] asymp 086



1(x+2)2

dx =int 2

11u2

du =int 2


x+2

21

= minus 12+2

minus (minus 11+2

)

= minus14

+ 13

= 112



43 Riemann Sum




lexi

le1


nsumi=1

f (ci)xi




xi = i2


= i2


n2


n2

= i2minusi

2minus2i+1)n2

= minus2i+1n2


limhrarrinfin

nsumi=1


nsumi=1

f ( i2

n2) middot (

2iminus 1

n2 )

= limhrarrinfin

nsumi=1

radic i2

n2 middot (

2iminus 1

n2 )

= limhrarrinfin

nsumi=1

i

nmiddot 2iminus 1

n2

= limhrarrinfin

nsumi=1

2i2

minusi

n3

13



= 1

n3 limhrarrinfin

nsumi=1

(2i2 minus i)

= 1

n3 limhrarrinfin

[2(n(n + 1)(2n + 1)

6 ) minus n(n + 1)

2 ]

= 1


6 minus n2 + n

2 ]

= 1

n3 limhrarrinfin

((2n2 + 2n)(2n + 1)

6 minus n2 + n

2 ]

= 1

n3 limhrarrinfin

(4n3 + 6n2 + 2n

6 minus n2 + n

2 ]

= 1

n3 limhrarrinfin


6

= 1

n3 limhrarrinfin

4n3 + 3n2 minus n

6n3

= limhrarrinfin

[4n3 + 3n2 minus n

6n3

]

= limhrarrinfin

4n3

6n3 +

3n2

6n3 +

n

6n3

= limhrarrinfin

2

3 +

1

2n +

1

6n2 =

2

3 + 0 + 0 =

2

3




i=1




nsumi=1



nsumi=1

f (ci)xi =int

b

a f (x)dx

14



Example 2


Solution

int 3minus2

xdx

Solution


n = 3minus(minus2)

n


nrarrinfin

nsum

i=1

f (ci)xi

= limnrarrinfin

nsum

i=1

1048616minus 2 +

5i

n

1048617middot 5

n

= limnrarrinfin

nsum

i=1

1048616minus10

n +

25i

n2

1048617

= limnrarrinfin

nsum

i=1

1048616minus10

n +

25

n2 middot n(n + 1)

2

1048617

= limnrarrinfin

nsumi=1

1048616minus10n

+ 25n2

+ 25n2n2

1048617

= limnrarrinfin

nsum

i=1

1048616minus10

n +

25n2

2n2 +

25n

2n21048617

= limnrarrinfin

nsum

i=1

1048616minus10

n +

25

2 +

25

2n

1048617 = minus0 + 125 + 0 = 125

Example 3


Solution

15



int 31 3x2dx

Solution

a = 1 b = 3 x = 3minus1n

= 2n

ci = 1 + i middot 2i = 1 + 2i

n


nrarrinfin

nsum

i=1

f (ci)xi

= limnrarrinfin

nsum

i=1

f (1 + 2i

n ) middot

2

n

= limnrarrinfin

nsum

i=1

3(1 + 2i

n )2 middot

2

n

= limnrarrinfin

nsum

i=1

3(4i2

n2 +

4i

n + 1) middot

2

n

= limnrarrinfin

nsum

i=1

3( 8i2

n3 + 8i

n2 + 2

n)

= limnrarrinfin

nsum

i=1

(24i2

n3 +

24i

n2 +

6

n)

= limnrarrinfin

nsum

i=1

104861624(n(n + 1)(2n + 1))

6n3 +

24n(n + 1)

2n2 +

6n

n

1048617

= limnrarrinfin

nsum

i=1

104861624(2n3 + 3n2 + n)

6n3 +

24n2 + 24n

2n2 +

6n

n

1048617

= limnrarrinfin

n

sumi=1

104861648n3

6n3 + 72n2

6n3 + 24n

6n3 + 24n2

2n2 + 24n

n2 + 61048617

16



=1048616

8 + 12

n +

4

n2 + 12 +

12

n + 6

1048617 = 26

Example 4


Solution

a = 4 b = 10 x = 10minus4n

= 6n


= 4 + 6in

f (x) = 6


nrarrinfin

nsumi=1

f (ci)xi

= limnrarrinfin

nsumi=1

f 983080

4 + 6i

n

983081middot 6

n

= limnrarrinfin

nsumi=1

(6) middot 6

n

= limn

rarrinfin(6n)

middot

6

n = 36

17



Example 5


a = 1 b = 2 x = 2minus1n

= 1n


= 1 + i

n

f (x) = x2 + 1


int 21 (x2 + 1)dx = lim

nrarrinfin

nsumi=1

f (ci)xi

= limnrarrinfin

n

sumi=1 f

9830801 +

i

n983081 middot

1

n

= limnrarrinfin

nsumi=1

9830809830801 +

i

n

9830812+ 1

983081middot 1

n

= limnrarrinfin

nsumi=1

9830809830801 +

i

n

9830819830801 +

i

n

983081+ 1

983081middot 1

n

= limnrarrinfin

nsumi=1

98308010486161 +

2i

n +

i2

n2

983081+ 1

983081middot 1

n

= limnrarrinfin

n

sumi=1

9830802 + 2i

n

+ i2

n2983081 middot

1

n

= limnrarrinfin

nsumi=1

9830802

n +

2i

n2 +

i2

n3

983081

= limnrarrinfin

983080 2

n middot (n) +

2

n2 middot n(n + 1)

2 +

1

n3 middot n(n + 1)(2n + 1)

6

983081

= limnrarrinfin

9830802 +

2n2 + 2n

2n2 +

(n2 + n)(2n + 1)

6n3

983081

= limnrarrinfin

9830802 +

2n2 + 2n

2n2 +

2n3 + 3n2 + n

6n3

983081


2 + 2n2

2n2

+ 2n

2n2

+ 2n3

6n3

+ 3n2

6n3

+ n

6n3983081

= limnrarrinfin

9830802 + 1 +

1

n +

1

3 +

1

2n +

1

6n2

983081 = 2 + 1 + 0 +

1

3 + 0 + 0 =

10

3

18



Example 6


a = 0 b = 3

x = 3minus0

n = 3

n


= 3n

f (x) = 5x


int 3


n

sumi=1

f (ci)

xi

= limnrarrinfin

nsumi=1

f 9830803i

n

983081middot 3

n

= limnrarrinfin

nsumi=1

59830803i

n

983081middot 3

n

= limnrarrinfin

nsumi=1

45i

n2

= limnrarrinfin

nsum

i=1

45

n2 middot n(n + 1)

2

= limnrarrinfin

nsumi=1

98308045n(n + 1)

2n2

983081

= limnrarrinfin

nsumi=1

98308045n2 + 45n

2n2

983081

= limnrarrinfin

nsumi=1

98308045n2

2n2 +

45n

2n2

983081

= limnrarrinfin

nsumi=1

98308045

2 +

45

2n

983081 =

45

2 + 0 =

45

2

19






int b




Example 1


Solution

int 20

4xdx = 41+1

x220

= 2x2|20 = 2(2)2 minus 2(0)2 = 8 minus 0 = 8

Example 2


Solution

int 21

x3 + x2dx =1048616

13+1

x3+1 + 12+1

x2+1)21

=1048616 14

x4 + 13

x3)21

=104861614

24 + 13

23) minus (14

14 + 13

13)

= 4 + 83 minus 1048616

14

+ 13

)

= 203 minus 1

12

= 7912

Example 3


2

dx

Solution

int 20 6x2dx = 6

2+1x2+1

21


= 16 minus 0= 16

20



Example 4


exdx

Solution

int 20



Example 5


x+radic x

2 dx

= [1

2 middot x2

2 + 1

2 middot 1

2 middot x

3

2

]10

= [x2

4 + 1

3 middot x

3

2 ]10

= [12

4 + 1

3 middot 1

3

2 ] minus [02

4 + 1

3 middot 0

3

2 ]14 minus 1

3 minus 0 = 7

12

Solution

int 10

x+radic x

2 dx =

int 10

x2

+radic x

2 dx =


Example 6


Solution

Note

f (x) = 2

xminus 3

xge

3

2



int 32


int 23

2

(2x minus 3)dx

=int 3

2


3

2

(2x minus 3)dx

= (minusx2 + 3x)|3

2

0 + (x2 minus 3x)|232

= [((32

)2 + 3 middot 32



)]= 10

4

21








1bminusa

int b

a f (x)dx


Example 7


Solution



= 14


= 1

4 middot (1

3x3

+ x)2minus2

= 14 middot [1

3(2)3 + 2] minus 1

3 middot [1

3(minus2)3 minus 2]

= 14 middot 14

3 minus 1

3 middot minus14

3

= 76

+ 76

= 146

= 73

Example 8


2

]

Solution

AV = 1π

2minus0int π

2

0 cos(x)dx

= 2π

int π2

0 cos(x)dx

= 2π middot 1048616 sin(x))|

π

2

0


2) minus sin(0)]


= 2π middot 1

= 2

π

22



Example 9


y = x2 + 2 y = 0 x = 0 x = 2

Solution


=1048616 x3

3 + 2x)

2

0

= [23

3 + 2(2)] minus [03

3 + 2(0)]= [8

3 + 4] minus [0 + 0]

= 203 minus 0

= 203

23



Example 10


y = ex y = 0 x = 0 x = 2

Solution

Area=int 20 (ex)dx

= [ex]|20

= e2 minus e0

= e2 minus 1

Example 11


0 (1 + cos(x))dx

Solution


1048616 x + sin(x))|

π

0


24







Example 1


Solution

f (x) = (x2 + 4x)6




Example 2

Evaluateint

(2x + 3)(x2 + 3x)6dx

Solution


int (2x + 3)(x2 + 3x)6dx =

int u6 middot du


6+1u6+1 + C = 1

7u7 + C = 1

7(x2 + 3x)7 + C

Example 3

Evaluateint

3(1 + 3x)3dx

Solution


int 3(1 + 3x)3dx =

int u3 middot du

25




3+1u3+1 + C = 1

4u4 + C = 1

4(1 + 3x)4 + C

Example 4

Evaluateint

x2(2 + x3)4dx

Solution



int 13

u4 middot du


int 13


5u5 + C = 1

15u5 + C = 1

15(2 + x3)5 + C

26



Example 5

Evaluateint

2tradic

t2 + 3dt

Solution


radic t2 + 3dt =


int u

1

2 middot du


1


3

2 + C = 23

(t2 + 3)3

2 + C = 23

(radic

t2 + 3)3 + C

Example 6

Evaluateint

πcos(πx)dx

Solution


int cos(u)du



Example 7

Evaluateint

xradic 1minus4x2 dx

Solution


= xdx

int x

radic 1minus4x2

dx = int minus1

8 middot du



1


1

2 + C = minus14

u1

2 + C = minus14

(1 minus 4x2)1

2 + C = minus14


27



Example 8

Evaluateint

2xe2x2

dx

Solution


= 2xdx

int 2xe2x

2

dx =int

12

eudu


eudu = 12

eu + C = 12

e2x2

+ C

Example 9

Evaluateint

2xsin(2x2)dx

Solution


du = 2xdx

int 2xsin2x2dx =

int 12

sin(u)du


int 12

sin(u)du =minus

1

2cos(u) + C =

minus1

2cos(2x2) + C

28







2int 1u

du = ln|x| + C

Example 1

Evaluateint

13x

dx

Solution


3du = dxint 13x

dx =int

13 middot 1

u du


int 13 middot 1

u du = 1

3ln|u| + C = 1

3ln|3x| + C

Example 2

Evaluateint

15x+3

dx

Solution


du = dx

int 13x

dx =int

15 middot 1

u du


int 15 middot 1

u du = 1

5ln|u| + C = 1

5ln|5x + 3| + C

29



Example 3

Evaluateint

x2

3minusx3 dx

Solution


du = x2dx

int x

2

3minusx3 dx =int

minus13 middot 1

u du


int minus1

3 middot 1

u du = minus1

3ln|u| + C = minus1

3ln|3 minus x3| + C

Example 4

Evaluateint

sinθ

cosθ dθ

Solution


cosθ dθ =

int minus 1

u du


int minus 1


Example 5

Evaluateint

cos(t)3+sin(t)

dt

Solution

Let u = 3 + sin(t)

rArr du = cos(t)dt

int cos(t)3+sin(t)

dθ =int

1u

du


int 1u

du = ln|u| + C = ln|3 + sin(t)| + C

30





Theorem 48



1int

duradic a2minusu2

= arcsin u

a +C

2int

du

a2+u2 = 1a

arctan u

a +C

3int

du

uradic u2minusa2

= 1a

arcsec |u|a

+C



Example 1

Evaluateint

41+9x2 middotdx

Solution


3du = dx


int 41+9x2 middotdx

=int

41+(3x)2

middotdx

= 13

int 4du1+u2

= 43

int du

1+u2

= 4

3arctan(u) + C

= 43

arctan(3x) + C

Example 2

Evaluateint

dx

4+(x+1)2

Solution



31



int dx

4+(x+1)2 =int

du

22+u2

= 12

arctan(u2

) + C

= 12

arctan(x+12

) + C

Example 3

Evaluateint

7radic 4minusx2

middotdx

Solution



int 7

radic 4minusx2

= int 7du

radic 22

minusu2


) + C

= 7arcsin(x2

) + C

Example 4

Evaluateint

dx


Solution



int dx


=int

du

uradic u2minus9

= 13

arcsec(u3

) + C

= 13

arcsec( |x|3

) + C

32





Example 8

Evaluateint


Solution





rArr minus12

du = (x minus 3)dx

Now substitute


=int


8radic 9+(xminus3)2

= minus12

int dvradic


9+(u)2

= 12

int vminus 1


= minus12

(2v1

2 ) + 8arcsin(u3

) + C



) + C


34






int b

a f (x)dx is


f (x)dx = bminusa


Example 1


int 20

2x2dx n = 4

Solution

x = 2minus04

= 24

= 12


= i2


2(4)[f (0) + 2f (1

2) + 2f (1) + 2f (3

2) + f (2)]

= 28

[2(0)2 + 2(2(12

)2) + 2(2(1)2) + 2(2( 32

)2) + 2(22)]

= 1

4

[0 + 2(1

2

) + 2(2) + 2( 18

4

) + 8]= 1

4[0 + 1 + 4 + 9 + 8]

= 14

[22]= 55



a f (x)dx is


int b


Example 2


int 2

0 2x2dx n = 4

Solution

35



x = 2minus04

= 24

= 12


= i2

int 20

2x2dx = 2minus03(4)

[f (0) + 4f (12

) + 2f (1) + 4f (32

) + f (2)]

= 212

[2(0)2 + 4(2(12

)2) + 2(2(1)2) + 4(2(32

)2) + 2(22)]= 1

6[0 + 4(1

2) + 2(2) + 4( 18

4 ) + 8]

= 16

[0 + 2 + 4 + 18 + 8]

= 16

[32]= 53333


int 2

0 2x2dx = 22+1

x2+120

= 23

x320

= 23

(2)3 minus 23

(0)3


= 163

Example 3


int π0 sin(x)dx

Solution

x = πminus04

= π4


= iπ4int π


[f (0) + 2f (π4

) + 2f (2π4

) + 2f (3π4

) + f (π)]

= π8

[sin(0) + 2sin(π4

) + 2sin(π2

) + 2sin(3π4

) + sin(π)]

= π8 [0 + 2(radic 22 ) + 2(1) + 2(radic 22 ) + 0]

= π8

[2radic

2 + 2]asymp 2003


int π0 sin(x)dx

Solution

x = πminus04

= π4

36





= 112

[111 + 4(0947) + 2(0816) + 4(0771) + 0625] asymp 086



1(x+2)2

dx =int 2

11u2

du =int 2


x+2

21

= minus 12+2

minus (minus 11+2

)

= minus14

+ 13

= 112



= 1

n3 limhrarrinfin

nsumi=1

(2i2 minus i)

= 1

n3 limhrarrinfin

[2(n(n + 1)(2n + 1)

6 ) minus n(n + 1)

2 ]

= 1


6 minus n2 + n

2 ]

= 1

n3 limhrarrinfin

((2n2 + 2n)(2n + 1)

6 minus n2 + n

2 ]

= 1

n3 limhrarrinfin

(4n3 + 6n2 + 2n

6 minus n2 + n

2 ]

= 1

n3 limhrarrinfin


6

= 1

n3 limhrarrinfin

4n3 + 3n2 minus n

6n3

= limhrarrinfin

[4n3 + 3n2 minus n

6n3

]

= limhrarrinfin

4n3

6n3 +

3n2

6n3 +

n

6n3

= limhrarrinfin

2

3 +

1

2n +

1

6n2 =

2

3 + 0 + 0 =

2

3




i=1




nsumi=1



nsumi=1

f (ci)xi =int

b

a f (x)dx

14



Example 2


Solution

int 3minus2

xdx

Solution


n = 3minus(minus2)

n


nrarrinfin

nsum

i=1

f (ci)xi

= limnrarrinfin

nsum

i=1

1048616minus 2 +

5i

n

1048617middot 5

n

= limnrarrinfin

nsum

i=1

1048616minus10

n +

25i

n2

1048617

= limnrarrinfin

nsum

i=1

1048616minus10

n +

25

n2 middot n(n + 1)

2

1048617

= limnrarrinfin

nsumi=1

1048616minus10n

+ 25n2

+ 25n2n2

1048617

= limnrarrinfin

nsum

i=1

1048616minus10

n +

25n2

2n2 +

25n

2n21048617

= limnrarrinfin

nsum

i=1

1048616minus10

n +

25

2 +

25

2n

1048617 = minus0 + 125 + 0 = 125

Example 3


Solution

15



int 31 3x2dx

Solution

a = 1 b = 3 x = 3minus1n

= 2n

ci = 1 + i middot 2i = 1 + 2i

n


nrarrinfin

nsum

i=1

f (ci)xi

= limnrarrinfin

nsum

i=1

f (1 + 2i

n ) middot

2

n

= limnrarrinfin

nsum

i=1

3(1 + 2i

n )2 middot

2

n

= limnrarrinfin

nsum

i=1

3(4i2

n2 +

4i

n + 1) middot

2

n

= limnrarrinfin

nsum

i=1

3( 8i2

n3 + 8i

n2 + 2

n)

= limnrarrinfin

nsum

i=1

(24i2

n3 +

24i

n2 +

6

n)

= limnrarrinfin

nsum

i=1

104861624(n(n + 1)(2n + 1))

6n3 +

24n(n + 1)

2n2 +

6n

n

1048617

= limnrarrinfin

nsum

i=1

104861624(2n3 + 3n2 + n)

6n3 +

24n2 + 24n

2n2 +

6n

n

1048617

= limnrarrinfin

n

sumi=1

104861648n3

6n3 + 72n2

6n3 + 24n

6n3 + 24n2

2n2 + 24n

n2 + 61048617

16



=1048616

8 + 12

n +

4

n2 + 12 +

12

n + 6

1048617 = 26

Example 4


Solution

a = 4 b = 10 x = 10minus4n

= 6n


= 4 + 6in

f (x) = 6


nrarrinfin

nsumi=1

f (ci)xi

= limnrarrinfin

nsumi=1

f 983080

4 + 6i

n

983081middot 6

n

= limnrarrinfin

nsumi=1

(6) middot 6

n

= limn

rarrinfin(6n)

middot

6

n = 36

17



Example 5


a = 1 b = 2 x = 2minus1n

= 1n


= 1 + i

n

f (x) = x2 + 1


int 21 (x2 + 1)dx = lim

nrarrinfin

nsumi=1

f (ci)xi

= limnrarrinfin

n

sumi=1 f

9830801 +

i

n983081 middot

1

n

= limnrarrinfin

nsumi=1

9830809830801 +

i

n

9830812+ 1

983081middot 1

n

= limnrarrinfin

nsumi=1

9830809830801 +

i

n

9830819830801 +

i

n

983081+ 1

983081middot 1

n

= limnrarrinfin

nsumi=1

98308010486161 +

2i

n +

i2

n2

983081+ 1

983081middot 1

n

= limnrarrinfin

n

sumi=1

9830802 + 2i

n

+ i2

n2983081 middot

1

n

= limnrarrinfin

nsumi=1

9830802

n +

2i

n2 +

i2

n3

983081

= limnrarrinfin

983080 2

n middot (n) +

2

n2 middot n(n + 1)

2 +

1

n3 middot n(n + 1)(2n + 1)

6

983081

= limnrarrinfin

9830802 +

2n2 + 2n

2n2 +

(n2 + n)(2n + 1)

6n3

983081

= limnrarrinfin

9830802 +

2n2 + 2n

2n2 +

2n3 + 3n2 + n

6n3

983081


2 + 2n2

2n2

+ 2n

2n2

+ 2n3

6n3

+ 3n2

6n3

+ n

6n3983081

= limnrarrinfin

9830802 + 1 +

1

n +

1

3 +

1

2n +

1

6n2

983081 = 2 + 1 + 0 +

1

3 + 0 + 0 =

10

3

18



Example 6


a = 0 b = 3

x = 3minus0

n = 3

n


= 3n

f (x) = 5x


int 3


n

sumi=1

f (ci)

xi

= limnrarrinfin

nsumi=1

f 9830803i

n

983081middot 3

n

= limnrarrinfin

nsumi=1

59830803i

n

983081middot 3

n

= limnrarrinfin

nsumi=1

45i

n2

= limnrarrinfin

nsum

i=1

45

n2 middot n(n + 1)

2

= limnrarrinfin

nsumi=1

98308045n(n + 1)

2n2

983081

= limnrarrinfin

nsumi=1

98308045n2 + 45n

2n2

983081

= limnrarrinfin

nsumi=1

98308045n2

2n2 +

45n

2n2

983081

= limnrarrinfin

nsumi=1

98308045

2 +

45

2n

983081 =

45

2 + 0 =

45

2

19






int b




Example 1


Solution

int 20

4xdx = 41+1

x220

= 2x2|20 = 2(2)2 minus 2(0)2 = 8 minus 0 = 8

Example 2


Solution

int 21

x3 + x2dx =1048616

13+1

x3+1 + 12+1

x2+1)21

=1048616 14

x4 + 13

x3)21

=104861614

24 + 13

23) minus (14

14 + 13

13)

= 4 + 83 minus 1048616

14

+ 13

)

= 203 minus 1

12

= 7912

Example 3


2

dx

Solution

int 20 6x2dx = 6

2+1x2+1

21


= 16 minus 0= 16

20



Example 4


exdx

Solution

int 20



Example 5


x+radic x

2 dx

= [1

2 middot x2

2 + 1

2 middot 1

2 middot x

3

2

]10

= [x2

4 + 1

3 middot x

3

2 ]10

= [12

4 + 1

3 middot 1

3

2 ] minus [02

4 + 1

3 middot 0

3

2 ]14 minus 1

3 minus 0 = 7

12

Solution

int 10

x+radic x

2 dx =

int 10

x2

+radic x

2 dx =


Example 6


Solution

Note

f (x) = 2

xminus 3

xge

3

2



int 32


int 23

2

(2x minus 3)dx

=int 3

2


3

2

(2x minus 3)dx

= (minusx2 + 3x)|3

2

0 + (x2 minus 3x)|232

= [((32

)2 + 3 middot 32



)]= 10

4

21








1bminusa

int b

a f (x)dx


Example 7


Solution



= 14


= 1

4 middot (1

3x3

+ x)2minus2

= 14 middot [1

3(2)3 + 2] minus 1

3 middot [1

3(minus2)3 minus 2]

= 14 middot 14

3 minus 1

3 middot minus14

3

= 76

+ 76

= 146

= 73

Example 8


2

]

Solution

AV = 1π

2minus0int π

2

0 cos(x)dx

= 2π

int π2

0 cos(x)dx

= 2π middot 1048616 sin(x))|

π

2

0


2) minus sin(0)]


= 2π middot 1

= 2

π

22



Example 9


y = x2 + 2 y = 0 x = 0 x = 2

Solution


=1048616 x3

3 + 2x)

2

0

= [23

3 + 2(2)] minus [03

3 + 2(0)]= [8

3 + 4] minus [0 + 0]

= 203 minus 0

= 203

23



Example 10


y = ex y = 0 x = 0 x = 2

Solution

Area=int 20 (ex)dx

= [ex]|20

= e2 minus e0

= e2 minus 1

Example 11


0 (1 + cos(x))dx

Solution


1048616 x + sin(x))|

π

0


24







Example 1


Solution

f (x) = (x2 + 4x)6




Example 2

Evaluateint

(2x + 3)(x2 + 3x)6dx

Solution


int (2x + 3)(x2 + 3x)6dx =

int u6 middot du


6+1u6+1 + C = 1

7u7 + C = 1

7(x2 + 3x)7 + C

Example 3

Evaluateint

3(1 + 3x)3dx

Solution


int 3(1 + 3x)3dx =

int u3 middot du

25




3+1u3+1 + C = 1

4u4 + C = 1

4(1 + 3x)4 + C

Example 4

Evaluateint

x2(2 + x3)4dx

Solution



int 13

u4 middot du


int 13


5u5 + C = 1

15u5 + C = 1

15(2 + x3)5 + C

26



Example 5

Evaluateint

2tradic

t2 + 3dt

Solution


radic t2 + 3dt =


int u

1

2 middot du


1


3

2 + C = 23

(t2 + 3)3

2 + C = 23

(radic

t2 + 3)3 + C

Example 6

Evaluateint

πcos(πx)dx

Solution


int cos(u)du



Example 7

Evaluateint

xradic 1minus4x2 dx

Solution


= xdx

int x

radic 1minus4x2

dx = int minus1

8 middot du



1


1

2 + C = minus14

u1

2 + C = minus14

(1 minus 4x2)1

2 + C = minus14


27



Example 8

Evaluateint

2xe2x2

dx

Solution


= 2xdx

int 2xe2x

2

dx =int

12

eudu


eudu = 12

eu + C = 12

e2x2

+ C

Example 9

Evaluateint

2xsin(2x2)dx

Solution


du = 2xdx

int 2xsin2x2dx =

int 12

sin(u)du


int 12

sin(u)du =minus

1

2cos(u) + C =

minus1

2cos(2x2) + C

28







2int 1u

du = ln|x| + C

Example 1

Evaluateint

13x

dx

Solution


3du = dxint 13x

dx =int

13 middot 1

u du


int 13 middot 1

u du = 1

3ln|u| + C = 1

3ln|3x| + C

Example 2

Evaluateint

15x+3

dx

Solution


du = dx

int 13x

dx =int

15 middot 1

u du


int 15 middot 1

u du = 1

5ln|u| + C = 1

5ln|5x + 3| + C

29



Example 3

Evaluateint

x2

3minusx3 dx

Solution


du = x2dx

int x

2

3minusx3 dx =int

minus13 middot 1

u du


int minus1

3 middot 1

u du = minus1

3ln|u| + C = minus1

3ln|3 minus x3| + C

Example 4

Evaluateint

sinθ

cosθ dθ

Solution


cosθ dθ =

int minus 1

u du


int minus 1


Example 5

Evaluateint

cos(t)3+sin(t)

dt

Solution

Let u = 3 + sin(t)

rArr du = cos(t)dt

int cos(t)3+sin(t)

dθ =int

1u

du


int 1u

du = ln|u| + C = ln|3 + sin(t)| + C

30





Theorem 48



1int

duradic a2minusu2

= arcsin u

a +C

2int

du

a2+u2 = 1a

arctan u

a +C

3int

du

uradic u2minusa2

= 1a

arcsec |u|a

+C



Example 1

Evaluateint

41+9x2 middotdx

Solution


3du = dx


int 41+9x2 middotdx

=int

41+(3x)2

middotdx

= 13

int 4du1+u2

= 43

int du

1+u2

= 4

3arctan(u) + C

= 43

arctan(3x) + C

Example 2

Evaluateint

dx

4+(x+1)2

Solution



31



int dx

4+(x+1)2 =int

du

22+u2

= 12

arctan(u2

) + C

= 12

arctan(x+12

) + C

Example 3

Evaluateint

7radic 4minusx2

middotdx

Solution



int 7

radic 4minusx2

= int 7du

radic 22

minusu2


) + C

= 7arcsin(x2

) + C

Example 4

Evaluateint

dx


Solution



int dx


=int

du

uradic u2minus9

= 13

arcsec(u3

) + C

= 13

arcsec( |x|3

) + C

32





Example 8

Evaluateint


Solution





rArr minus12

du = (x minus 3)dx

Now substitute


=int


8radic 9+(xminus3)2

= minus12

int dvradic


9+(u)2

= 12

int vminus 1


= minus12

(2v1

2 ) + 8arcsin(u3

) + C



) + C


34






int b

a f (x)dx is


f (x)dx = bminusa


Example 1


int 20

2x2dx n = 4

Solution

x = 2minus04

= 24

= 12


= i2


2(4)[f (0) + 2f (1

2) + 2f (1) + 2f (3

2) + f (2)]

= 28

[2(0)2 + 2(2(12

)2) + 2(2(1)2) + 2(2( 32

)2) + 2(22)]

= 1

4

[0 + 2(1

2

) + 2(2) + 2( 18

4

) + 8]= 1

4[0 + 1 + 4 + 9 + 8]

= 14

[22]= 55



a f (x)dx is


int b


Example 2


int 2

0 2x2dx n = 4

Solution

35



x = 2minus04

= 24

= 12


= i2

int 20

2x2dx = 2minus03(4)

[f (0) + 4f (12

) + 2f (1) + 4f (32

) + f (2)]

= 212

[2(0)2 + 4(2(12

)2) + 2(2(1)2) + 4(2(32

)2) + 2(22)]= 1

6[0 + 4(1

2) + 2(2) + 4( 18

4 ) + 8]

= 16

[0 + 2 + 4 + 18 + 8]

= 16

[32]= 53333


int 2

0 2x2dx = 22+1

x2+120

= 23

x320

= 23

(2)3 minus 23

(0)3


= 163

Example 3


int π0 sin(x)dx

Solution

x = πminus04

= π4


= iπ4int π


[f (0) + 2f (π4

) + 2f (2π4

) + 2f (3π4

) + f (π)]

= π8

[sin(0) + 2sin(π4

) + 2sin(π2

) + 2sin(3π4

) + sin(π)]

= π8 [0 + 2(radic 22 ) + 2(1) + 2(radic 22 ) + 0]

= π8

[2radic

2 + 2]asymp 2003


int π0 sin(x)dx

Solution

x = πminus04

= π4

36





= 112

[111 + 4(0947) + 2(0816) + 4(0771) + 0625] asymp 086



1(x+2)2

dx =int 2

11u2

du =int 2


x+2

21

= minus 12+2

minus (minus 11+2

)

= minus14

+ 13

= 112



Example 2


Solution

int 3minus2

xdx

Solution


n = 3minus(minus2)

n


nrarrinfin

nsum

i=1

f (ci)xi

= limnrarrinfin

nsum

i=1

1048616minus 2 +

5i

n

1048617middot 5

n

= limnrarrinfin

nsum

i=1

1048616minus10

n +

25i

n2

1048617

= limnrarrinfin

nsum

i=1

1048616minus10

n +

25

n2 middot n(n + 1)

2

1048617

= limnrarrinfin

nsumi=1

1048616minus10n

+ 25n2

+ 25n2n2

1048617

= limnrarrinfin

nsum

i=1

1048616minus10

n +

25n2

2n2 +

25n

2n21048617

= limnrarrinfin

nsum

i=1

1048616minus10

n +

25

2 +

25

2n

1048617 = minus0 + 125 + 0 = 125

Example 3


Solution

15



int 31 3x2dx

Solution

a = 1 b = 3 x = 3minus1n

= 2n

ci = 1 + i middot 2i = 1 + 2i

n


nrarrinfin

nsum

i=1

f (ci)xi

= limnrarrinfin

nsum

i=1

f (1 + 2i

n ) middot

2

n

= limnrarrinfin

nsum

i=1

3(1 + 2i

n )2 middot

2

n

= limnrarrinfin

nsum

i=1

3(4i2

n2 +

4i

n + 1) middot

2

n

= limnrarrinfin

nsum

i=1

3( 8i2

n3 + 8i

n2 + 2

n)

= limnrarrinfin

nsum

i=1

(24i2

n3 +

24i

n2 +

6

n)

= limnrarrinfin

nsum

i=1

104861624(n(n + 1)(2n + 1))

6n3 +

24n(n + 1)

2n2 +

6n

n

1048617

= limnrarrinfin

nsum

i=1

104861624(2n3 + 3n2 + n)

6n3 +

24n2 + 24n

2n2 +

6n

n

1048617

= limnrarrinfin

n

sumi=1

104861648n3

6n3 + 72n2

6n3 + 24n

6n3 + 24n2

2n2 + 24n

n2 + 61048617

16



=1048616

8 + 12

n +

4

n2 + 12 +

12

n + 6

1048617 = 26

Example 4


Solution

a = 4 b = 10 x = 10minus4n

= 6n


= 4 + 6in

f (x) = 6


nrarrinfin

nsumi=1

f (ci)xi

= limnrarrinfin

nsumi=1

f 983080

4 + 6i

n

983081middot 6

n

= limnrarrinfin

nsumi=1

(6) middot 6

n

= limn

rarrinfin(6n)

middot

6

n = 36

17



Example 5


a = 1 b = 2 x = 2minus1n

= 1n


= 1 + i

n

f (x) = x2 + 1


int 21 (x2 + 1)dx = lim

nrarrinfin

nsumi=1

f (ci)xi

= limnrarrinfin

n

sumi=1 f

9830801 +

i

n983081 middot

1

n

= limnrarrinfin

nsumi=1

9830809830801 +

i

n

9830812+ 1

983081middot 1

n

= limnrarrinfin

nsumi=1

9830809830801 +

i

n

9830819830801 +

i

n

983081+ 1

983081middot 1

n

= limnrarrinfin

nsumi=1

98308010486161 +

2i

n +

i2

n2

983081+ 1

983081middot 1

n

= limnrarrinfin

n

sumi=1

9830802 + 2i

n

+ i2

n2983081 middot

1

n

= limnrarrinfin

nsumi=1

9830802

n +

2i

n2 +

i2

n3

983081

= limnrarrinfin

983080 2

n middot (n) +

2

n2 middot n(n + 1)

2 +

1

n3 middot n(n + 1)(2n + 1)

6

983081

= limnrarrinfin

9830802 +

2n2 + 2n

2n2 +

(n2 + n)(2n + 1)

6n3

983081

= limnrarrinfin

9830802 +

2n2 + 2n

2n2 +

2n3 + 3n2 + n

6n3

983081


2 + 2n2

2n2

+ 2n

2n2

+ 2n3

6n3

+ 3n2

6n3

+ n

6n3983081

= limnrarrinfin

9830802 + 1 +

1

n +

1

3 +

1

2n +

1

6n2

983081 = 2 + 1 + 0 +

1

3 + 0 + 0 =

10

3

18



Example 6


a = 0 b = 3

x = 3minus0

n = 3

n


= 3n

f (x) = 5x


int 3


n

sumi=1

f (ci)

xi

= limnrarrinfin

nsumi=1

f 9830803i

n

983081middot 3

n

= limnrarrinfin

nsumi=1

59830803i

n

983081middot 3

n

= limnrarrinfin

nsumi=1

45i

n2

= limnrarrinfin

nsum

i=1

45

n2 middot n(n + 1)

2

= limnrarrinfin

nsumi=1

98308045n(n + 1)

2n2

983081

= limnrarrinfin

nsumi=1

98308045n2 + 45n

2n2

983081

= limnrarrinfin

nsumi=1

98308045n2

2n2 +

45n

2n2

983081

= limnrarrinfin

nsumi=1

98308045

2 +

45

2n

983081 =

45

2 + 0 =

45

2

19






int b




Example 1


Solution

int 20

4xdx = 41+1

x220

= 2x2|20 = 2(2)2 minus 2(0)2 = 8 minus 0 = 8

Example 2


Solution

int 21

x3 + x2dx =1048616

13+1

x3+1 + 12+1

x2+1)21

=1048616 14

x4 + 13

x3)21

=104861614

24 + 13

23) minus (14

14 + 13

13)

= 4 + 83 minus 1048616

14

+ 13

)

= 203 minus 1

12

= 7912

Example 3


2

dx

Solution

int 20 6x2dx = 6

2+1x2+1

21


= 16 minus 0= 16

20



Example 4


exdx

Solution

int 20



Example 5


x+radic x

2 dx

= [1

2 middot x2

2 + 1

2 middot 1

2 middot x

3

2

]10

= [x2

4 + 1

3 middot x

3

2 ]10

= [12

4 + 1

3 middot 1

3

2 ] minus [02

4 + 1

3 middot 0

3

2 ]14 minus 1

3 minus 0 = 7

12

Solution

int 10

x+radic x

2 dx =

int 10

x2

+radic x

2 dx =


Example 6


Solution

Note

f (x) = 2

xminus 3

xge

3

2



int 32


int 23

2

(2x minus 3)dx

=int 3

2


3

2

(2x minus 3)dx

= (minusx2 + 3x)|3

2

0 + (x2 minus 3x)|232

= [((32

)2 + 3 middot 32



)]= 10

4

21








1bminusa

int b

a f (x)dx


Example 7


Solution



= 14


= 1

4 middot (1

3x3

+ x)2minus2

= 14 middot [1

3(2)3 + 2] minus 1

3 middot [1

3(minus2)3 minus 2]

= 14 middot 14

3 minus 1

3 middot minus14

3

= 76

+ 76

= 146

= 73

Example 8


2

]

Solution

AV = 1π

2minus0int π

2

0 cos(x)dx

= 2π

int π2

0 cos(x)dx

= 2π middot 1048616 sin(x))|

π

2

0


2) minus sin(0)]


= 2π middot 1

= 2

π

22



Example 9


y = x2 + 2 y = 0 x = 0 x = 2

Solution


=1048616 x3

3 + 2x)

2

0

= [23

3 + 2(2)] minus [03

3 + 2(0)]= [8

3 + 4] minus [0 + 0]

= 203 minus 0

= 203

23



Example 10


y = ex y = 0 x = 0 x = 2

Solution

Area=int 20 (ex)dx

= [ex]|20

= e2 minus e0

= e2 minus 1

Example 11


0 (1 + cos(x))dx

Solution


1048616 x + sin(x))|

π

0


24







Example 1


Solution

f (x) = (x2 + 4x)6




Example 2

Evaluateint

(2x + 3)(x2 + 3x)6dx

Solution


int (2x + 3)(x2 + 3x)6dx =

int u6 middot du


6+1u6+1 + C = 1

7u7 + C = 1

7(x2 + 3x)7 + C

Example 3

Evaluateint

3(1 + 3x)3dx

Solution


int 3(1 + 3x)3dx =

int u3 middot du

25




3+1u3+1 + C = 1

4u4 + C = 1

4(1 + 3x)4 + C

Example 4

Evaluateint

x2(2 + x3)4dx

Solution



int 13

u4 middot du


int 13


5u5 + C = 1

15u5 + C = 1

15(2 + x3)5 + C

26



Example 5

Evaluateint

2tradic

t2 + 3dt

Solution


radic t2 + 3dt =


int u

1

2 middot du


1


3

2 + C = 23

(t2 + 3)3

2 + C = 23

(radic

t2 + 3)3 + C

Example 6

Evaluateint

πcos(πx)dx

Solution


int cos(u)du



Example 7

Evaluateint

xradic 1minus4x2 dx

Solution


= xdx

int x

radic 1minus4x2

dx = int minus1

8 middot du



1


1

2 + C = minus14

u1

2 + C = minus14

(1 minus 4x2)1

2 + C = minus14


27



Example 8

Evaluateint

2xe2x2

dx

Solution


= 2xdx

int 2xe2x

2

dx =int

12

eudu


eudu = 12

eu + C = 12

e2x2

+ C

Example 9

Evaluateint

2xsin(2x2)dx

Solution


du = 2xdx

int 2xsin2x2dx =

int 12

sin(u)du


int 12

sin(u)du =minus

1

2cos(u) + C =

minus1

2cos(2x2) + C

28







2int 1u

du = ln|x| + C

Example 1

Evaluateint

13x

dx

Solution


3du = dxint 13x

dx =int

13 middot 1

u du


int 13 middot 1

u du = 1

3ln|u| + C = 1

3ln|3x| + C

Example 2

Evaluateint

15x+3

dx

Solution


du = dx

int 13x

dx =int

15 middot 1

u du


int 15 middot 1

u du = 1

5ln|u| + C = 1

5ln|5x + 3| + C

29



Example 3

Evaluateint

x2

3minusx3 dx

Solution


du = x2dx

int x

2

3minusx3 dx =int

minus13 middot 1

u du


int minus1

3 middot 1

u du = minus1

3ln|u| + C = minus1

3ln|3 minus x3| + C

Example 4

Evaluateint

sinθ

cosθ dθ

Solution


cosθ dθ =

int minus 1

u du


int minus 1


Example 5

Evaluateint

cos(t)3+sin(t)

dt

Solution

Let u = 3 + sin(t)

rArr du = cos(t)dt

int cos(t)3+sin(t)

dθ =int

1u

du


int 1u

du = ln|u| + C = ln|3 + sin(t)| + C

30





Theorem 48



1int

duradic a2minusu2

= arcsin u

a +C

2int

du

a2+u2 = 1a

arctan u

a +C

3int

du

uradic u2minusa2

= 1a

arcsec |u|a

+C



Example 1

Evaluateint

41+9x2 middotdx

Solution


3du = dx


int 41+9x2 middotdx

=int

41+(3x)2

middotdx

= 13

int 4du1+u2

= 43

int du

1+u2

= 4

3arctan(u) + C

= 43

arctan(3x) + C

Example 2

Evaluateint

dx

4+(x+1)2

Solution



31



int dx

4+(x+1)2 =int

du

22+u2

= 12

arctan(u2

) + C

= 12

arctan(x+12

) + C

Example 3

Evaluateint

7radic 4minusx2

middotdx

Solution



int 7

radic 4minusx2

= int 7du

radic 22

minusu2


) + C

= 7arcsin(x2

) + C

Example 4

Evaluateint

dx


Solution



int dx


=int

du

uradic u2minus9

= 13

arcsec(u3

) + C

= 13

arcsec( |x|3

) + C

32





Example 8

Evaluateint


Solution





rArr minus12

du = (x minus 3)dx

Now substitute


=int


8radic 9+(xminus3)2

= minus12

int dvradic


9+(u)2

= 12

int vminus 1


= minus12

(2v1

2 ) + 8arcsin(u3

) + C



) + C


34






int b

a f (x)dx is


f (x)dx = bminusa


Example 1


int 20

2x2dx n = 4

Solution

x = 2minus04

= 24

= 12


= i2


2(4)[f (0) + 2f (1

2) + 2f (1) + 2f (3

2) + f (2)]

= 28

[2(0)2 + 2(2(12

)2) + 2(2(1)2) + 2(2( 32

)2) + 2(22)]

= 1

4

[0 + 2(1

2

) + 2(2) + 2( 18

4

) + 8]= 1

4[0 + 1 + 4 + 9 + 8]

= 14

[22]= 55



a f (x)dx is


int b


Example 2


int 2

0 2x2dx n = 4

Solution

35



x = 2minus04

= 24

= 12


= i2

int 20

2x2dx = 2minus03(4)

[f (0) + 4f (12

) + 2f (1) + 4f (32

) + f (2)]

= 212

[2(0)2 + 4(2(12

)2) + 2(2(1)2) + 4(2(32

)2) + 2(22)]= 1

6[0 + 4(1

2) + 2(2) + 4( 18

4 ) + 8]

= 16

[0 + 2 + 4 + 18 + 8]

= 16

[32]= 53333


int 2

0 2x2dx = 22+1

x2+120

= 23

x320

= 23

(2)3 minus 23

(0)3


= 163

Example 3


int π0 sin(x)dx

Solution

x = πminus04

= π4


= iπ4int π


[f (0) + 2f (π4

) + 2f (2π4

) + 2f (3π4

) + f (π)]

= π8

[sin(0) + 2sin(π4

) + 2sin(π2

) + 2sin(3π4

) + sin(π)]

= π8 [0 + 2(radic 22 ) + 2(1) + 2(radic 22 ) + 0]

= π8

[2radic

2 + 2]asymp 2003


int π0 sin(x)dx

Solution

x = πminus04

= π4

36





= 112

[111 + 4(0947) + 2(0816) + 4(0771) + 0625] asymp 086



1(x+2)2

dx =int 2

11u2

du =int 2


x+2

21

= minus 12+2

minus (minus 11+2

)

= minus14

+ 13

= 112



int 31 3x2dx

Solution

a = 1 b = 3 x = 3minus1n

= 2n

ci = 1 + i middot 2i = 1 + 2i

n


nrarrinfin

nsum

i=1

f (ci)xi

= limnrarrinfin

nsum

i=1

f (1 + 2i

n ) middot

2

n

= limnrarrinfin

nsum

i=1

3(1 + 2i

n )2 middot

2

n

= limnrarrinfin

nsum

i=1

3(4i2

n2 +

4i

n + 1) middot

2

n

= limnrarrinfin

nsum

i=1

3( 8i2

n3 + 8i

n2 + 2

n)

= limnrarrinfin

nsum

i=1

(24i2

n3 +

24i

n2 +

6

n)

= limnrarrinfin

nsum

i=1

104861624(n(n + 1)(2n + 1))

6n3 +

24n(n + 1)

2n2 +

6n

n

1048617

= limnrarrinfin

nsum

i=1

104861624(2n3 + 3n2 + n)

6n3 +

24n2 + 24n

2n2 +

6n

n

1048617

= limnrarrinfin

n

sumi=1

104861648n3

6n3 + 72n2

6n3 + 24n

6n3 + 24n2

2n2 + 24n

n2 + 61048617

16



=1048616

8 + 12

n +

4

n2 + 12 +

12

n + 6

1048617 = 26

Example 4


Solution

a = 4 b = 10 x = 10minus4n

= 6n


= 4 + 6in

f (x) = 6


nrarrinfin

nsumi=1

f (ci)xi

= limnrarrinfin

nsumi=1

f 983080

4 + 6i

n

983081middot 6

n

= limnrarrinfin

nsumi=1

(6) middot 6

n

= limn

rarrinfin(6n)

middot

6

n = 36

17



Example 5


a = 1 b = 2 x = 2minus1n

= 1n


= 1 + i

n

f (x) = x2 + 1


int 21 (x2 + 1)dx = lim

nrarrinfin

nsumi=1

f (ci)xi

= limnrarrinfin

n

sumi=1 f

9830801 +

i

n983081 middot

1

n

= limnrarrinfin

nsumi=1

9830809830801 +

i

n

9830812+ 1

983081middot 1

n

= limnrarrinfin

nsumi=1

9830809830801 +

i

n

9830819830801 +

i

n

983081+ 1

983081middot 1

n

= limnrarrinfin

nsumi=1

98308010486161 +

2i

n +

i2

n2

983081+ 1

983081middot 1

n

= limnrarrinfin

n

sumi=1

9830802 + 2i

n

+ i2

n2983081 middot

1

n

= limnrarrinfin

nsumi=1

9830802

n +

2i

n2 +

i2

n3

983081

= limnrarrinfin

983080 2

n middot (n) +

2

n2 middot n(n + 1)

2 +

1

n3 middot n(n + 1)(2n + 1)

6

983081

= limnrarrinfin

9830802 +

2n2 + 2n

2n2 +

(n2 + n)(2n + 1)

6n3

983081

= limnrarrinfin

9830802 +

2n2 + 2n

2n2 +

2n3 + 3n2 + n

6n3

983081


2 + 2n2

2n2

+ 2n

2n2

+ 2n3

6n3

+ 3n2

6n3

+ n

6n3983081

= limnrarrinfin

9830802 + 1 +

1

n +

1

3 +

1

2n +

1

6n2

983081 = 2 + 1 + 0 +

1

3 + 0 + 0 =

10

3

18



Example 6


a = 0 b = 3

x = 3minus0

n = 3

n


= 3n

f (x) = 5x


int 3


n

sumi=1

f (ci)

xi

= limnrarrinfin

nsumi=1

f 9830803i

n

983081middot 3

n

= limnrarrinfin

nsumi=1

59830803i

n

983081middot 3

n

= limnrarrinfin

nsumi=1

45i

n2

= limnrarrinfin

nsum

i=1

45

n2 middot n(n + 1)

2

= limnrarrinfin

nsumi=1

98308045n(n + 1)

2n2

983081

= limnrarrinfin

nsumi=1

98308045n2 + 45n

2n2

983081

= limnrarrinfin

nsumi=1

98308045n2

2n2 +

45n

2n2

983081

= limnrarrinfin

nsumi=1

98308045

2 +

45

2n

983081 =

45

2 + 0 =

45

2

19






int b




Example 1


Solution

int 20

4xdx = 41+1

x220

= 2x2|20 = 2(2)2 minus 2(0)2 = 8 minus 0 = 8

Example 2


Solution

int 21

x3 + x2dx =1048616

13+1

x3+1 + 12+1

x2+1)21

=1048616 14

x4 + 13

x3)21

=104861614

24 + 13

23) minus (14

14 + 13

13)

= 4 + 83 minus 1048616

14

+ 13

)

= 203 minus 1

12

= 7912

Example 3


2

dx

Solution

int 20 6x2dx = 6

2+1x2+1

21


= 16 minus 0= 16

20



Example 4


exdx

Solution

int 20



Example 5


x+radic x

2 dx

= [1

2 middot x2

2 + 1

2 middot 1

2 middot x

3

2

]10

= [x2

4 + 1

3 middot x

3

2 ]10

= [12

4 + 1

3 middot 1

3

2 ] minus [02

4 + 1

3 middot 0

3

2 ]14 minus 1

3 minus 0 = 7

12

Solution

int 10

x+radic x

2 dx =

int 10

x2

+radic x

2 dx =


Example 6


Solution

Note

f (x) = 2

xminus 3

xge

3

2



int 32


int 23

2

(2x minus 3)dx

=int 3

2


3

2

(2x minus 3)dx

= (minusx2 + 3x)|3

2

0 + (x2 minus 3x)|232

= [((32

)2 + 3 middot 32



)]= 10

4

21








1bminusa

int b

a f (x)dx


Example 7


Solution



= 14


= 1

4 middot (1

3x3

+ x)2minus2

= 14 middot [1

3(2)3 + 2] minus 1

3 middot [1

3(minus2)3 minus 2]

= 14 middot 14

3 minus 1

3 middot minus14

3

= 76

+ 76

= 146

= 73

Example 8


2

]

Solution

AV = 1π

2minus0int π

2

0 cos(x)dx

= 2π

int π2

0 cos(x)dx

= 2π middot 1048616 sin(x))|

π

2

0


2) minus sin(0)]


= 2π middot 1

= 2

π

22



Example 9


y = x2 + 2 y = 0 x = 0 x = 2

Solution


=1048616 x3

3 + 2x)

2

0

= [23

3 + 2(2)] minus [03

3 + 2(0)]= [8

3 + 4] minus [0 + 0]

= 203 minus 0

= 203

23



Example 10


y = ex y = 0 x = 0 x = 2

Solution

Area=int 20 (ex)dx

= [ex]|20

= e2 minus e0

= e2 minus 1

Example 11


0 (1 + cos(x))dx

Solution


1048616 x + sin(x))|

π

0


24







Example 1


Solution

f (x) = (x2 + 4x)6




Example 2

Evaluateint

(2x + 3)(x2 + 3x)6dx

Solution


int (2x + 3)(x2 + 3x)6dx =

int u6 middot du


6+1u6+1 + C = 1

7u7 + C = 1

7(x2 + 3x)7 + C

Example 3

Evaluateint

3(1 + 3x)3dx

Solution


int 3(1 + 3x)3dx =

int u3 middot du

25




3+1u3+1 + C = 1

4u4 + C = 1

4(1 + 3x)4 + C

Example 4

Evaluateint

x2(2 + x3)4dx

Solution



int 13

u4 middot du


int 13


5u5 + C = 1

15u5 + C = 1

15(2 + x3)5 + C

26



Example 5

Evaluateint

2tradic

t2 + 3dt

Solution


radic t2 + 3dt =


int u

1

2 middot du


1


3

2 + C = 23

(t2 + 3)3

2 + C = 23

(radic

t2 + 3)3 + C

Example 6

Evaluateint

πcos(πx)dx

Solution


int cos(u)du



Example 7

Evaluateint

xradic 1minus4x2 dx

Solution


= xdx

int x

radic 1minus4x2

dx = int minus1

8 middot du



1


1

2 + C = minus14

u1

2 + C = minus14

(1 minus 4x2)1

2 + C = minus14


27



Example 8

Evaluateint

2xe2x2

dx

Solution


= 2xdx

int 2xe2x

2

dx =int

12

eudu


eudu = 12

eu + C = 12

e2x2

+ C

Example 9

Evaluateint

2xsin(2x2)dx

Solution


du = 2xdx

int 2xsin2x2dx =

int 12

sin(u)du


int 12

sin(u)du =minus

1

2cos(u) + C =

minus1

2cos(2x2) + C

28







2int 1u

du = ln|x| + C

Example 1

Evaluateint

13x

dx

Solution


3du = dxint 13x

dx =int

13 middot 1

u du


int 13 middot 1

u du = 1

3ln|u| + C = 1

3ln|3x| + C

Example 2

Evaluateint

15x+3

dx

Solution


du = dx

int 13x

dx =int

15 middot 1

u du


int 15 middot 1

u du = 1

5ln|u| + C = 1

5ln|5x + 3| + C

29



Example 3

Evaluateint

x2

3minusx3 dx

Solution


du = x2dx

int x

2

3minusx3 dx =int

minus13 middot 1

u du


int minus1

3 middot 1

u du = minus1

3ln|u| + C = minus1

3ln|3 minus x3| + C

Example 4

Evaluateint

sinθ

cosθ dθ

Solution


cosθ dθ =

int minus 1

u du


int minus 1


Example 5

Evaluateint

cos(t)3+sin(t)

dt

Solution

Let u = 3 + sin(t)

rArr du = cos(t)dt

int cos(t)3+sin(t)

dθ =int

1u

du


int 1u

du = ln|u| + C = ln|3 + sin(t)| + C

30





Theorem 48



1int

duradic a2minusu2

= arcsin u

a +C

2int

du

a2+u2 = 1a

arctan u

a +C

3int

du

uradic u2minusa2

= 1a

arcsec |u|a

+C



Example 1

Evaluateint

41+9x2 middotdx

Solution


3du = dx


int 41+9x2 middotdx

=int

41+(3x)2

middotdx

= 13

int 4du1+u2

= 43

int du

1+u2

= 4

3arctan(u) + C

= 43

arctan(3x) + C

Example 2

Evaluateint

dx

4+(x+1)2

Solution



31



int dx

4+(x+1)2 =int

du

22+u2

= 12

arctan(u2

) + C

= 12

arctan(x+12

) + C

Example 3

Evaluateint

7radic 4minusx2

middotdx

Solution



int 7

radic 4minusx2

= int 7du

radic 22

minusu2


) + C

= 7arcsin(x2

) + C

Example 4

Evaluateint

dx


Solution



int dx


=int

du

uradic u2minus9

= 13

arcsec(u3

) + C

= 13

arcsec( |x|3

) + C

32





Example 8

Evaluateint


Solution





rArr minus12

du = (x minus 3)dx

Now substitute


=int


8radic 9+(xminus3)2

= minus12

int dvradic


9+(u)2

= 12

int vminus 1


= minus12

(2v1

2 ) + 8arcsin(u3

) + C



) + C


34






int b

a f (x)dx is


f (x)dx = bminusa


Example 1


int 20

2x2dx n = 4

Solution

x = 2minus04

= 24

= 12


= i2


2(4)[f (0) + 2f (1

2) + 2f (1) + 2f (3

2) + f (2)]

= 28

[2(0)2 + 2(2(12

)2) + 2(2(1)2) + 2(2( 32

)2) + 2(22)]

= 1

4

[0 + 2(1

2

) + 2(2) + 2( 18

4

) + 8]= 1

4[0 + 1 + 4 + 9 + 8]

= 14

[22]= 55



a f (x)dx is


int b


Example 2


int 2

0 2x2dx n = 4

Solution

35



x = 2minus04

= 24

= 12


= i2

int 20

2x2dx = 2minus03(4)

[f (0) + 4f (12

) + 2f (1) + 4f (32

) + f (2)]

= 212

[2(0)2 + 4(2(12

)2) + 2(2(1)2) + 4(2(32

)2) + 2(22)]= 1

6[0 + 4(1

2) + 2(2) + 4( 18

4 ) + 8]

= 16

[0 + 2 + 4 + 18 + 8]

= 16

[32]= 53333


int 2

0 2x2dx = 22+1

x2+120

= 23

x320

= 23

(2)3 minus 23

(0)3


= 163

Example 3


int π0 sin(x)dx

Solution

x = πminus04

= π4


= iπ4int π


[f (0) + 2f (π4

) + 2f (2π4

) + 2f (3π4

) + f (π)]

= π8

[sin(0) + 2sin(π4

) + 2sin(π2

) + 2sin(3π4

) + sin(π)]

= π8 [0 + 2(radic 22 ) + 2(1) + 2(radic 22 ) + 0]

= π8

[2radic

2 + 2]asymp 2003


int π0 sin(x)dx

Solution

x = πminus04

= π4

36





= 112

[111 + 4(0947) + 2(0816) + 4(0771) + 0625] asymp 086



1(x+2)2

dx =int 2

11u2

du =int 2


x+2

21

= minus 12+2

minus (minus 11+2

)

= minus14

+ 13

= 112



=1048616

8 + 12

n +

4

n2 + 12 +

12

n + 6

1048617 = 26

Example 4


Solution

a = 4 b = 10 x = 10minus4n

= 6n


= 4 + 6in

f (x) = 6


nrarrinfin

nsumi=1

f (ci)xi

= limnrarrinfin

nsumi=1

f 983080

4 + 6i

n

983081middot 6

n

= limnrarrinfin

nsumi=1

(6) middot 6

n

= limn

rarrinfin(6n)

middot

6

n = 36

17



Example 5


a = 1 b = 2 x = 2minus1n

= 1n


= 1 + i

n

f (x) = x2 + 1


int 21 (x2 + 1)dx = lim

nrarrinfin

nsumi=1

f (ci)xi

= limnrarrinfin

n

sumi=1 f

9830801 +

i

n983081 middot

1

n

= limnrarrinfin

nsumi=1

9830809830801 +

i

n

9830812+ 1

983081middot 1

n

= limnrarrinfin

nsumi=1

9830809830801 +

i

n

9830819830801 +

i

n

983081+ 1

983081middot 1

n

= limnrarrinfin

nsumi=1

98308010486161 +

2i

n +

i2

n2

983081+ 1

983081middot 1

n

= limnrarrinfin

n

sumi=1

9830802 + 2i

n

+ i2

n2983081 middot

1

n

= limnrarrinfin

nsumi=1

9830802

n +

2i

n2 +

i2

n3

983081

= limnrarrinfin

983080 2

n middot (n) +

2

n2 middot n(n + 1)

2 +

1

n3 middot n(n + 1)(2n + 1)

6

983081

= limnrarrinfin

9830802 +

2n2 + 2n

2n2 +

(n2 + n)(2n + 1)

6n3

983081

= limnrarrinfin

9830802 +

2n2 + 2n

2n2 +

2n3 + 3n2 + n

6n3

983081


2 + 2n2

2n2

+ 2n

2n2

+ 2n3

6n3

+ 3n2

6n3

+ n

6n3983081

= limnrarrinfin

9830802 + 1 +

1

n +

1

3 +

1

2n +

1

6n2

983081 = 2 + 1 + 0 +

1

3 + 0 + 0 =

10

3

18



Example 6


a = 0 b = 3

x = 3minus0

n = 3

n


= 3n

f (x) = 5x


int 3


n

sumi=1

f (ci)

xi

= limnrarrinfin

nsumi=1

f 9830803i

n

983081middot 3

n

= limnrarrinfin

nsumi=1

59830803i

n

983081middot 3

n

= limnrarrinfin

nsumi=1

45i

n2

= limnrarrinfin

nsum

i=1

45

n2 middot n(n + 1)

2

= limnrarrinfin

nsumi=1

98308045n(n + 1)

2n2

983081

= limnrarrinfin

nsumi=1

98308045n2 + 45n

2n2

983081

= limnrarrinfin

nsumi=1

98308045n2

2n2 +

45n

2n2

983081

= limnrarrinfin

nsumi=1

98308045

2 +

45

2n

983081 =

45

2 + 0 =

45

2

19






int b




Example 1


Solution

int 20

4xdx = 41+1

x220

= 2x2|20 = 2(2)2 minus 2(0)2 = 8 minus 0 = 8

Example 2


Solution

int 21

x3 + x2dx =1048616

13+1

x3+1 + 12+1

x2+1)21

=1048616 14

x4 + 13

x3)21

=104861614

24 + 13

23) minus (14

14 + 13

13)

= 4 + 83 minus 1048616

14

+ 13

)

= 203 minus 1

12

= 7912

Example 3


2

dx

Solution

int 20 6x2dx = 6

2+1x2+1

21


= 16 minus 0= 16

20



Example 4


exdx

Solution

int 20



Example 5


x+radic x

2 dx

= [1

2 middot x2

2 + 1

2 middot 1

2 middot x

3

2

]10

= [x2

4 + 1

3 middot x

3

2 ]10

= [12

4 + 1

3 middot 1

3

2 ] minus [02

4 + 1

3 middot 0

3

2 ]14 minus 1

3 minus 0 = 7

12

Solution

int 10

x+radic x

2 dx =

int 10

x2

+radic x

2 dx =


Example 6


Solution

Note

f (x) = 2

xminus 3

xge

3

2



int 32


int 23

2

(2x minus 3)dx

=int 3

2


3

2

(2x minus 3)dx

= (minusx2 + 3x)|3

2

0 + (x2 minus 3x)|232

= [((32

)2 + 3 middot 32



)]= 10

4

21








1bminusa

int b

a f (x)dx


Example 7


Solution



= 14


= 1

4 middot (1

3x3

+ x)2minus2

= 14 middot [1

3(2)3 + 2] minus 1

3 middot [1

3(minus2)3 minus 2]

= 14 middot 14

3 minus 1

3 middot minus14

3

= 76

+ 76

= 146

= 73

Example 8


2

]

Solution

AV = 1π

2minus0int π

2

0 cos(x)dx

= 2π

int π2

0 cos(x)dx

= 2π middot 1048616 sin(x))|

π

2

0


2) minus sin(0)]


= 2π middot 1

= 2

π

22



Example 9


y = x2 + 2 y = 0 x = 0 x = 2

Solution


=1048616 x3

3 + 2x)

2

0

= [23

3 + 2(2)] minus [03

3 + 2(0)]= [8

3 + 4] minus [0 + 0]

= 203 minus 0

= 203

23



Example 10


y = ex y = 0 x = 0 x = 2

Solution

Area=int 20 (ex)dx

= [ex]|20

= e2 minus e0

= e2 minus 1

Example 11


0 (1 + cos(x))dx

Solution


1048616 x + sin(x))|

π

0


24







Example 1


Solution

f (x) = (x2 + 4x)6




Example 2

Evaluateint

(2x + 3)(x2 + 3x)6dx

Solution


int (2x + 3)(x2 + 3x)6dx =

int u6 middot du


6+1u6+1 + C = 1

7u7 + C = 1

7(x2 + 3x)7 + C

Example 3

Evaluateint

3(1 + 3x)3dx

Solution


int 3(1 + 3x)3dx =

int u3 middot du

25




3+1u3+1 + C = 1

4u4 + C = 1

4(1 + 3x)4 + C

Example 4

Evaluateint

x2(2 + x3)4dx

Solution



int 13

u4 middot du


int 13


5u5 + C = 1

15u5 + C = 1

15(2 + x3)5 + C

26



Example 5

Evaluateint

2tradic

t2 + 3dt

Solution


radic t2 + 3dt =


int u

1

2 middot du


1


3

2 + C = 23

(t2 + 3)3

2 + C = 23

(radic

t2 + 3)3 + C

Example 6

Evaluateint

πcos(πx)dx

Solution


int cos(u)du



Example 7

Evaluateint

xradic 1minus4x2 dx

Solution


= xdx

int x

radic 1minus4x2

dx = int minus1

8 middot du



1


1

2 + C = minus14

u1

2 + C = minus14

(1 minus 4x2)1

2 + C = minus14


27



Example 8

Evaluateint

2xe2x2

dx

Solution


= 2xdx

int 2xe2x

2

dx =int

12

eudu


eudu = 12

eu + C = 12

e2x2

+ C

Example 9

Evaluateint

2xsin(2x2)dx

Solution


du = 2xdx

int 2xsin2x2dx =

int 12

sin(u)du


int 12

sin(u)du =minus

1

2cos(u) + C =

minus1

2cos(2x2) + C

28







2int 1u

du = ln|x| + C

Example 1

Evaluateint

13x

dx

Solution


3du = dxint 13x

dx =int

13 middot 1

u du


int 13 middot 1

u du = 1

3ln|u| + C = 1

3ln|3x| + C

Example 2

Evaluateint

15x+3

dx

Solution


du = dx

int 13x

dx =int

15 middot 1

u du


int 15 middot 1

u du = 1

5ln|u| + C = 1

5ln|5x + 3| + C

29



Example 3

Evaluateint

x2

3minusx3 dx

Solution


du = x2dx

int x

2

3minusx3 dx =int

minus13 middot 1

u du


int minus1

3 middot 1

u du = minus1

3ln|u| + C = minus1

3ln|3 minus x3| + C

Example 4

Evaluateint

sinθ

cosθ dθ

Solution


cosθ dθ =

int minus 1

u du


int minus 1


Example 5

Evaluateint

cos(t)3+sin(t)

dt

Solution

Let u = 3 + sin(t)

rArr du = cos(t)dt

int cos(t)3+sin(t)

dθ =int

1u

du


int 1u

du = ln|u| + C = ln|3 + sin(t)| + C

30





Theorem 48



1int

duradic a2minusu2

= arcsin u

a +C

2int

du

a2+u2 = 1a

arctan u

a +C

3int

du

uradic u2minusa2

= 1a

arcsec |u|a

+C



Example 1

Evaluateint

41+9x2 middotdx

Solution


3du = dx


int 41+9x2 middotdx

=int

41+(3x)2

middotdx

= 13

int 4du1+u2

= 43

int du

1+u2

= 4

3arctan(u) + C

= 43

arctan(3x) + C

Example 2

Evaluateint

dx

4+(x+1)2

Solution



31



int dx

4+(x+1)2 =int

du

22+u2

= 12

arctan(u2

) + C

= 12

arctan(x+12

) + C

Example 3

Evaluateint

7radic 4minusx2

middotdx

Solution



int 7

radic 4minusx2

= int 7du

radic 22

minusu2


) + C

= 7arcsin(x2

) + C

Example 4

Evaluateint

dx


Solution



int dx


=int

du

uradic u2minus9

= 13

arcsec(u3

) + C

= 13

arcsec( |x|3

) + C

32





Example 8

Evaluateint


Solution





rArr minus12

du = (x minus 3)dx

Now substitute


=int


8radic 9+(xminus3)2

= minus12

int dvradic


9+(u)2

= 12

int vminus 1


= minus12

(2v1

2 ) + 8arcsin(u3

) + C



) + C


34






int b

a f (x)dx is


f (x)dx = bminusa


Example 1


int 20

2x2dx n = 4

Solution

x = 2minus04

= 24

= 12


= i2


2(4)[f (0) + 2f (1

2) + 2f (1) + 2f (3

2) + f (2)]

= 28

[2(0)2 + 2(2(12

)2) + 2(2(1)2) + 2(2( 32

)2) + 2(22)]

= 1

4

[0 + 2(1

2

) + 2(2) + 2( 18

4

) + 8]= 1

4[0 + 1 + 4 + 9 + 8]

= 14

[22]= 55



a f (x)dx is


int b


Example 2


int 2

0 2x2dx n = 4

Solution

35



x = 2minus04

= 24

= 12


= i2

int 20

2x2dx = 2minus03(4)

[f (0) + 4f (12

) + 2f (1) + 4f (32

) + f (2)]

= 212

[2(0)2 + 4(2(12

)2) + 2(2(1)2) + 4(2(32

)2) + 2(22)]= 1

6[0 + 4(1

2) + 2(2) + 4( 18

4 ) + 8]

= 16

[0 + 2 + 4 + 18 + 8]

= 16

[32]= 53333


int 2

0 2x2dx = 22+1

x2+120

= 23

x320

= 23

(2)3 minus 23

(0)3


= 163

Example 3


int π0 sin(x)dx

Solution

x = πminus04

= π4


= iπ4int π


[f (0) + 2f (π4

) + 2f (2π4

) + 2f (3π4

) + f (π)]

= π8

[sin(0) + 2sin(π4

) + 2sin(π2

) + 2sin(3π4

) + sin(π)]

= π8 [0 + 2(radic 22 ) + 2(1) + 2(radic 22 ) + 0]

= π8

[2radic

2 + 2]asymp 2003


int π0 sin(x)dx

Solution

x = πminus04

= π4

36





= 112

[111 + 4(0947) + 2(0816) + 4(0771) + 0625] asymp 086



1(x+2)2

dx =int 2

11u2

du =int 2


x+2

21

= minus 12+2

minus (minus 11+2

)

= minus14

+ 13

= 112



Example 5


a = 1 b = 2 x = 2minus1n

= 1n


= 1 + i

n

f (x) = x2 + 1


int 21 (x2 + 1)dx = lim

nrarrinfin

nsumi=1

f (ci)xi

= limnrarrinfin

n

sumi=1 f

9830801 +

i

n983081 middot

1

n

= limnrarrinfin

nsumi=1

9830809830801 +

i

n

9830812+ 1

983081middot 1

n

= limnrarrinfin

nsumi=1

9830809830801 +

i

n

9830819830801 +

i

n

983081+ 1

983081middot 1

n

= limnrarrinfin

nsumi=1

98308010486161 +

2i

n +

i2

n2

983081+ 1

983081middot 1

n

= limnrarrinfin

n

sumi=1

9830802 + 2i

n

+ i2

n2983081 middot

1

n

= limnrarrinfin

nsumi=1

9830802

n +

2i

n2 +

i2

n3

983081

= limnrarrinfin

983080 2

n middot (n) +

2

n2 middot n(n + 1)

2 +

1

n3 middot n(n + 1)(2n + 1)

6

983081

= limnrarrinfin

9830802 +

2n2 + 2n

2n2 +

(n2 + n)(2n + 1)

6n3

983081

= limnrarrinfin

9830802 +

2n2 + 2n

2n2 +

2n3 + 3n2 + n

6n3

983081


2 + 2n2

2n2

+ 2n

2n2

+ 2n3

6n3

+ 3n2

6n3

+ n

6n3983081

= limnrarrinfin

9830802 + 1 +

1

n +

1

3 +

1

2n +

1

6n2

983081 = 2 + 1 + 0 +

1

3 + 0 + 0 =

10

3

18



Example 6


a = 0 b = 3

x = 3minus0

n = 3

n


= 3n

f (x) = 5x


int 3


n

sumi=1

f (ci)

xi

= limnrarrinfin

nsumi=1

f 9830803i

n

983081middot 3

n

= limnrarrinfin

nsumi=1

59830803i

n

983081middot 3

n

= limnrarrinfin

nsumi=1

45i

n2

= limnrarrinfin

nsum

i=1

45

n2 middot n(n + 1)

2

= limnrarrinfin

nsumi=1

98308045n(n + 1)

2n2

983081

= limnrarrinfin

nsumi=1

98308045n2 + 45n

2n2

983081

= limnrarrinfin

nsumi=1

98308045n2

2n2 +

45n

2n2

983081

= limnrarrinfin

nsumi=1

98308045

2 +

45

2n

983081 =

45

2 + 0 =

45

2

19






int b




Example 1


Solution

int 20

4xdx = 41+1

x220

= 2x2|20 = 2(2)2 minus 2(0)2 = 8 minus 0 = 8

Example 2


Solution

int 21

x3 + x2dx =1048616

13+1

x3+1 + 12+1

x2+1)21

=1048616 14

x4 + 13

x3)21

=104861614

24 + 13

23) minus (14

14 + 13

13)

= 4 + 83 minus 1048616

14

+ 13

)

= 203 minus 1

12

= 7912

Example 3


2

dx

Solution

int 20 6x2dx = 6

2+1x2+1

21


= 16 minus 0= 16

20



Example 4


exdx

Solution

int 20



Example 5


x+radic x

2 dx

= [1

2 middot x2

2 + 1

2 middot 1

2 middot x

3

2

]10

= [x2

4 + 1

3 middot x

3

2 ]10

= [12

4 + 1

3 middot 1

3

2 ] minus [02

4 + 1

3 middot 0

3

2 ]14 minus 1

3 minus 0 = 7

12

Solution

int 10

x+radic x

2 dx =

int 10

x2

+radic x

2 dx =


Example 6


Solution

Note

f (x) = 2

xminus 3

xge

3

2



int 32


int 23

2

(2x minus 3)dx

=int 3

2


3

2

(2x minus 3)dx

= (minusx2 + 3x)|3

2

0 + (x2 minus 3x)|232

= [((32

)2 + 3 middot 32



)]= 10

4

21








1bminusa

int b

a f (x)dx


Example 7


Solution



= 14


= 1

4 middot (1

3x3

+ x)2minus2

= 14 middot [1

3(2)3 + 2] minus 1

3 middot [1

3(minus2)3 minus 2]

= 14 middot 14

3 minus 1

3 middot minus14

3

= 76

+ 76

= 146

= 73

Example 8


2

]

Solution

AV = 1π

2minus0int π

2

0 cos(x)dx

= 2π

int π2

0 cos(x)dx

= 2π middot 1048616 sin(x))|

π

2

0


2) minus sin(0)]


= 2π middot 1

= 2

π

22



Example 9


y = x2 + 2 y = 0 x = 0 x = 2

Solution


=1048616 x3

3 + 2x)

2

0

= [23

3 + 2(2)] minus [03

3 + 2(0)]= [8

3 + 4] minus [0 + 0]

= 203 minus 0

= 203

23



Example 10


y = ex y = 0 x = 0 x = 2

Solution

Area=int 20 (ex)dx

= [ex]|20

= e2 minus e0

= e2 minus 1

Example 11


0 (1 + cos(x))dx

Solution


1048616 x + sin(x))|

π

0


24







Example 1


Solution

f (x) = (x2 + 4x)6




Example 2

Evaluateint

(2x + 3)(x2 + 3x)6dx

Solution


int (2x + 3)(x2 + 3x)6dx =

int u6 middot du


6+1u6+1 + C = 1

7u7 + C = 1

7(x2 + 3x)7 + C

Example 3

Evaluateint

3(1 + 3x)3dx

Solution


int 3(1 + 3x)3dx =

int u3 middot du

25




3+1u3+1 + C = 1

4u4 + C = 1

4(1 + 3x)4 + C

Example 4

Evaluateint

x2(2 + x3)4dx

Solution



int 13

u4 middot du


int 13


5u5 + C = 1

15u5 + C = 1

15(2 + x3)5 + C

26



Example 5

Evaluateint

2tradic

t2 + 3dt

Solution


radic t2 + 3dt =


int u

1

2 middot du


1


3

2 + C = 23

(t2 + 3)3

2 + C = 23

(radic

t2 + 3)3 + C

Example 6

Evaluateint

πcos(πx)dx

Solution


int cos(u)du



Example 7

Evaluateint

xradic 1minus4x2 dx

Solution


= xdx

int x

radic 1minus4x2

dx = int minus1

8 middot du



1


1

2 + C = minus14

u1

2 + C = minus14

(1 minus 4x2)1

2 + C = minus14


27



Example 8

Evaluateint

2xe2x2

dx

Solution


= 2xdx

int 2xe2x

2

dx =int

12

eudu


eudu = 12

eu + C = 12

e2x2

+ C

Example 9

Evaluateint

2xsin(2x2)dx

Solution


du = 2xdx

int 2xsin2x2dx =

int 12

sin(u)du


int 12

sin(u)du =minus

1

2cos(u) + C =

minus1

2cos(2x2) + C

28







2int 1u

du = ln|x| + C

Example 1

Evaluateint

13x

dx

Solution


3du = dxint 13x

dx =int

13 middot 1

u du


int 13 middot 1

u du = 1

3ln|u| + C = 1

3ln|3x| + C

Example 2

Evaluateint

15x+3

dx

Solution


du = dx

int 13x

dx =int

15 middot 1

u du


int 15 middot 1

u du = 1

5ln|u| + C = 1

5ln|5x + 3| + C

29



Example 3

Evaluateint

x2

3minusx3 dx

Solution


du = x2dx

int x

2

3minusx3 dx =int

minus13 middot 1

u du


int minus1

3 middot 1

u du = minus1

3ln|u| + C = minus1

3ln|3 minus x3| + C

Example 4

Evaluateint

sinθ

cosθ dθ

Solution


cosθ dθ =

int minus 1

u du


int minus 1


Example 5

Evaluateint

cos(t)3+sin(t)

dt

Solution

Let u = 3 + sin(t)

rArr du = cos(t)dt

int cos(t)3+sin(t)

dθ =int

1u

du


int 1u

du = ln|u| + C = ln|3 + sin(t)| + C

30





Theorem 48



1int

duradic a2minusu2

= arcsin u

a +C

2int

du

a2+u2 = 1a

arctan u

a +C

3int

du

uradic u2minusa2

= 1a

arcsec |u|a

+C



Example 1

Evaluateint

41+9x2 middotdx

Solution


3du = dx


int 41+9x2 middotdx

=int

41+(3x)2

middotdx

= 13

int 4du1+u2

= 43

int du

1+u2

= 4

3arctan(u) + C

= 43

arctan(3x) + C

Example 2

Evaluateint

dx

4+(x+1)2

Solution



31



int dx

4+(x+1)2 =int

du

22+u2

= 12

arctan(u2

) + C

= 12

arctan(x+12

) + C

Example 3

Evaluateint

7radic 4minusx2

middotdx

Solution



int 7

radic 4minusx2

= int 7du

radic 22

minusu2


) + C

= 7arcsin(x2

) + C

Example 4

Evaluateint

dx


Solution



int dx


=int

du

uradic u2minus9

= 13

arcsec(u3

) + C

= 13

arcsec( |x|3

) + C

32





Example 8

Evaluateint


Solution





rArr minus12

du = (x minus 3)dx

Now substitute


=int


8radic 9+(xminus3)2

= minus12

int dvradic


9+(u)2

= 12

int vminus 1


= minus12

(2v1

2 ) + 8arcsin(u3

) + C



) + C


34






int b

a f (x)dx is


f (x)dx = bminusa


Example 1


int 20

2x2dx n = 4

Solution

x = 2minus04

= 24

= 12


= i2


2(4)[f (0) + 2f (1

2) + 2f (1) + 2f (3

2) + f (2)]

= 28

[2(0)2 + 2(2(12

)2) + 2(2(1)2) + 2(2( 32

)2) + 2(22)]

= 1

4

[0 + 2(1

2

) + 2(2) + 2( 18

4

) + 8]= 1

4[0 + 1 + 4 + 9 + 8]

= 14

[22]= 55



a f (x)dx is


int b


Example 2


int 2

0 2x2dx n = 4

Solution

35



x = 2minus04

= 24

= 12


= i2

int 20

2x2dx = 2minus03(4)

[f (0) + 4f (12

) + 2f (1) + 4f (32

) + f (2)]

= 212

[2(0)2 + 4(2(12

)2) + 2(2(1)2) + 4(2(32

)2) + 2(22)]= 1

6[0 + 4(1

2) + 2(2) + 4( 18

4 ) + 8]

= 16

[0 + 2 + 4 + 18 + 8]

= 16

[32]= 53333


int 2

0 2x2dx = 22+1

x2+120

= 23

x320

= 23

(2)3 minus 23

(0)3


= 163

Example 3


int π0 sin(x)dx

Solution

x = πminus04

= π4


= iπ4int π


[f (0) + 2f (π4

) + 2f (2π4

) + 2f (3π4

) + f (π)]

= π8

[sin(0) + 2sin(π4

) + 2sin(π2

) + 2sin(3π4

) + sin(π)]

= π8 [0 + 2(radic 22 ) + 2(1) + 2(radic 22 ) + 0]

= π8

[2radic

2 + 2]asymp 2003


int π0 sin(x)dx

Solution

x = πminus04

= π4

36





= 112

[111 + 4(0947) + 2(0816) + 4(0771) + 0625] asymp 086



1(x+2)2

dx =int 2

11u2

du =int 2


x+2

21

= minus 12+2

minus (minus 11+2

)

= minus14

+ 13

= 112



Example 6


a = 0 b = 3

x = 3minus0

n = 3

n


= 3n

f (x) = 5x


int 3


n

sumi=1

f (ci)

xi

= limnrarrinfin

nsumi=1

f 9830803i

n

983081middot 3

n

= limnrarrinfin

nsumi=1

59830803i

n

983081middot 3

n

= limnrarrinfin

nsumi=1

45i

n2

= limnrarrinfin

nsum

i=1

45

n2 middot n(n + 1)

2

= limnrarrinfin

nsumi=1

98308045n(n + 1)

2n2

983081

= limnrarrinfin

nsumi=1

98308045n2 + 45n

2n2

983081

= limnrarrinfin

nsumi=1

98308045n2

2n2 +

45n

2n2

983081

= limnrarrinfin

nsumi=1

98308045

2 +

45

2n

983081 =

45

2 + 0 =

45

2

19






int b




Example 1


Solution

int 20

4xdx = 41+1

x220

= 2x2|20 = 2(2)2 minus 2(0)2 = 8 minus 0 = 8

Example 2


Solution

int 21

x3 + x2dx =1048616

13+1

x3+1 + 12+1

x2+1)21

=1048616 14

x4 + 13

x3)21

=104861614

24 + 13

23) minus (14

14 + 13

13)

= 4 + 83 minus 1048616

14

+ 13

)

= 203 minus 1

12

= 7912

Example 3


2

dx

Solution

int 20 6x2dx = 6

2+1x2+1

21


= 16 minus 0= 16

20



Example 4


exdx

Solution

int 20



Example 5


x+radic x

2 dx

= [1

2 middot x2

2 + 1

2 middot 1

2 middot x

3

2

]10

= [x2

4 + 1

3 middot x

3

2 ]10

= [12

4 + 1

3 middot 1

3

2 ] minus [02

4 + 1

3 middot 0

3

2 ]14 minus 1

3 minus 0 = 7

12

Solution

int 10

x+radic x

2 dx =

int 10

x2

+radic x

2 dx =


Example 6


Solution

Note

f (x) = 2

xminus 3

xge

3

2



int 32


int 23

2

(2x minus 3)dx

=int 3

2


3

2

(2x minus 3)dx

= (minusx2 + 3x)|3

2

0 + (x2 minus 3x)|232

= [((32

)2 + 3 middot 32



)]= 10

4

21








1bminusa

int b

a f (x)dx


Example 7


Solution



= 14


= 1

4 middot (1

3x3

+ x)2minus2

= 14 middot [1

3(2)3 + 2] minus 1

3 middot [1

3(minus2)3 minus 2]

= 14 middot 14

3 minus 1

3 middot minus14

3

= 76

+ 76

= 146

= 73

Example 8


2

]

Solution

AV = 1π

2minus0int π

2

0 cos(x)dx

= 2π

int π2

0 cos(x)dx

= 2π middot 1048616 sin(x))|

π

2

0


2) minus sin(0)]


= 2π middot 1

= 2

π

22



Example 9


y = x2 + 2 y = 0 x = 0 x = 2

Solution


=1048616 x3

3 + 2x)

2

0

= [23

3 + 2(2)] minus [03

3 + 2(0)]= [8

3 + 4] minus [0 + 0]

= 203 minus 0

= 203

23



Example 10


y = ex y = 0 x = 0 x = 2

Solution

Area=int 20 (ex)dx

= [ex]|20

= e2 minus e0

= e2 minus 1

Example 11


0 (1 + cos(x))dx

Solution


1048616 x + sin(x))|

π

0


24







Example 1


Solution

f (x) = (x2 + 4x)6




Example 2

Evaluateint

(2x + 3)(x2 + 3x)6dx

Solution


int (2x + 3)(x2 + 3x)6dx =

int u6 middot du


6+1u6+1 + C = 1

7u7 + C = 1

7(x2 + 3x)7 + C

Example 3

Evaluateint

3(1 + 3x)3dx

Solution


int 3(1 + 3x)3dx =

int u3 middot du

25




3+1u3+1 + C = 1

4u4 + C = 1

4(1 + 3x)4 + C

Example 4

Evaluateint

x2(2 + x3)4dx

Solution



int 13

u4 middot du


int 13


5u5 + C = 1

15u5 + C = 1

15(2 + x3)5 + C

26



Example 5

Evaluateint

2tradic

t2 + 3dt

Solution


radic t2 + 3dt =


int u

1

2 middot du


1


3

2 + C = 23

(t2 + 3)3

2 + C = 23

(radic

t2 + 3)3 + C

Example 6

Evaluateint

πcos(πx)dx

Solution


int cos(u)du



Example 7

Evaluateint

xradic 1minus4x2 dx

Solution


= xdx

int x

radic 1minus4x2

dx = int minus1

8 middot du



1


1

2 + C = minus14

u1

2 + C = minus14

(1 minus 4x2)1

2 + C = minus14


27



Example 8

Evaluateint

2xe2x2

dx

Solution


= 2xdx

int 2xe2x

2

dx =int

12

eudu


eudu = 12

eu + C = 12

e2x2

+ C

Example 9

Evaluateint

2xsin(2x2)dx

Solution


du = 2xdx

int 2xsin2x2dx =

int 12

sin(u)du


int 12

sin(u)du =minus

1

2cos(u) + C =

minus1

2cos(2x2) + C

28







2int 1u

du = ln|x| + C

Example 1

Evaluateint

13x

dx

Solution


3du = dxint 13x

dx =int

13 middot 1

u du


int 13 middot 1

u du = 1

3ln|u| + C = 1

3ln|3x| + C

Example 2

Evaluateint

15x+3

dx

Solution


du = dx

int 13x

dx =int

15 middot 1

u du


int 15 middot 1

u du = 1

5ln|u| + C = 1

5ln|5x + 3| + C

29



Example 3

Evaluateint

x2

3minusx3 dx

Solution


du = x2dx

int x

2

3minusx3 dx =int

minus13 middot 1

u du


int minus1

3 middot 1

u du = minus1

3ln|u| + C = minus1

3ln|3 minus x3| + C

Example 4

Evaluateint

sinθ

cosθ dθ

Solution


cosθ dθ =

int minus 1

u du


int minus 1


Example 5

Evaluateint

cos(t)3+sin(t)

dt

Solution

Let u = 3 + sin(t)

rArr du = cos(t)dt

int cos(t)3+sin(t)

dθ =int

1u

du


int 1u

du = ln|u| + C = ln|3 + sin(t)| + C

30





Theorem 48



1int

duradic a2minusu2

= arcsin u

a +C

2int

du

a2+u2 = 1a

arctan u

a +C

3int

du

uradic u2minusa2

= 1a

arcsec |u|a

+C



Example 1

Evaluateint

41+9x2 middotdx

Solution


3du = dx


int 41+9x2 middotdx

=int

41+(3x)2

middotdx

= 13

int 4du1+u2

= 43

int du

1+u2

= 4

3arctan(u) + C

= 43

arctan(3x) + C

Example 2

Evaluateint

dx

4+(x+1)2

Solution



31



int dx

4+(x+1)2 =int

du

22+u2

= 12

arctan(u2

) + C

= 12

arctan(x+12

) + C

Example 3

Evaluateint

7radic 4minusx2

middotdx

Solution



int 7

radic 4minusx2

= int 7du

radic 22

minusu2


) + C

= 7arcsin(x2

) + C

Example 4

Evaluateint

dx


Solution



int dx


=int

du

uradic u2minus9

= 13

arcsec(u3

) + C

= 13

arcsec( |x|3

) + C

32





Example 8

Evaluateint


Solution





rArr minus12

du = (x minus 3)dx

Now substitute


=int


8radic 9+(xminus3)2

= minus12

int dvradic


9+(u)2

= 12

int vminus 1


= minus12

(2v1

2 ) + 8arcsin(u3

) + C



) + C


34






int b

a f (x)dx is


f (x)dx = bminusa


Example 1


int 20

2x2dx n = 4

Solution

x = 2minus04

= 24

= 12


= i2


2(4)[f (0) + 2f (1

2) + 2f (1) + 2f (3

2) + f (2)]

= 28

[2(0)2 + 2(2(12

)2) + 2(2(1)2) + 2(2( 32

)2) + 2(22)]

= 1

4

[0 + 2(1

2

) + 2(2) + 2( 18

4

) + 8]= 1

4[0 + 1 + 4 + 9 + 8]

= 14

[22]= 55



a f (x)dx is


int b


Example 2


int 2

0 2x2dx n = 4

Solution

35



x = 2minus04

= 24

= 12


= i2

int 20

2x2dx = 2minus03(4)

[f (0) + 4f (12

) + 2f (1) + 4f (32

) + f (2)]

= 212

[2(0)2 + 4(2(12

)2) + 2(2(1)2) + 4(2(32

)2) + 2(22)]= 1

6[0 + 4(1

2) + 2(2) + 4( 18

4 ) + 8]

= 16

[0 + 2 + 4 + 18 + 8]

= 16

[32]= 53333


int 2

0 2x2dx = 22+1

x2+120

= 23

x320

= 23

(2)3 minus 23

(0)3


= 163

Example 3


int π0 sin(x)dx

Solution

x = πminus04

= π4


= iπ4int π


[f (0) + 2f (π4

) + 2f (2π4

) + 2f (3π4

) + f (π)]

= π8

[sin(0) + 2sin(π4

) + 2sin(π2

) + 2sin(3π4

) + sin(π)]

= π8 [0 + 2(radic 22 ) + 2(1) + 2(radic 22 ) + 0]

= π8

[2radic

2 + 2]asymp 2003


int π0 sin(x)dx

Solution

x = πminus04

= π4

36





= 112

[111 + 4(0947) + 2(0816) + 4(0771) + 0625] asymp 086



1(x+2)2

dx =int 2

11u2

du =int 2


x+2

21

= minus 12+2

minus (minus 11+2

)

= minus14

+ 13

= 112






int b




Example 1


Solution

int 20

4xdx = 41+1

x220

= 2x2|20 = 2(2)2 minus 2(0)2 = 8 minus 0 = 8

Example 2


Solution

int 21

x3 + x2dx =1048616

13+1

x3+1 + 12+1

x2+1)21

=1048616 14

x4 + 13

x3)21

=104861614

24 + 13

23) minus (14

14 + 13

13)

= 4 + 83 minus 1048616

14

+ 13

)

= 203 minus 1

12

= 7912

Example 3


2

dx

Solution

int 20 6x2dx = 6

2+1x2+1

21


= 16 minus 0= 16

20



Example 4


exdx

Solution

int 20



Example 5


x+radic x

2 dx

= [1

2 middot x2

2 + 1

2 middot 1

2 middot x

3

2

]10

= [x2

4 + 1

3 middot x

3

2 ]10

= [12

4 + 1

3 middot 1

3

2 ] minus [02

4 + 1

3 middot 0

3

2 ]14 minus 1

3 minus 0 = 7

12

Solution

int 10

x+radic x

2 dx =

int 10

x2

+radic x

2 dx =


Example 6


Solution

Note

f (x) = 2

xminus 3

xge

3

2



int 32


int 23

2

(2x minus 3)dx

=int 3

2


3

2

(2x minus 3)dx

= (minusx2 + 3x)|3

2

0 + (x2 minus 3x)|232

= [((32

)2 + 3 middot 32



)]= 10

4

21








1bminusa

int b

a f (x)dx


Example 7


Solution



= 14


= 1

4 middot (1

3x3

+ x)2minus2

= 14 middot [1

3(2)3 + 2] minus 1

3 middot [1

3(minus2)3 minus 2]

= 14 middot 14

3 minus 1

3 middot minus14

3

= 76

+ 76

= 146

= 73

Example 8


2

]

Solution

AV = 1π

2minus0int π

2

0 cos(x)dx

= 2π

int π2

0 cos(x)dx

= 2π middot 1048616 sin(x))|

π

2

0


2) minus sin(0)]


= 2π middot 1

= 2

π

22



Example 9


y = x2 + 2 y = 0 x = 0 x = 2

Solution


=1048616 x3

3 + 2x)

2

0

= [23

3 + 2(2)] minus [03

3 + 2(0)]= [8

3 + 4] minus [0 + 0]

= 203 minus 0

= 203

23



Example 10


y = ex y = 0 x = 0 x = 2

Solution

Area=int 20 (ex)dx

= [ex]|20

= e2 minus e0

= e2 minus 1

Example 11


0 (1 + cos(x))dx

Solution


1048616 x + sin(x))|

π

0


24







Example 1


Solution

f (x) = (x2 + 4x)6




Example 2

Evaluateint

(2x + 3)(x2 + 3x)6dx

Solution


int (2x + 3)(x2 + 3x)6dx =

int u6 middot du


6+1u6+1 + C = 1

7u7 + C = 1

7(x2 + 3x)7 + C

Example 3

Evaluateint

3(1 + 3x)3dx

Solution


int 3(1 + 3x)3dx =

int u3 middot du

25




3+1u3+1 + C = 1

4u4 + C = 1

4(1 + 3x)4 + C

Example 4

Evaluateint

x2(2 + x3)4dx

Solution



int 13

u4 middot du


int 13


5u5 + C = 1

15u5 + C = 1

15(2 + x3)5 + C

26



Example 5

Evaluateint

2tradic

t2 + 3dt

Solution


radic t2 + 3dt =


int u

1

2 middot du


1


3

2 + C = 23

(t2 + 3)3

2 + C = 23

(radic

t2 + 3)3 + C

Example 6

Evaluateint

πcos(πx)dx

Solution


int cos(u)du



Example 7

Evaluateint

xradic 1minus4x2 dx

Solution


= xdx

int x

radic 1minus4x2

dx = int minus1

8 middot du



1


1

2 + C = minus14

u1

2 + C = minus14

(1 minus 4x2)1

2 + C = minus14


27



Example 8

Evaluateint

2xe2x2

dx

Solution


= 2xdx

int 2xe2x

2

dx =int

12

eudu


eudu = 12

eu + C = 12

e2x2

+ C

Example 9

Evaluateint

2xsin(2x2)dx

Solution


du = 2xdx

int 2xsin2x2dx =

int 12

sin(u)du


int 12

sin(u)du =minus

1

2cos(u) + C =

minus1

2cos(2x2) + C

28







2int 1u

du = ln|x| + C

Example 1

Evaluateint

13x

dx

Solution


3du = dxint 13x

dx =int

13 middot 1

u du


int 13 middot 1

u du = 1

3ln|u| + C = 1

3ln|3x| + C

Example 2

Evaluateint

15x+3

dx

Solution


du = dx

int 13x

dx =int

15 middot 1

u du


int 15 middot 1

u du = 1

5ln|u| + C = 1

5ln|5x + 3| + C

29



Example 3

Evaluateint

x2

3minusx3 dx

Solution


du = x2dx

int x

2

3minusx3 dx =int

minus13 middot 1

u du


int minus1

3 middot 1

u du = minus1

3ln|u| + C = minus1

3ln|3 minus x3| + C

Example 4

Evaluateint

sinθ

cosθ dθ

Solution


cosθ dθ =

int minus 1

u du


int minus 1


Example 5

Evaluateint

cos(t)3+sin(t)

dt

Solution

Let u = 3 + sin(t)

rArr du = cos(t)dt

int cos(t)3+sin(t)

dθ =int

1u

du


int 1u

du = ln|u| + C = ln|3 + sin(t)| + C

30





Theorem 48



1int

duradic a2minusu2

= arcsin u

a +C

2int

du

a2+u2 = 1a

arctan u

a +C

3int

du

uradic u2minusa2

= 1a

arcsec |u|a

+C



Example 1

Evaluateint

41+9x2 middotdx

Solution


3du = dx


int 41+9x2 middotdx

=int

41+(3x)2

middotdx

= 13

int 4du1+u2

= 43

int du

1+u2

= 4

3arctan(u) + C

= 43

arctan(3x) + C

Example 2

Evaluateint

dx

4+(x+1)2

Solution



31



int dx

4+(x+1)2 =int

du

22+u2

= 12

arctan(u2

) + C

= 12

arctan(x+12

) + C

Example 3

Evaluateint

7radic 4minusx2

middotdx

Solution



int 7

radic 4minusx2

= int 7du

radic 22

minusu2


) + C

= 7arcsin(x2

) + C

Example 4

Evaluateint

dx


Solution



int dx


=int

du

uradic u2minus9

= 13

arcsec(u3

) + C

= 13

arcsec( |x|3

) + C

32





Example 8

Evaluateint


Solution





rArr minus12

du = (x minus 3)dx

Now substitute


=int


8radic 9+(xminus3)2

= minus12

int dvradic


9+(u)2

= 12

int vminus 1


= minus12

(2v1

2 ) + 8arcsin(u3

) + C



) + C


34






int b

a f (x)dx is


f (x)dx = bminusa


Example 1


int 20

2x2dx n = 4

Solution

x = 2minus04

= 24

= 12


= i2


2(4)[f (0) + 2f (1

2) + 2f (1) + 2f (3

2) + f (2)]

= 28

[2(0)2 + 2(2(12

)2) + 2(2(1)2) + 2(2( 32

)2) + 2(22)]

= 1

4

[0 + 2(1

2

) + 2(2) + 2( 18

4

) + 8]= 1

4[0 + 1 + 4 + 9 + 8]

= 14

[22]= 55



a f (x)dx is


int b


Example 2


int 2

0 2x2dx n = 4

Solution

35



x = 2minus04

= 24

= 12


= i2

int 20

2x2dx = 2minus03(4)

[f (0) + 4f (12

) + 2f (1) + 4f (32

) + f (2)]

= 212

[2(0)2 + 4(2(12

)2) + 2(2(1)2) + 4(2(32

)2) + 2(22)]= 1

6[0 + 4(1

2) + 2(2) + 4( 18

4 ) + 8]

= 16

[0 + 2 + 4 + 18 + 8]

= 16

[32]= 53333


int 2

0 2x2dx = 22+1

x2+120

= 23

x320

= 23

(2)3 minus 23

(0)3


= 163

Example 3


int π0 sin(x)dx

Solution

x = πminus04

= π4


= iπ4int π


[f (0) + 2f (π4

) + 2f (2π4

) + 2f (3π4

) + f (π)]

= π8

[sin(0) + 2sin(π4

) + 2sin(π2

) + 2sin(3π4

) + sin(π)]

= π8 [0 + 2(radic 22 ) + 2(1) + 2(radic 22 ) + 0]

= π8

[2radic

2 + 2]asymp 2003


int π0 sin(x)dx

Solution

x = πminus04

= π4

36





= 112

[111 + 4(0947) + 2(0816) + 4(0771) + 0625] asymp 086



1(x+2)2

dx =int 2

11u2

du =int 2


x+2

21

= minus 12+2

minus (minus 11+2

)

= minus14

+ 13

= 112



Example 4


exdx

Solution

int 20



Example 5


x+radic x

2 dx

= [1

2 middot x2

2 + 1

2 middot 1

2 middot x

3

2

]10

= [x2

4 + 1

3 middot x

3

2 ]10

= [12

4 + 1

3 middot 1

3

2 ] minus [02

4 + 1

3 middot 0

3

2 ]14 minus 1

3 minus 0 = 7

12

Solution

int 10

x+radic x

2 dx =

int 10

x2

+radic x

2 dx =


Example 6


Solution

Note

f (x) = 2

xminus 3

xge

3

2



int 32


int 23

2

(2x minus 3)dx

=int 3

2


3

2

(2x minus 3)dx

= (minusx2 + 3x)|3

2

0 + (x2 minus 3x)|232

= [((32

)2 + 3 middot 32



)]= 10

4

21








1bminusa

int b

a f (x)dx


Example 7


Solution



= 14


= 1

4 middot (1

3x3

+ x)2minus2

= 14 middot [1

3(2)3 + 2] minus 1

3 middot [1

3(minus2)3 minus 2]

= 14 middot 14

3 minus 1

3 middot minus14

3

= 76

+ 76

= 146

= 73

Example 8


2

]

Solution

AV = 1π

2minus0int π

2

0 cos(x)dx

= 2π

int π2

0 cos(x)dx

= 2π middot 1048616 sin(x))|

π

2

0


2) minus sin(0)]


= 2π middot 1

= 2

π

22



Example 9


y = x2 + 2 y = 0 x = 0 x = 2

Solution


=1048616 x3

3 + 2x)

2

0

= [23

3 + 2(2)] minus [03

3 + 2(0)]= [8

3 + 4] minus [0 + 0]

= 203 minus 0

= 203

23



Example 10


y = ex y = 0 x = 0 x = 2

Solution

Area=int 20 (ex)dx

= [ex]|20

= e2 minus e0

= e2 minus 1

Example 11


0 (1 + cos(x))dx

Solution


1048616 x + sin(x))|

π

0


24







Example 1


Solution

f (x) = (x2 + 4x)6




Example 2

Evaluateint

(2x + 3)(x2 + 3x)6dx

Solution


int (2x + 3)(x2 + 3x)6dx =

int u6 middot du


6+1u6+1 + C = 1

7u7 + C = 1

7(x2 + 3x)7 + C

Example 3

Evaluateint

3(1 + 3x)3dx

Solution


int 3(1 + 3x)3dx =

int u3 middot du

25




3+1u3+1 + C = 1

4u4 + C = 1

4(1 + 3x)4 + C

Example 4

Evaluateint

x2(2 + x3)4dx

Solution



int 13

u4 middot du


int 13


5u5 + C = 1

15u5 + C = 1

15(2 + x3)5 + C

26



Example 5

Evaluateint

2tradic

t2 + 3dt

Solution


radic t2 + 3dt =


int u

1

2 middot du


1


3

2 + C = 23

(t2 + 3)3

2 + C = 23

(radic

t2 + 3)3 + C

Example 6

Evaluateint

πcos(πx)dx

Solution


int cos(u)du



Example 7

Evaluateint

xradic 1minus4x2 dx

Solution


= xdx

int x

radic 1minus4x2

dx = int minus1

8 middot du



1


1

2 + C = minus14

u1

2 + C = minus14

(1 minus 4x2)1

2 + C = minus14


27



Example 8

Evaluateint

2xe2x2

dx

Solution


= 2xdx

int 2xe2x

2

dx =int

12

eudu


eudu = 12

eu + C = 12

e2x2

+ C

Example 9

Evaluateint

2xsin(2x2)dx

Solution


du = 2xdx

int 2xsin2x2dx =

int 12

sin(u)du


int 12

sin(u)du =minus

1

2cos(u) + C =

minus1

2cos(2x2) + C

28







2int 1u

du = ln|x| + C

Example 1

Evaluateint

13x

dx

Solution


3du = dxint 13x

dx =int

13 middot 1

u du


int 13 middot 1

u du = 1

3ln|u| + C = 1

3ln|3x| + C

Example 2

Evaluateint

15x+3

dx

Solution


du = dx

int 13x

dx =int

15 middot 1

u du


int 15 middot 1

u du = 1

5ln|u| + C = 1

5ln|5x + 3| + C

29



Example 3

Evaluateint

x2

3minusx3 dx

Solution


du = x2dx

int x

2

3minusx3 dx =int

minus13 middot 1

u du


int minus1

3 middot 1

u du = minus1

3ln|u| + C = minus1

3ln|3 minus x3| + C

Example 4

Evaluateint

sinθ

cosθ dθ

Solution


cosθ dθ =

int minus 1

u du


int minus 1


Example 5

Evaluateint

cos(t)3+sin(t)

dt

Solution

Let u = 3 + sin(t)

rArr du = cos(t)dt

int cos(t)3+sin(t)

dθ =int

1u

du


int 1u

du = ln|u| + C = ln|3 + sin(t)| + C

30





Theorem 48



1int

duradic a2minusu2

= arcsin u

a +C

2int

du

a2+u2 = 1a

arctan u

a +C

3int

du

uradic u2minusa2

= 1a

arcsec |u|a

+C



Example 1

Evaluateint

41+9x2 middotdx

Solution


3du = dx


int 41+9x2 middotdx

=int

41+(3x)2

middotdx

= 13

int 4du1+u2

= 43

int du

1+u2

= 4

3arctan(u) + C

= 43

arctan(3x) + C

Example 2

Evaluateint

dx

4+(x+1)2

Solution



31



int dx

4+(x+1)2 =int

du

22+u2

= 12

arctan(u2

) + C

= 12

arctan(x+12

) + C

Example 3

Evaluateint

7radic 4minusx2

middotdx

Solution



int 7

radic 4minusx2

= int 7du

radic 22

minusu2


) + C

= 7arcsin(x2

) + C

Example 4

Evaluateint

dx


Solution



int dx


=int

du

uradic u2minus9

= 13

arcsec(u3

) + C

= 13

arcsec( |x|3

) + C

32





Example 8

Evaluateint


Solution





rArr minus12

du = (x minus 3)dx

Now substitute


=int


8radic 9+(xminus3)2

= minus12

int dvradic


9+(u)2

= 12

int vminus 1


= minus12

(2v1

2 ) + 8arcsin(u3

) + C



) + C


34






int b

a f (x)dx is


f (x)dx = bminusa


Example 1


int 20

2x2dx n = 4

Solution

x = 2minus04

= 24

= 12


= i2


2(4)[f (0) + 2f (1

2) + 2f (1) + 2f (3

2) + f (2)]

= 28

[2(0)2 + 2(2(12

)2) + 2(2(1)2) + 2(2( 32

)2) + 2(22)]

= 1

4

[0 + 2(1

2

) + 2(2) + 2( 18

4

) + 8]= 1

4[0 + 1 + 4 + 9 + 8]

= 14

[22]= 55



a f (x)dx is


int b


Example 2


int 2

0 2x2dx n = 4

Solution

35



x = 2minus04

= 24

= 12


= i2

int 20

2x2dx = 2minus03(4)

[f (0) + 4f (12

) + 2f (1) + 4f (32

) + f (2)]

= 212

[2(0)2 + 4(2(12

)2) + 2(2(1)2) + 4(2(32

)2) + 2(22)]= 1

6[0 + 4(1

2) + 2(2) + 4( 18

4 ) + 8]

= 16

[0 + 2 + 4 + 18 + 8]

= 16

[32]= 53333


int 2

0 2x2dx = 22+1

x2+120

= 23

x320

= 23

(2)3 minus 23

(0)3


= 163

Example 3


int π0 sin(x)dx

Solution

x = πminus04

= π4


= iπ4int π


[f (0) + 2f (π4

) + 2f (2π4

) + 2f (3π4

) + f (π)]

= π8

[sin(0) + 2sin(π4

) + 2sin(π2

) + 2sin(3π4

) + sin(π)]

= π8 [0 + 2(radic 22 ) + 2(1) + 2(radic 22 ) + 0]

= π8

[2radic

2 + 2]asymp 2003


int π0 sin(x)dx

Solution

x = πminus04

= π4

36





= 112

[111 + 4(0947) + 2(0816) + 4(0771) + 0625] asymp 086



1(x+2)2

dx =int 2

11u2

du =int 2


x+2

21

= minus 12+2

minus (minus 11+2

)

= minus14

+ 13

= 112








1bminusa

int b

a f (x)dx


Example 7


Solution



= 14


= 1

4 middot (1

3x3

+ x)2minus2

= 14 middot [1

3(2)3 + 2] minus 1

3 middot [1

3(minus2)3 minus 2]

= 14 middot 14

3 minus 1

3 middot minus14

3

= 76

+ 76

= 146

= 73

Example 8


2

]

Solution

AV = 1π

2minus0int π

2

0 cos(x)dx

= 2π

int π2

0 cos(x)dx

= 2π middot 1048616 sin(x))|

π

2

0


2) minus sin(0)]


= 2π middot 1

= 2

π

22



Example 9


y = x2 + 2 y = 0 x = 0 x = 2

Solution


=1048616 x3

3 + 2x)

2

0

= [23

3 + 2(2)] minus [03

3 + 2(0)]= [8

3 + 4] minus [0 + 0]

= 203 minus 0

= 203

23



Example 10


y = ex y = 0 x = 0 x = 2

Solution

Area=int 20 (ex)dx

= [ex]|20

= e2 minus e0

= e2 minus 1

Example 11


0 (1 + cos(x))dx

Solution


1048616 x + sin(x))|

π

0


24







Example 1


Solution

f (x) = (x2 + 4x)6




Example 2

Evaluateint

(2x + 3)(x2 + 3x)6dx

Solution


int (2x + 3)(x2 + 3x)6dx =

int u6 middot du


6+1u6+1 + C = 1

7u7 + C = 1

7(x2 + 3x)7 + C

Example 3

Evaluateint

3(1 + 3x)3dx

Solution


int 3(1 + 3x)3dx =

int u3 middot du

25




3+1u3+1 + C = 1

4u4 + C = 1

4(1 + 3x)4 + C

Example 4

Evaluateint

x2(2 + x3)4dx

Solution



int 13

u4 middot du


int 13


5u5 + C = 1

15u5 + C = 1

15(2 + x3)5 + C

26



Example 5

Evaluateint

2tradic

t2 + 3dt

Solution


radic t2 + 3dt =


int u

1

2 middot du


1


3

2 + C = 23

(t2 + 3)3

2 + C = 23

(radic

t2 + 3)3 + C

Example 6

Evaluateint

πcos(πx)dx

Solution


int cos(u)du



Example 7

Evaluateint

xradic 1minus4x2 dx

Solution


= xdx

int x

radic 1minus4x2

dx = int minus1

8 middot du



1


1

2 + C = minus14

u1

2 + C = minus14

(1 minus 4x2)1

2 + C = minus14


27



Example 8

Evaluateint

2xe2x2

dx

Solution


= 2xdx

int 2xe2x

2

dx =int

12

eudu


eudu = 12

eu + C = 12

e2x2

+ C

Example 9

Evaluateint

2xsin(2x2)dx

Solution


du = 2xdx

int 2xsin2x2dx =

int 12

sin(u)du


int 12

sin(u)du =minus

1

2cos(u) + C =

minus1

2cos(2x2) + C

28







2int 1u

du = ln|x| + C

Example 1

Evaluateint

13x

dx

Solution


3du = dxint 13x

dx =int

13 middot 1

u du


int 13 middot 1

u du = 1

3ln|u| + C = 1

3ln|3x| + C

Example 2

Evaluateint

15x+3

dx

Solution


du = dx

int 13x

dx =int

15 middot 1

u du


int 15 middot 1

u du = 1

5ln|u| + C = 1

5ln|5x + 3| + C

29



Example 3

Evaluateint

x2

3minusx3 dx

Solution


du = x2dx

int x

2

3minusx3 dx =int

minus13 middot 1

u du


int minus1

3 middot 1

u du = minus1

3ln|u| + C = minus1

3ln|3 minus x3| + C

Example 4

Evaluateint

sinθ

cosθ dθ

Solution


cosθ dθ =

int minus 1

u du


int minus 1


Example 5

Evaluateint

cos(t)3+sin(t)

dt

Solution

Let u = 3 + sin(t)

rArr du = cos(t)dt

int cos(t)3+sin(t)

dθ =int

1u

du


int 1u

du = ln|u| + C = ln|3 + sin(t)| + C

30





Theorem 48



1int

duradic a2minusu2

= arcsin u

a +C

2int

du

a2+u2 = 1a

arctan u

a +C

3int

du

uradic u2minusa2

= 1a

arcsec |u|a

+C



Example 1

Evaluateint

41+9x2 middotdx

Solution


3du = dx


int 41+9x2 middotdx

=int

41+(3x)2

middotdx

= 13

int 4du1+u2

= 43

int du

1+u2

= 4

3arctan(u) + C

= 43

arctan(3x) + C

Example 2

Evaluateint

dx

4+(x+1)2

Solution



31



int dx

4+(x+1)2 =int

du

22+u2

= 12

arctan(u2

) + C

= 12

arctan(x+12

) + C

Example 3

Evaluateint

7radic 4minusx2

middotdx

Solution



int 7

radic 4minusx2

= int 7du

radic 22

minusu2


) + C

= 7arcsin(x2

) + C

Example 4

Evaluateint

dx


Solution



int dx


=int

du

uradic u2minus9

= 13

arcsec(u3

) + C

= 13

arcsec( |x|3

) + C

32





Example 8

Evaluateint


Solution





rArr minus12

du = (x minus 3)dx

Now substitute


=int


8radic 9+(xminus3)2

= minus12

int dvradic


9+(u)2

= 12

int vminus 1


= minus12

(2v1

2 ) + 8arcsin(u3

) + C



) + C


34






int b

a f (x)dx is


f (x)dx = bminusa


Example 1


int 20

2x2dx n = 4

Solution

x = 2minus04

= 24

= 12


= i2


2(4)[f (0) + 2f (1

2) + 2f (1) + 2f (3

2) + f (2)]

= 28

[2(0)2 + 2(2(12

)2) + 2(2(1)2) + 2(2( 32

)2) + 2(22)]

= 1

4

[0 + 2(1

2

) + 2(2) + 2( 18

4

) + 8]= 1

4[0 + 1 + 4 + 9 + 8]

= 14

[22]= 55



a f (x)dx is


int b


Example 2


int 2

0 2x2dx n = 4

Solution

35



x = 2minus04

= 24

= 12


= i2

int 20

2x2dx = 2minus03(4)

[f (0) + 4f (12

) + 2f (1) + 4f (32

) + f (2)]

= 212

[2(0)2 + 4(2(12

)2) + 2(2(1)2) + 4(2(32

)2) + 2(22)]= 1

6[0 + 4(1

2) + 2(2) + 4( 18

4 ) + 8]

= 16

[0 + 2 + 4 + 18 + 8]

= 16

[32]= 53333


int 2

0 2x2dx = 22+1

x2+120

= 23

x320

= 23

(2)3 minus 23

(0)3


= 163

Example 3


int π0 sin(x)dx

Solution

x = πminus04

= π4


= iπ4int π


[f (0) + 2f (π4

) + 2f (2π4

) + 2f (3π4

) + f (π)]

= π8

[sin(0) + 2sin(π4

) + 2sin(π2

) + 2sin(3π4

) + sin(π)]

= π8 [0 + 2(radic 22 ) + 2(1) + 2(radic 22 ) + 0]

= π8

[2radic

2 + 2]asymp 2003


int π0 sin(x)dx

Solution

x = πminus04

= π4

36





= 112

[111 + 4(0947) + 2(0816) + 4(0771) + 0625] asymp 086



1(x+2)2

dx =int 2

11u2

du =int 2


x+2

21

= minus 12+2

minus (minus 11+2

)

= minus14

+ 13

= 112



Example 9


y = x2 + 2 y = 0 x = 0 x = 2

Solution


=1048616 x3

3 + 2x)

2

0

= [23

3 + 2(2)] minus [03

3 + 2(0)]= [8

3 + 4] minus [0 + 0]

= 203 minus 0

= 203

23



Example 10


y = ex y = 0 x = 0 x = 2

Solution

Area=int 20 (ex)dx

= [ex]|20

= e2 minus e0

= e2 minus 1

Example 11


0 (1 + cos(x))dx

Solution


1048616 x + sin(x))|

π

0


24







Example 1


Solution

f (x) = (x2 + 4x)6




Example 2

Evaluateint

(2x + 3)(x2 + 3x)6dx

Solution


int (2x + 3)(x2 + 3x)6dx =

int u6 middot du


6+1u6+1 + C = 1

7u7 + C = 1

7(x2 + 3x)7 + C

Example 3

Evaluateint

3(1 + 3x)3dx

Solution


int 3(1 + 3x)3dx =

int u3 middot du

25




3+1u3+1 + C = 1

4u4 + C = 1

4(1 + 3x)4 + C

Example 4

Evaluateint

x2(2 + x3)4dx

Solution



int 13

u4 middot du


int 13


5u5 + C = 1

15u5 + C = 1

15(2 + x3)5 + C

26



Example 5

Evaluateint

2tradic

t2 + 3dt

Solution


radic t2 + 3dt =


int u

1

2 middot du


1


3

2 + C = 23

(t2 + 3)3

2 + C = 23

(radic

t2 + 3)3 + C

Example 6

Evaluateint

πcos(πx)dx

Solution


int cos(u)du



Example 7

Evaluateint

xradic 1minus4x2 dx

Solution


= xdx

int x

radic 1minus4x2

dx = int minus1

8 middot du



1


1

2 + C = minus14

u1

2 + C = minus14

(1 minus 4x2)1

2 + C = minus14


27



Example 8

Evaluateint

2xe2x2

dx

Solution


= 2xdx

int 2xe2x

2

dx =int

12

eudu


eudu = 12

eu + C = 12

e2x2

+ C

Example 9

Evaluateint

2xsin(2x2)dx

Solution


du = 2xdx

int 2xsin2x2dx =

int 12

sin(u)du


int 12

sin(u)du =minus

1

2cos(u) + C =

minus1

2cos(2x2) + C

28







2int 1u

du = ln|x| + C

Example 1

Evaluateint

13x

dx

Solution


3du = dxint 13x

dx =int

13 middot 1

u du


int 13 middot 1

u du = 1

3ln|u| + C = 1

3ln|3x| + C

Example 2

Evaluateint

15x+3

dx

Solution


du = dx

int 13x

dx =int

15 middot 1

u du


int 15 middot 1

u du = 1

5ln|u| + C = 1

5ln|5x + 3| + C

29



Example 3

Evaluateint

x2

3minusx3 dx

Solution


du = x2dx

int x

2

3minusx3 dx =int

minus13 middot 1

u du


int minus1

3 middot 1

u du = minus1

3ln|u| + C = minus1

3ln|3 minus x3| + C

Example 4

Evaluateint

sinθ

cosθ dθ

Solution


cosθ dθ =

int minus 1

u du


int minus 1


Example 5

Evaluateint

cos(t)3+sin(t)

dt

Solution

Let u = 3 + sin(t)

rArr du = cos(t)dt

int cos(t)3+sin(t)

dθ =int

1u

du


int 1u

du = ln|u| + C = ln|3 + sin(t)| + C

30





Theorem 48



1int

duradic a2minusu2

= arcsin u

a +C

2int

du

a2+u2 = 1a

arctan u

a +C

3int

du

uradic u2minusa2

= 1a

arcsec |u|a

+C



Example 1

Evaluateint

41+9x2 middotdx

Solution


3du = dx


int 41+9x2 middotdx

=int

41+(3x)2

middotdx

= 13

int 4du1+u2

= 43

int du

1+u2

= 4

3arctan(u) + C

= 43

arctan(3x) + C

Example 2

Evaluateint

dx

4+(x+1)2

Solution



31



int dx

4+(x+1)2 =int

du

22+u2

= 12

arctan(u2

) + C

= 12

arctan(x+12

) + C

Example 3

Evaluateint

7radic 4minusx2

middotdx

Solution



int 7

radic 4minusx2

= int 7du

radic 22

minusu2


) + C

= 7arcsin(x2

) + C

Example 4

Evaluateint

dx


Solution



int dx


=int

du

uradic u2minus9

= 13

arcsec(u3

) + C

= 13

arcsec( |x|3

) + C

32





Example 8

Evaluateint


Solution





rArr minus12

du = (x minus 3)dx

Now substitute


=int


8radic 9+(xminus3)2

= minus12

int dvradic


9+(u)2

= 12

int vminus 1


= minus12

(2v1

2 ) + 8arcsin(u3

) + C



) + C


34






int b

a f (x)dx is


f (x)dx = bminusa


Example 1


int 20

2x2dx n = 4

Solution

x = 2minus04

= 24

= 12


= i2


2(4)[f (0) + 2f (1

2) + 2f (1) + 2f (3

2) + f (2)]

= 28

[2(0)2 + 2(2(12

)2) + 2(2(1)2) + 2(2( 32

)2) + 2(22)]

= 1

4

[0 + 2(1

2

) + 2(2) + 2( 18

4

) + 8]= 1

4[0 + 1 + 4 + 9 + 8]

= 14

[22]= 55



a f (x)dx is


int b


Example 2


int 2

0 2x2dx n = 4

Solution

35



x = 2minus04

= 24

= 12


= i2

int 20

2x2dx = 2minus03(4)

[f (0) + 4f (12

) + 2f (1) + 4f (32

) + f (2)]

= 212

[2(0)2 + 4(2(12

)2) + 2(2(1)2) + 4(2(32

)2) + 2(22)]= 1

6[0 + 4(1

2) + 2(2) + 4( 18

4 ) + 8]

= 16

[0 + 2 + 4 + 18 + 8]

= 16

[32]= 53333


int 2

0 2x2dx = 22+1

x2+120

= 23

x320

= 23

(2)3 minus 23

(0)3


= 163

Example 3


int π0 sin(x)dx

Solution

x = πminus04

= π4


= iπ4int π


[f (0) + 2f (π4

) + 2f (2π4

) + 2f (3π4

) + f (π)]

= π8

[sin(0) + 2sin(π4

) + 2sin(π2

) + 2sin(3π4

) + sin(π)]

= π8 [0 + 2(radic 22 ) + 2(1) + 2(radic 22 ) + 0]

= π8

[2radic

2 + 2]asymp 2003


int π0 sin(x)dx

Solution

x = πminus04

= π4

36





= 112

[111 + 4(0947) + 2(0816) + 4(0771) + 0625] asymp 086



1(x+2)2

dx =int 2

11u2

du =int 2


x+2

21

= minus 12+2

minus (minus 11+2

)

= minus14

+ 13

= 112



Example 10


y = ex y = 0 x = 0 x = 2

Solution

Area=int 20 (ex)dx

= [ex]|20

= e2 minus e0

= e2 minus 1

Example 11


0 (1 + cos(x))dx

Solution


1048616 x + sin(x))|

π

0


24







Example 1


Solution

f (x) = (x2 + 4x)6




Example 2

Evaluateint

(2x + 3)(x2 + 3x)6dx

Solution


int (2x + 3)(x2 + 3x)6dx =

int u6 middot du


6+1u6+1 + C = 1

7u7 + C = 1

7(x2 + 3x)7 + C

Example 3

Evaluateint

3(1 + 3x)3dx

Solution


int 3(1 + 3x)3dx =

int u3 middot du

25




3+1u3+1 + C = 1

4u4 + C = 1

4(1 + 3x)4 + C

Example 4

Evaluateint

x2(2 + x3)4dx

Solution



int 13

u4 middot du


int 13


5u5 + C = 1

15u5 + C = 1

15(2 + x3)5 + C

26



Example 5

Evaluateint

2tradic

t2 + 3dt

Solution


radic t2 + 3dt =


int u

1

2 middot du


1


3

2 + C = 23

(t2 + 3)3

2 + C = 23

(radic

t2 + 3)3 + C

Example 6

Evaluateint

πcos(πx)dx

Solution


int cos(u)du



Example 7

Evaluateint

xradic 1minus4x2 dx

Solution


= xdx

int x

radic 1minus4x2

dx = int minus1

8 middot du



1


1

2 + C = minus14

u1

2 + C = minus14

(1 minus 4x2)1

2 + C = minus14


27



Example 8

Evaluateint

2xe2x2

dx

Solution


= 2xdx

int 2xe2x

2

dx =int

12

eudu


eudu = 12

eu + C = 12

e2x2

+ C

Example 9

Evaluateint

2xsin(2x2)dx

Solution


du = 2xdx

int 2xsin2x2dx =

int 12

sin(u)du


int 12

sin(u)du =minus

1

2cos(u) + C =

minus1

2cos(2x2) + C

28







2int 1u

du = ln|x| + C

Example 1

Evaluateint

13x

dx

Solution


3du = dxint 13x

dx =int

13 middot 1

u du


int 13 middot 1

u du = 1

3ln|u| + C = 1

3ln|3x| + C

Example 2

Evaluateint

15x+3

dx

Solution


du = dx

int 13x

dx =int

15 middot 1

u du


int 15 middot 1

u du = 1

5ln|u| + C = 1

5ln|5x + 3| + C

29



Example 3

Evaluateint

x2

3minusx3 dx

Solution


du = x2dx

int x

2

3minusx3 dx =int

minus13 middot 1

u du


int minus1

3 middot 1

u du = minus1

3ln|u| + C = minus1

3ln|3 minus x3| + C

Example 4

Evaluateint

sinθ

cosθ dθ

Solution


cosθ dθ =

int minus 1

u du


int minus 1


Example 5

Evaluateint

cos(t)3+sin(t)

dt

Solution

Let u = 3 + sin(t)

rArr du = cos(t)dt

int cos(t)3+sin(t)

dθ =int

1u

du


int 1u

du = ln|u| + C = ln|3 + sin(t)| + C

30





Theorem 48



1int

duradic a2minusu2

= arcsin u

a +C

2int

du

a2+u2 = 1a

arctan u

a +C

3int

du

uradic u2minusa2

= 1a

arcsec |u|a

+C



Example 1

Evaluateint

41+9x2 middotdx

Solution


3du = dx


int 41+9x2 middotdx

=int

41+(3x)2

middotdx

= 13

int 4du1+u2

= 43

int du

1+u2

= 4

3arctan(u) + C

= 43

arctan(3x) + C

Example 2

Evaluateint

dx

4+(x+1)2

Solution



31



int dx

4+(x+1)2 =int

du

22+u2

= 12

arctan(u2

) + C

= 12

arctan(x+12

) + C

Example 3

Evaluateint

7radic 4minusx2

middotdx

Solution



int 7

radic 4minusx2

= int 7du

radic 22

minusu2


) + C

= 7arcsin(x2

) + C

Example 4

Evaluateint

dx


Solution



int dx


=int

du

uradic u2minus9

= 13

arcsec(u3

) + C

= 13

arcsec( |x|3

) + C

32





Example 8

Evaluateint


Solution





rArr minus12

du = (x minus 3)dx

Now substitute


=int


8radic 9+(xminus3)2

= minus12

int dvradic


9+(u)2

= 12

int vminus 1


= minus12

(2v1

2 ) + 8arcsin(u3

) + C



) + C


34






int b

a f (x)dx is


f (x)dx = bminusa


Example 1


int 20

2x2dx n = 4

Solution

x = 2minus04

= 24

= 12


= i2


2(4)[f (0) + 2f (1

2) + 2f (1) + 2f (3

2) + f (2)]

= 28

[2(0)2 + 2(2(12

)2) + 2(2(1)2) + 2(2( 32

)2) + 2(22)]

= 1

4

[0 + 2(1

2

) + 2(2) + 2( 18

4

) + 8]= 1

4[0 + 1 + 4 + 9 + 8]

= 14

[22]= 55



a f (x)dx is


int b


Example 2


int 2

0 2x2dx n = 4

Solution

35



x = 2minus04

= 24

= 12


= i2

int 20

2x2dx = 2minus03(4)

[f (0) + 4f (12

) + 2f (1) + 4f (32

) + f (2)]

= 212

[2(0)2 + 4(2(12

)2) + 2(2(1)2) + 4(2(32

)2) + 2(22)]= 1

6[0 + 4(1

2) + 2(2) + 4( 18

4 ) + 8]

= 16

[0 + 2 + 4 + 18 + 8]

= 16

[32]= 53333


int 2

0 2x2dx = 22+1

x2+120

= 23

x320

= 23

(2)3 minus 23

(0)3


= 163

Example 3


int π0 sin(x)dx

Solution

x = πminus04

= π4


= iπ4int π


[f (0) + 2f (π4

) + 2f (2π4

) + 2f (3π4

) + f (π)]

= π8

[sin(0) + 2sin(π4

) + 2sin(π2

) + 2sin(3π4

) + sin(π)]

= π8 [0 + 2(radic 22 ) + 2(1) + 2(radic 22 ) + 0]

= π8

[2radic

2 + 2]asymp 2003


int π0 sin(x)dx

Solution

x = πminus04

= π4

36





= 112

[111 + 4(0947) + 2(0816) + 4(0771) + 0625] asymp 086



1(x+2)2

dx =int 2

11u2

du =int 2


x+2

21

= minus 12+2

minus (minus 11+2

)

= minus14

+ 13

= 112







Example 1


Solution

f (x) = (x2 + 4x)6




Example 2

Evaluateint

(2x + 3)(x2 + 3x)6dx

Solution


int (2x + 3)(x2 + 3x)6dx =

int u6 middot du


6+1u6+1 + C = 1

7u7 + C = 1

7(x2 + 3x)7 + C

Example 3

Evaluateint

3(1 + 3x)3dx

Solution


int 3(1 + 3x)3dx =

int u3 middot du

25




3+1u3+1 + C = 1

4u4 + C = 1

4(1 + 3x)4 + C

Example 4

Evaluateint

x2(2 + x3)4dx

Solution



int 13

u4 middot du


int 13


5u5 + C = 1

15u5 + C = 1

15(2 + x3)5 + C

26



Example 5

Evaluateint

2tradic

t2 + 3dt

Solution


radic t2 + 3dt =


int u

1

2 middot du


1


3

2 + C = 23

(t2 + 3)3

2 + C = 23

(radic

t2 + 3)3 + C

Example 6

Evaluateint

πcos(πx)dx

Solution


int cos(u)du



Example 7

Evaluateint

xradic 1minus4x2 dx

Solution


= xdx

int x

radic 1minus4x2

dx = int minus1

8 middot du



1


1

2 + C = minus14

u1

2 + C = minus14

(1 minus 4x2)1

2 + C = minus14


27



Example 8

Evaluateint

2xe2x2

dx

Solution


= 2xdx

int 2xe2x

2

dx =int

12

eudu


eudu = 12

eu + C = 12

e2x2

+ C

Example 9

Evaluateint

2xsin(2x2)dx

Solution


du = 2xdx

int 2xsin2x2dx =

int 12

sin(u)du


int 12

sin(u)du =minus

1

2cos(u) + C =

minus1

2cos(2x2) + C

28







2int 1u

du = ln|x| + C

Example 1

Evaluateint

13x

dx

Solution


3du = dxint 13x

dx =int

13 middot 1

u du


int 13 middot 1

u du = 1

3ln|u| + C = 1

3ln|3x| + C

Example 2

Evaluateint

15x+3

dx

Solution


du = dx

int 13x

dx =int

15 middot 1

u du


int 15 middot 1

u du = 1

5ln|u| + C = 1

5ln|5x + 3| + C

29



Example 3

Evaluateint

x2

3minusx3 dx

Solution


du = x2dx

int x

2

3minusx3 dx =int

minus13 middot 1

u du


int minus1

3 middot 1

u du = minus1

3ln|u| + C = minus1

3ln|3 minus x3| + C

Example 4

Evaluateint

sinθ

cosθ dθ

Solution


cosθ dθ =

int minus 1

u du


int minus 1


Example 5

Evaluateint

cos(t)3+sin(t)

dt

Solution

Let u = 3 + sin(t)

rArr du = cos(t)dt

int cos(t)3+sin(t)

dθ =int

1u

du


int 1u

du = ln|u| + C = ln|3 + sin(t)| + C

30





Theorem 48



1int

duradic a2minusu2

= arcsin u

a +C

2int

du

a2+u2 = 1a

arctan u

a +C

3int

du

uradic u2minusa2

= 1a

arcsec |u|a

+C



Example 1

Evaluateint

41+9x2 middotdx

Solution


3du = dx


int 41+9x2 middotdx

=int

41+(3x)2

middotdx

= 13

int 4du1+u2

= 43

int du

1+u2

= 4

3arctan(u) + C

= 43

arctan(3x) + C

Example 2

Evaluateint

dx

4+(x+1)2

Solution



31



int dx

4+(x+1)2 =int

du

22+u2

= 12

arctan(u2

) + C

= 12

arctan(x+12

) + C

Example 3

Evaluateint

7radic 4minusx2

middotdx

Solution



int 7

radic 4minusx2

= int 7du

radic 22

minusu2


) + C

= 7arcsin(x2

) + C

Example 4

Evaluateint

dx


Solution



int dx


=int

du

uradic u2minus9

= 13

arcsec(u3

) + C

= 13

arcsec( |x|3

) + C

32





Example 8

Evaluateint


Solution





rArr minus12

du = (x minus 3)dx

Now substitute


=int


8radic 9+(xminus3)2

= minus12

int dvradic


9+(u)2

= 12

int vminus 1


= minus12

(2v1

2 ) + 8arcsin(u3

) + C



) + C


34






int b

a f (x)dx is


f (x)dx = bminusa


Example 1


int 20

2x2dx n = 4

Solution

x = 2minus04

= 24

= 12


= i2


2(4)[f (0) + 2f (1

2) + 2f (1) + 2f (3

2) + f (2)]

= 28

[2(0)2 + 2(2(12

)2) + 2(2(1)2) + 2(2( 32

)2) + 2(22)]

= 1

4

[0 + 2(1

2

) + 2(2) + 2( 18

4

) + 8]= 1

4[0 + 1 + 4 + 9 + 8]

= 14

[22]= 55



a f (x)dx is


int b


Example 2


int 2

0 2x2dx n = 4

Solution

35



x = 2minus04

= 24

= 12


= i2

int 20

2x2dx = 2minus03(4)

[f (0) + 4f (12

) + 2f (1) + 4f (32

) + f (2)]

= 212

[2(0)2 + 4(2(12

)2) + 2(2(1)2) + 4(2(32

)2) + 2(22)]= 1

6[0 + 4(1

2) + 2(2) + 4( 18

4 ) + 8]

= 16

[0 + 2 + 4 + 18 + 8]

= 16

[32]= 53333


int 2

0 2x2dx = 22+1

x2+120

= 23

x320

= 23

(2)3 minus 23

(0)3


= 163

Example 3


int π0 sin(x)dx

Solution

x = πminus04

= π4


= iπ4int π


[f (0) + 2f (π4

) + 2f (2π4

) + 2f (3π4

) + f (π)]

= π8

[sin(0) + 2sin(π4

) + 2sin(π2

) + 2sin(3π4

) + sin(π)]

= π8 [0 + 2(radic 22 ) + 2(1) + 2(radic 22 ) + 0]

= π8

[2radic

2 + 2]asymp 2003


int π0 sin(x)dx

Solution

x = πminus04

= π4

36





= 112

[111 + 4(0947) + 2(0816) + 4(0771) + 0625] asymp 086



1(x+2)2

dx =int 2

11u2

du =int 2


x+2

21

= minus 12+2

minus (minus 11+2

)

= minus14

+ 13

= 112




3+1u3+1 + C = 1

4u4 + C = 1

4(1 + 3x)4 + C

Example 4

Evaluateint

x2(2 + x3)4dx

Solution



int 13

u4 middot du


int 13


5u5 + C = 1

15u5 + C = 1

15(2 + x3)5 + C

26



Example 5

Evaluateint

2tradic

t2 + 3dt

Solution


radic t2 + 3dt =


int u

1

2 middot du


1


3

2 + C = 23

(t2 + 3)3

2 + C = 23

(radic

t2 + 3)3 + C

Example 6

Evaluateint

πcos(πx)dx

Solution


int cos(u)du



Example 7

Evaluateint

xradic 1minus4x2 dx

Solution


= xdx

int x

radic 1minus4x2

dx = int minus1

8 middot du



1


1

2 + C = minus14

u1

2 + C = minus14

(1 minus 4x2)1

2 + C = minus14


27



Example 8

Evaluateint

2xe2x2

dx

Solution


= 2xdx

int 2xe2x

2

dx =int

12

eudu


eudu = 12

eu + C = 12

e2x2

+ C

Example 9

Evaluateint

2xsin(2x2)dx

Solution


du = 2xdx

int 2xsin2x2dx =

int 12

sin(u)du


int 12

sin(u)du =minus

1

2cos(u) + C =

minus1

2cos(2x2) + C

28







2int 1u

du = ln|x| + C

Example 1

Evaluateint

13x

dx

Solution


3du = dxint 13x

dx =int

13 middot 1

u du


int 13 middot 1

u du = 1

3ln|u| + C = 1

3ln|3x| + C

Example 2

Evaluateint

15x+3

dx

Solution


du = dx

int 13x

dx =int

15 middot 1

u du


int 15 middot 1

u du = 1

5ln|u| + C = 1

5ln|5x + 3| + C

29



Example 3

Evaluateint

x2

3minusx3 dx

Solution


du = x2dx

int x

2

3minusx3 dx =int

minus13 middot 1

u du


int minus1

3 middot 1

u du = minus1

3ln|u| + C = minus1

3ln|3 minus x3| + C

Example 4

Evaluateint

sinθ

cosθ dθ

Solution


cosθ dθ =

int minus 1

u du


int minus 1


Example 5

Evaluateint

cos(t)3+sin(t)

dt

Solution

Let u = 3 + sin(t)

rArr du = cos(t)dt

int cos(t)3+sin(t)

dθ =int

1u

du


int 1u

du = ln|u| + C = ln|3 + sin(t)| + C

30





Theorem 48



1int

duradic a2minusu2

= arcsin u

a +C

2int

du

a2+u2 = 1a

arctan u

a +C

3int

du

uradic u2minusa2

= 1a

arcsec |u|a

+C



Example 1

Evaluateint

41+9x2 middotdx

Solution


3du = dx


int 41+9x2 middotdx

=int

41+(3x)2

middotdx

= 13

int 4du1+u2

= 43

int du

1+u2

= 4

3arctan(u) + C

= 43

arctan(3x) + C

Example 2

Evaluateint

dx

4+(x+1)2

Solution



31



int dx

4+(x+1)2 =int

du

22+u2

= 12

arctan(u2

) + C

= 12

arctan(x+12

) + C

Example 3

Evaluateint

7radic 4minusx2

middotdx

Solution



int 7

radic 4minusx2

= int 7du

radic 22

minusu2


) + C

= 7arcsin(x2

) + C

Example 4

Evaluateint

dx


Solution



int dx


=int

du

uradic u2minus9

= 13

arcsec(u3

) + C

= 13

arcsec( |x|3

) + C

32





Example 8

Evaluateint


Solution





rArr minus12

du = (x minus 3)dx

Now substitute


=int


8radic 9+(xminus3)2

= minus12

int dvradic


9+(u)2

= 12

int vminus 1


= minus12

(2v1

2 ) + 8arcsin(u3

) + C



) + C


34






int b

a f (x)dx is


f (x)dx = bminusa


Example 1


int 20

2x2dx n = 4

Solution

x = 2minus04

= 24

= 12


= i2


2(4)[f (0) + 2f (1

2) + 2f (1) + 2f (3

2) + f (2)]

= 28

[2(0)2 + 2(2(12

)2) + 2(2(1)2) + 2(2( 32

)2) + 2(22)]

= 1

4

[0 + 2(1

2

) + 2(2) + 2( 18

4

) + 8]= 1

4[0 + 1 + 4 + 9 + 8]

= 14

[22]= 55



a f (x)dx is


int b


Example 2


int 2

0 2x2dx n = 4

Solution

35



x = 2minus04

= 24

= 12


= i2

int 20

2x2dx = 2minus03(4)

[f (0) + 4f (12

) + 2f (1) + 4f (32

) + f (2)]

= 212

[2(0)2 + 4(2(12

)2) + 2(2(1)2) + 4(2(32

)2) + 2(22)]= 1

6[0 + 4(1

2) + 2(2) + 4( 18

4 ) + 8]

= 16

[0 + 2 + 4 + 18 + 8]

= 16

[32]= 53333


int 2

0 2x2dx = 22+1

x2+120

= 23

x320

= 23

(2)3 minus 23

(0)3


= 163

Example 3


int π0 sin(x)dx

Solution

x = πminus04

= π4


= iπ4int π


[f (0) + 2f (π4

) + 2f (2π4

) + 2f (3π4

) + f (π)]

= π8

[sin(0) + 2sin(π4

) + 2sin(π2

) + 2sin(3π4

) + sin(π)]

= π8 [0 + 2(radic 22 ) + 2(1) + 2(radic 22 ) + 0]

= π8

[2radic

2 + 2]asymp 2003


int π0 sin(x)dx

Solution

x = πminus04

= π4

36





= 112

[111 + 4(0947) + 2(0816) + 4(0771) + 0625] asymp 086



1(x+2)2

dx =int 2

11u2

du =int 2


x+2

21

= minus 12+2

minus (minus 11+2

)

= minus14

+ 13

= 112



Example 5

Evaluateint

2tradic

t2 + 3dt

Solution


radic t2 + 3dt =


int u

1

2 middot du


1


3

2 + C = 23

(t2 + 3)3

2 + C = 23

(radic

t2 + 3)3 + C

Example 6

Evaluateint

πcos(πx)dx

Solution


int cos(u)du



Example 7

Evaluateint

xradic 1minus4x2 dx

Solution


= xdx

int x

radic 1minus4x2

dx = int minus1

8 middot du



1


1

2 + C = minus14

u1

2 + C = minus14

(1 minus 4x2)1

2 + C = minus14


27



Example 8

Evaluateint

2xe2x2

dx

Solution


= 2xdx

int 2xe2x

2

dx =int

12

eudu


eudu = 12

eu + C = 12

e2x2

+ C

Example 9

Evaluateint

2xsin(2x2)dx

Solution


du = 2xdx

int 2xsin2x2dx =

int 12

sin(u)du


int 12

sin(u)du =minus

1

2cos(u) + C =

minus1

2cos(2x2) + C

28







2int 1u

du = ln|x| + C

Example 1

Evaluateint

13x

dx

Solution


3du = dxint 13x

dx =int

13 middot 1

u du


int 13 middot 1

u du = 1

3ln|u| + C = 1

3ln|3x| + C

Example 2

Evaluateint

15x+3

dx

Solution


du = dx

int 13x

dx =int

15 middot 1

u du


int 15 middot 1

u du = 1

5ln|u| + C = 1

5ln|5x + 3| + C

29



Example 3

Evaluateint

x2

3minusx3 dx

Solution


du = x2dx

int x

2

3minusx3 dx =int

minus13 middot 1

u du


int minus1

3 middot 1

u du = minus1

3ln|u| + C = minus1

3ln|3 minus x3| + C

Example 4

Evaluateint

sinθ

cosθ dθ

Solution


cosθ dθ =

int minus 1

u du


int minus 1


Example 5

Evaluateint

cos(t)3+sin(t)

dt

Solution

Let u = 3 + sin(t)

rArr du = cos(t)dt

int cos(t)3+sin(t)

dθ =int

1u

du


int 1u

du = ln|u| + C = ln|3 + sin(t)| + C

30





Theorem 48



1int

duradic a2minusu2

= arcsin u

a +C

2int

du

a2+u2 = 1a

arctan u

a +C

3int

du

uradic u2minusa2

= 1a

arcsec |u|a

+C



Example 1

Evaluateint

41+9x2 middotdx

Solution


3du = dx


int 41+9x2 middotdx

=int

41+(3x)2

middotdx

= 13

int 4du1+u2

= 43

int du

1+u2

= 4

3arctan(u) + C

= 43

arctan(3x) + C

Example 2

Evaluateint

dx

4+(x+1)2

Solution



31



int dx

4+(x+1)2 =int

du

22+u2

= 12

arctan(u2

) + C

= 12

arctan(x+12

) + C

Example 3

Evaluateint

7radic 4minusx2

middotdx

Solution



int 7

radic 4minusx2

= int 7du

radic 22

minusu2


) + C

= 7arcsin(x2

) + C

Example 4

Evaluateint

dx


Solution



int dx


=int

du

uradic u2minus9

= 13

arcsec(u3

) + C

= 13

arcsec( |x|3

) + C

32





Example 8

Evaluateint


Solution





rArr minus12

du = (x minus 3)dx

Now substitute


=int


8radic 9+(xminus3)2

= minus12

int dvradic


9+(u)2

= 12

int vminus 1


= minus12

(2v1

2 ) + 8arcsin(u3

) + C



) + C


34






int b

a f (x)dx is


f (x)dx = bminusa


Example 1


int 20

2x2dx n = 4

Solution

x = 2minus04

= 24

= 12


= i2


2(4)[f (0) + 2f (1

2) + 2f (1) + 2f (3

2) + f (2)]

= 28

[2(0)2 + 2(2(12

)2) + 2(2(1)2) + 2(2( 32

)2) + 2(22)]

= 1

4

[0 + 2(1

2

) + 2(2) + 2( 18

4

) + 8]= 1

4[0 + 1 + 4 + 9 + 8]

= 14

[22]= 55



a f (x)dx is


int b


Example 2


int 2

0 2x2dx n = 4

Solution

35



x = 2minus04

= 24

= 12


= i2

int 20

2x2dx = 2minus03(4)

[f (0) + 4f (12

) + 2f (1) + 4f (32

) + f (2)]

= 212

[2(0)2 + 4(2(12

)2) + 2(2(1)2) + 4(2(32

)2) + 2(22)]= 1

6[0 + 4(1

2) + 2(2) + 4( 18

4 ) + 8]

= 16

[0 + 2 + 4 + 18 + 8]

= 16

[32]= 53333


int 2

0 2x2dx = 22+1

x2+120

= 23

x320

= 23

(2)3 minus 23

(0)3


= 163

Example 3


int π0 sin(x)dx

Solution

x = πminus04

= π4


= iπ4int π


[f (0) + 2f (π4

) + 2f (2π4

) + 2f (3π4

) + f (π)]

= π8

[sin(0) + 2sin(π4

) + 2sin(π2

) + 2sin(3π4

) + sin(π)]

= π8 [0 + 2(radic 22 ) + 2(1) + 2(radic 22 ) + 0]

= π8

[2radic

2 + 2]asymp 2003


int π0 sin(x)dx

Solution

x = πminus04

= π4

36





= 112

[111 + 4(0947) + 2(0816) + 4(0771) + 0625] asymp 086



1(x+2)2

dx =int 2

11u2

du =int 2


x+2

21

= minus 12+2

minus (minus 11+2

)

= minus14

+ 13

= 112



Example 8

Evaluateint

2xe2x2

dx

Solution


= 2xdx

int 2xe2x

2

dx =int

12

eudu


eudu = 12

eu + C = 12

e2x2

+ C

Example 9

Evaluateint

2xsin(2x2)dx

Solution


du = 2xdx

int 2xsin2x2dx =

int 12

sin(u)du


int 12

sin(u)du =minus

1

2cos(u) + C =

minus1

2cos(2x2) + C

28







2int 1u

du = ln|x| + C

Example 1

Evaluateint

13x

dx

Solution


3du = dxint 13x

dx =int

13 middot 1

u du


int 13 middot 1

u du = 1

3ln|u| + C = 1

3ln|3x| + C

Example 2

Evaluateint

15x+3

dx

Solution


du = dx

int 13x

dx =int

15 middot 1

u du


int 15 middot 1

u du = 1

5ln|u| + C = 1

5ln|5x + 3| + C

29



Example 3

Evaluateint

x2

3minusx3 dx

Solution


du = x2dx

int x

2

3minusx3 dx =int

minus13 middot 1

u du


int minus1

3 middot 1

u du = minus1

3ln|u| + C = minus1

3ln|3 minus x3| + C

Example 4

Evaluateint

sinθ

cosθ dθ

Solution


cosθ dθ =

int minus 1

u du


int minus 1


Example 5

Evaluateint

cos(t)3+sin(t)

dt

Solution

Let u = 3 + sin(t)

rArr du = cos(t)dt

int cos(t)3+sin(t)

dθ =int

1u

du


int 1u

du = ln|u| + C = ln|3 + sin(t)| + C

30





Theorem 48



1int

duradic a2minusu2

= arcsin u

a +C

2int

du

a2+u2 = 1a

arctan u

a +C

3int

du

uradic u2minusa2

= 1a

arcsec |u|a

+C



Example 1

Evaluateint

41+9x2 middotdx

Solution


3du = dx


int 41+9x2 middotdx

=int

41+(3x)2

middotdx

= 13

int 4du1+u2

= 43

int du

1+u2

= 4

3arctan(u) + C

= 43

arctan(3x) + C

Example 2

Evaluateint

dx

4+(x+1)2

Solution



31



int dx

4+(x+1)2 =int

du

22+u2

= 12

arctan(u2

) + C

= 12

arctan(x+12

) + C

Example 3

Evaluateint

7radic 4minusx2

middotdx

Solution



int 7

radic 4minusx2

= int 7du

radic 22

minusu2


) + C

= 7arcsin(x2

) + C

Example 4

Evaluateint

dx


Solution



int dx


=int

du

uradic u2minus9

= 13

arcsec(u3

) + C

= 13

arcsec( |x|3

) + C

32





Example 8

Evaluateint


Solution





rArr minus12

du = (x minus 3)dx

Now substitute


=int


8radic 9+(xminus3)2

= minus12

int dvradic


9+(u)2

= 12

int vminus 1


= minus12

(2v1

2 ) + 8arcsin(u3

) + C



) + C


34






int b

a f (x)dx is


f (x)dx = bminusa


Example 1


int 20

2x2dx n = 4

Solution

x = 2minus04

= 24

= 12


= i2


2(4)[f (0) + 2f (1

2) + 2f (1) + 2f (3

2) + f (2)]

= 28

[2(0)2 + 2(2(12

)2) + 2(2(1)2) + 2(2( 32

)2) + 2(22)]

= 1

4

[0 + 2(1

2

) + 2(2) + 2( 18

4

) + 8]= 1

4[0 + 1 + 4 + 9 + 8]

= 14

[22]= 55



a f (x)dx is


int b


Example 2


int 2

0 2x2dx n = 4

Solution

35



x = 2minus04

= 24

= 12


= i2

int 20

2x2dx = 2minus03(4)

[f (0) + 4f (12

) + 2f (1) + 4f (32

) + f (2)]

= 212

[2(0)2 + 4(2(12

)2) + 2(2(1)2) + 4(2(32

)2) + 2(22)]= 1

6[0 + 4(1

2) + 2(2) + 4( 18

4 ) + 8]

= 16

[0 + 2 + 4 + 18 + 8]

= 16

[32]= 53333


int 2

0 2x2dx = 22+1

x2+120

= 23

x320

= 23

(2)3 minus 23

(0)3


= 163

Example 3


int π0 sin(x)dx

Solution

x = πminus04

= π4


= iπ4int π


[f (0) + 2f (π4

) + 2f (2π4

) + 2f (3π4

) + f (π)]

= π8

[sin(0) + 2sin(π4

) + 2sin(π2

) + 2sin(3π4

) + sin(π)]

= π8 [0 + 2(radic 22 ) + 2(1) + 2(radic 22 ) + 0]

= π8

[2radic

2 + 2]asymp 2003


int π0 sin(x)dx

Solution

x = πminus04

= π4

36





= 112

[111 + 4(0947) + 2(0816) + 4(0771) + 0625] asymp 086



1(x+2)2

dx =int 2

11u2

du =int 2


x+2

21

= minus 12+2

minus (minus 11+2

)

= minus14

+ 13

= 112







2int 1u

du = ln|x| + C

Example 1

Evaluateint

13x

dx

Solution


3du = dxint 13x

dx =int

13 middot 1

u du


int 13 middot 1

u du = 1

3ln|u| + C = 1

3ln|3x| + C

Example 2

Evaluateint

15x+3

dx

Solution


du = dx

int 13x

dx =int

15 middot 1

u du


int 15 middot 1

u du = 1

5ln|u| + C = 1

5ln|5x + 3| + C

29



Example 3

Evaluateint

x2

3minusx3 dx

Solution


du = x2dx

int x

2

3minusx3 dx =int

minus13 middot 1

u du


int minus1

3 middot 1

u du = minus1

3ln|u| + C = minus1

3ln|3 minus x3| + C

Example 4

Evaluateint

sinθ

cosθ dθ

Solution


cosθ dθ =

int minus 1

u du


int minus 1


Example 5

Evaluateint

cos(t)3+sin(t)

dt

Solution

Let u = 3 + sin(t)

rArr du = cos(t)dt

int cos(t)3+sin(t)

dθ =int

1u

du


int 1u

du = ln|u| + C = ln|3 + sin(t)| + C

30





Theorem 48



1int

duradic a2minusu2

= arcsin u

a +C

2int

du

a2+u2 = 1a

arctan u

a +C

3int

du

uradic u2minusa2

= 1a

arcsec |u|a

+C



Example 1

Evaluateint

41+9x2 middotdx

Solution


3du = dx


int 41+9x2 middotdx

=int

41+(3x)2

middotdx

= 13

int 4du1+u2

= 43

int du

1+u2

= 4

3arctan(u) + C

= 43

arctan(3x) + C

Example 2

Evaluateint

dx

4+(x+1)2

Solution



31



int dx

4+(x+1)2 =int

du

22+u2

= 12

arctan(u2

) + C

= 12

arctan(x+12

) + C

Example 3

Evaluateint

7radic 4minusx2

middotdx

Solution



int 7

radic 4minusx2

= int 7du

radic 22

minusu2


) + C

= 7arcsin(x2

) + C

Example 4

Evaluateint

dx


Solution



int dx


=int

du

uradic u2minus9

= 13

arcsec(u3

) + C

= 13

arcsec( |x|3

) + C

32





Example 8

Evaluateint


Solution





rArr minus12

du = (x minus 3)dx

Now substitute


=int


8radic 9+(xminus3)2

= minus12

int dvradic


9+(u)2

= 12

int vminus 1


= minus12

(2v1

2 ) + 8arcsin(u3

) + C



) + C


34






int b

a f (x)dx is


f (x)dx = bminusa


Example 1


int 20

2x2dx n = 4

Solution

x = 2minus04

= 24

= 12


= i2


2(4)[f (0) + 2f (1

2) + 2f (1) + 2f (3

2) + f (2)]

= 28

[2(0)2 + 2(2(12

)2) + 2(2(1)2) + 2(2( 32

)2) + 2(22)]

= 1

4

[0 + 2(1

2

) + 2(2) + 2( 18

4

) + 8]= 1

4[0 + 1 + 4 + 9 + 8]

= 14

[22]= 55



a f (x)dx is


int b


Example 2


int 2

0 2x2dx n = 4

Solution

35



x = 2minus04

= 24

= 12


= i2

int 20

2x2dx = 2minus03(4)

[f (0) + 4f (12

) + 2f (1) + 4f (32

) + f (2)]

= 212

[2(0)2 + 4(2(12

)2) + 2(2(1)2) + 4(2(32

)2) + 2(22)]= 1

6[0 + 4(1

2) + 2(2) + 4( 18

4 ) + 8]

= 16

[0 + 2 + 4 + 18 + 8]

= 16

[32]= 53333


int 2

0 2x2dx = 22+1

x2+120

= 23

x320

= 23

(2)3 minus 23

(0)3


= 163

Example 3


int π0 sin(x)dx

Solution

x = πminus04

= π4


= iπ4int π


[f (0) + 2f (π4

) + 2f (2π4

) + 2f (3π4

) + f (π)]

= π8

[sin(0) + 2sin(π4

) + 2sin(π2

) + 2sin(3π4

) + sin(π)]

= π8 [0 + 2(radic 22 ) + 2(1) + 2(radic 22 ) + 0]

= π8

[2radic

2 + 2]asymp 2003


int π0 sin(x)dx

Solution

x = πminus04

= π4

36





= 112

[111 + 4(0947) + 2(0816) + 4(0771) + 0625] asymp 086



1(x+2)2

dx =int 2

11u2

du =int 2


x+2

21

= minus 12+2

minus (minus 11+2

)

= minus14

+ 13

= 112



Example 3

Evaluateint

x2

3minusx3 dx

Solution


du = x2dx

int x

2

3minusx3 dx =int

minus13 middot 1

u du


int minus1

3 middot 1

u du = minus1

3ln|u| + C = minus1

3ln|3 minus x3| + C

Example 4

Evaluateint

sinθ

cosθ dθ

Solution


cosθ dθ =

int minus 1

u du


int minus 1


Example 5

Evaluateint

cos(t)3+sin(t)

dt

Solution

Let u = 3 + sin(t)

rArr du = cos(t)dt

int cos(t)3+sin(t)

dθ =int

1u

du


int 1u

du = ln|u| + C = ln|3 + sin(t)| + C

30





Theorem 48



1int

duradic a2minusu2

= arcsin u

a +C

2int

du

a2+u2 = 1a

arctan u

a +C

3int

du

uradic u2minusa2

= 1a

arcsec |u|a

+C



Example 1

Evaluateint

41+9x2 middotdx

Solution


3du = dx


int 41+9x2 middotdx

=int

41+(3x)2

middotdx

= 13

int 4du1+u2

= 43

int du

1+u2

= 4

3arctan(u) + C

= 43

arctan(3x) + C

Example 2

Evaluateint

dx

4+(x+1)2

Solution



31



int dx

4+(x+1)2 =int

du

22+u2

= 12

arctan(u2

) + C

= 12

arctan(x+12

) + C

Example 3

Evaluateint

7radic 4minusx2

middotdx

Solution



int 7

radic 4minusx2

= int 7du

radic 22

minusu2


) + C

= 7arcsin(x2

) + C

Example 4

Evaluateint

dx


Solution



int dx


=int

du

uradic u2minus9

= 13

arcsec(u3

) + C

= 13

arcsec( |x|3

) + C

32





Example 8

Evaluateint


Solution





rArr minus12

du = (x minus 3)dx

Now substitute


=int


8radic 9+(xminus3)2

= minus12

int dvradic


9+(u)2

= 12

int vminus 1


= minus12

(2v1

2 ) + 8arcsin(u3

) + C



) + C


34






int b

a f (x)dx is


f (x)dx = bminusa


Example 1


int 20

2x2dx n = 4

Solution

x = 2minus04

= 24

= 12


= i2


2(4)[f (0) + 2f (1

2) + 2f (1) + 2f (3

2) + f (2)]

= 28

[2(0)2 + 2(2(12

)2) + 2(2(1)2) + 2(2( 32

)2) + 2(22)]

= 1

4

[0 + 2(1

2

) + 2(2) + 2( 18

4

) + 8]= 1

4[0 + 1 + 4 + 9 + 8]

= 14

[22]= 55



a f (x)dx is


int b


Example 2


int 2

0 2x2dx n = 4

Solution

35



x = 2minus04

= 24

= 12


= i2

int 20

2x2dx = 2minus03(4)

[f (0) + 4f (12

) + 2f (1) + 4f (32

) + f (2)]

= 212

[2(0)2 + 4(2(12

)2) + 2(2(1)2) + 4(2(32

)2) + 2(22)]= 1

6[0 + 4(1

2) + 2(2) + 4( 18

4 ) + 8]

= 16

[0 + 2 + 4 + 18 + 8]

= 16

[32]= 53333


int 2

0 2x2dx = 22+1

x2+120

= 23

x320

= 23

(2)3 minus 23

(0)3


= 163

Example 3


int π0 sin(x)dx

Solution

x = πminus04

= π4


= iπ4int π


[f (0) + 2f (π4

) + 2f (2π4

) + 2f (3π4

) + f (π)]

= π8

[sin(0) + 2sin(π4

) + 2sin(π2

) + 2sin(3π4

) + sin(π)]

= π8 [0 + 2(radic 22 ) + 2(1) + 2(radic 22 ) + 0]

= π8

[2radic

2 + 2]asymp 2003


int π0 sin(x)dx

Solution

x = πminus04

= π4

36





= 112

[111 + 4(0947) + 2(0816) + 4(0771) + 0625] asymp 086



1(x+2)2

dx =int 2

11u2

du =int 2


x+2

21

= minus 12+2

minus (minus 11+2

)

= minus14

+ 13

= 112





Theorem 48



1int

duradic a2minusu2

= arcsin u

a +C

2int

du

a2+u2 = 1a

arctan u

a +C

3int

du

uradic u2minusa2

= 1a

arcsec |u|a

+C



Example 1

Evaluateint

41+9x2 middotdx

Solution


3du = dx


int 41+9x2 middotdx

=int

41+(3x)2

middotdx

= 13

int 4du1+u2

= 43

int du

1+u2

= 4

3arctan(u) + C

= 43

arctan(3x) + C

Example 2

Evaluateint

dx

4+(x+1)2

Solution



31



int dx

4+(x+1)2 =int

du

22+u2

= 12

arctan(u2

) + C

= 12

arctan(x+12

) + C

Example 3

Evaluateint

7radic 4minusx2

middotdx

Solution



int 7

radic 4minusx2

= int 7du

radic 22

minusu2


) + C

= 7arcsin(x2

) + C

Example 4

Evaluateint

dx


Solution



int dx


=int

du

uradic u2minus9

= 13

arcsec(u3

) + C

= 13

arcsec( |x|3

) + C

32





Example 8

Evaluateint


Solution





rArr minus12

du = (x minus 3)dx

Now substitute


=int


8radic 9+(xminus3)2

= minus12

int dvradic


9+(u)2

= 12

int vminus 1


= minus12

(2v1

2 ) + 8arcsin(u3

) + C



) + C


34






int b

a f (x)dx is


f (x)dx = bminusa


Example 1


int 20

2x2dx n = 4

Solution

x = 2minus04

= 24

= 12


= i2


2(4)[f (0) + 2f (1

2) + 2f (1) + 2f (3

2) + f (2)]

= 28

[2(0)2 + 2(2(12

)2) + 2(2(1)2) + 2(2( 32

)2) + 2(22)]

= 1

4

[0 + 2(1

2

) + 2(2) + 2( 18

4

) + 8]= 1

4[0 + 1 + 4 + 9 + 8]

= 14

[22]= 55



a f (x)dx is


int b


Example 2


int 2

0 2x2dx n = 4

Solution

35



x = 2minus04

= 24

= 12


= i2

int 20

2x2dx = 2minus03(4)

[f (0) + 4f (12

) + 2f (1) + 4f (32

) + f (2)]

= 212

[2(0)2 + 4(2(12

)2) + 2(2(1)2) + 4(2(32

)2) + 2(22)]= 1

6[0 + 4(1

2) + 2(2) + 4( 18

4 ) + 8]

= 16

[0 + 2 + 4 + 18 + 8]

= 16

[32]= 53333


int 2

0 2x2dx = 22+1

x2+120

= 23

x320

= 23

(2)3 minus 23

(0)3


= 163

Example 3


int π0 sin(x)dx

Solution

x = πminus04

= π4


= iπ4int π


[f (0) + 2f (π4

) + 2f (2π4

) + 2f (3π4

) + f (π)]

= π8

[sin(0) + 2sin(π4

) + 2sin(π2

) + 2sin(3π4

) + sin(π)]

= π8 [0 + 2(radic 22 ) + 2(1) + 2(radic 22 ) + 0]

= π8

[2radic

2 + 2]asymp 2003


int π0 sin(x)dx

Solution

x = πminus04

= π4

36





= 112

[111 + 4(0947) + 2(0816) + 4(0771) + 0625] asymp 086



1(x+2)2

dx =int 2

11u2

du =int 2


x+2

21

= minus 12+2

minus (minus 11+2

)

= minus14

+ 13

= 112



int dx

4+(x+1)2 =int

du

22+u2

= 12

arctan(u2

) + C

= 12

arctan(x+12

) + C

Example 3

Evaluateint

7radic 4minusx2

middotdx

Solution



int 7

radic 4minusx2

= int 7du

radic 22

minusu2


) + C

= 7arcsin(x2

) + C

Example 4

Evaluateint

dx


Solution



int dx


=int

du

uradic u2minus9

= 13

arcsec(u3

) + C

= 13

arcsec( |x|3

) + C

32





Example 8

Evaluateint


Solution





rArr minus12

du = (x minus 3)dx

Now substitute


=int


8radic 9+(xminus3)2

= minus12

int dvradic


9+(u)2

= 12

int vminus 1


= minus12

(2v1

2 ) + 8arcsin(u3

) + C



) + C


34






int b

a f (x)dx is


f (x)dx = bminusa


Example 1


int 20

2x2dx n = 4

Solution

x = 2minus04

= 24

= 12


= i2


2(4)[f (0) + 2f (1

2) + 2f (1) + 2f (3

2) + f (2)]

= 28

[2(0)2 + 2(2(12

)2) + 2(2(1)2) + 2(2( 32

)2) + 2(22)]

= 1

4

[0 + 2(1

2

) + 2(2) + 2( 18

4

) + 8]= 1

4[0 + 1 + 4 + 9 + 8]

= 14

[22]= 55



a f (x)dx is


int b


Example 2


int 2

0 2x2dx n = 4

Solution

35



x = 2minus04

= 24

= 12


= i2

int 20

2x2dx = 2minus03(4)

[f (0) + 4f (12

) + 2f (1) + 4f (32

) + f (2)]

= 212

[2(0)2 + 4(2(12

)2) + 2(2(1)2) + 4(2(32

)2) + 2(22)]= 1

6[0 + 4(1

2) + 2(2) + 4( 18

4 ) + 8]

= 16

[0 + 2 + 4 + 18 + 8]

= 16

[32]= 53333


int 2

0 2x2dx = 22+1

x2+120

= 23

x320

= 23

(2)3 minus 23

(0)3


= 163

Example 3


int π0 sin(x)dx

Solution

x = πminus04

= π4


= iπ4int π


[f (0) + 2f (π4

) + 2f (2π4

) + 2f (3π4

) + f (π)]

= π8

[sin(0) + 2sin(π4

) + 2sin(π2

) + 2sin(3π4

) + sin(π)]

= π8 [0 + 2(radic 22 ) + 2(1) + 2(radic 22 ) + 0]

= π8

[2radic

2 + 2]asymp 2003


int π0 sin(x)dx

Solution

x = πminus04

= π4

36





= 112

[111 + 4(0947) + 2(0816) + 4(0771) + 0625] asymp 086



1(x+2)2

dx =int 2

11u2

du =int 2


x+2

21

= minus 12+2

minus (minus 11+2

)

= minus14

+ 13

= 112





Example 8

Evaluateint


Solution





rArr minus12

du = (x minus 3)dx

Now substitute


=int


8radic 9+(xminus3)2

= minus12

int dvradic


9+(u)2

= 12

int vminus 1


= minus12

(2v1

2 ) + 8arcsin(u3

) + C



) + C


34






int b

a f (x)dx is


f (x)dx = bminusa


Example 1


int 20

2x2dx n = 4

Solution

x = 2minus04

= 24

= 12


= i2


2(4)[f (0) + 2f (1

2) + 2f (1) + 2f (3

2) + f (2)]

= 28

[2(0)2 + 2(2(12

)2) + 2(2(1)2) + 2(2( 32

)2) + 2(22)]

= 1

4

[0 + 2(1

2

) + 2(2) + 2( 18

4

) + 8]= 1

4[0 + 1 + 4 + 9 + 8]

= 14

[22]= 55



a f (x)dx is


int b


Example 2


int 2

0 2x2dx n = 4

Solution

35



x = 2minus04

= 24

= 12


= i2

int 20

2x2dx = 2minus03(4)

[f (0) + 4f (12

) + 2f (1) + 4f (32

) + f (2)]

= 212

[2(0)2 + 4(2(12

)2) + 2(2(1)2) + 4(2(32

)2) + 2(22)]= 1

6[0 + 4(1

2) + 2(2) + 4( 18

4 ) + 8]

= 16

[0 + 2 + 4 + 18 + 8]

= 16

[32]= 53333


int 2

0 2x2dx = 22+1

x2+120

= 23

x320

= 23

(2)3 minus 23

(0)3


= 163

Example 3


int π0 sin(x)dx

Solution

x = πminus04

= π4


= iπ4int π


[f (0) + 2f (π4

) + 2f (2π4

) + 2f (3π4

) + f (π)]

= π8

[sin(0) + 2sin(π4

) + 2sin(π2

) + 2sin(3π4

) + sin(π)]

= π8 [0 + 2(radic 22 ) + 2(1) + 2(radic 22 ) + 0]

= π8

[2radic

2 + 2]asymp 2003


int π0 sin(x)dx

Solution

x = πminus04

= π4

36





= 112

[111 + 4(0947) + 2(0816) + 4(0771) + 0625] asymp 086



1(x+2)2

dx =int 2

11u2

du =int 2


x+2

21

= minus 12+2

minus (minus 11+2

)

= minus14

+ 13

= 112






int b

a f (x)dx is


f (x)dx = bminusa


Example 1


int 20

2x2dx n = 4

Solution

x = 2minus04

= 24

= 12


= i2


2(4)[f (0) + 2f (1

2) + 2f (1) + 2f (3

2) + f (2)]

= 28

[2(0)2 + 2(2(12

)2) + 2(2(1)2) + 2(2( 32

)2) + 2(22)]

= 1

4

[0 + 2(1

2

) + 2(2) + 2( 18

4

) + 8]= 1

4[0 + 1 + 4 + 9 + 8]

= 14

[22]= 55



a f (x)dx is


int b


Example 2


int 2

0 2x2dx n = 4

Solution

35



x = 2minus04

= 24

= 12


= i2

int 20

2x2dx = 2minus03(4)

[f (0) + 4f (12

) + 2f (1) + 4f (32

) + f (2)]

= 212

[2(0)2 + 4(2(12

)2) + 2(2(1)2) + 4(2(32

)2) + 2(22)]= 1

6[0 + 4(1

2) + 2(2) + 4( 18

4 ) + 8]

= 16

[0 + 2 + 4 + 18 + 8]

= 16

[32]= 53333


int 2

0 2x2dx = 22+1

x2+120

= 23

x320

= 23

(2)3 minus 23

(0)3


= 163

Example 3


int π0 sin(x)dx

Solution

x = πminus04

= π4


= iπ4int π


[f (0) + 2f (π4

) + 2f (2π4

) + 2f (3π4

) + f (π)]

= π8

[sin(0) + 2sin(π4

) + 2sin(π2

) + 2sin(3π4

) + sin(π)]

= π8 [0 + 2(radic 22 ) + 2(1) + 2(radic 22 ) + 0]

= π8

[2radic

2 + 2]asymp 2003


int π0 sin(x)dx

Solution

x = πminus04

= π4

36





= 112

[111 + 4(0947) + 2(0816) + 4(0771) + 0625] asymp 086



1(x+2)2

dx =int 2

11u2

du =int 2


x+2

21

= minus 12+2

minus (minus 11+2

)

= minus14

+ 13

= 112



x = 2minus04

= 24

= 12


= i2

int 20

2x2dx = 2minus03(4)

[f (0) + 4f (12

) + 2f (1) + 4f (32

) + f (2)]

= 212

[2(0)2 + 4(2(12

)2) + 2(2(1)2) + 4(2(32

)2) + 2(22)]= 1

6[0 + 4(1

2) + 2(2) + 4( 18

4 ) + 8]

= 16

[0 + 2 + 4 + 18 + 8]

= 16

[32]= 53333


int 2

0 2x2dx = 22+1

x2+120

= 23

x320

= 23

(2)3 minus 23

(0)3


= 163

Example 3


int π0 sin(x)dx

Solution

x = πminus04

= π4


= iπ4int π


[f (0) + 2f (π4

) + 2f (2π4

) + 2f (3π4

) + f (π)]

= π8

[sin(0) + 2sin(π4

) + 2sin(π2

) + 2sin(3π4

) + sin(π)]

= π8 [0 + 2(radic 22 ) + 2(1) + 2(radic 22 ) + 0]

= π8

[2radic

2 + 2]asymp 2003


int π0 sin(x)dx

Solution

x = πminus04

= π4

36





= 112

[111 + 4(0947) + 2(0816) + 4(0771) + 0625] asymp 086



1(x+2)2

dx =int 2

11u2

du =int 2


x+2

21

= minus 12+2

minus (minus 11+2

)

= minus14

+ 13

= 112





= 112

[111 + 4(0947) + 2(0816) + 4(0771) + 0625] asymp 086



1(x+2)2

dx =int 2

11u2

du =int 2


x+2

21

= minus 12+2

minus (minus 11+2

)

= minus14

+ 13

= 112

math 152 calculus ii chapter 4

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