chapter 3 integral calculus

8/10/2019 Chapter 3 Integral Calculus

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Integral Calculus

An antiderivative of f`(x) = f(x)

The indefinite integral: `( ) ( )f x dx f x c You need to remember all the integral identities from higher.

1

1

1cos( ) sin( )

1sin( ) cos( )

n naax dx x c n

ax b dx ax b c a

ax b dx ax b c a


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A definite integral is where limits are given.

This gives the area under the curve of f`(x) between these limits.

Practise can be had by completing page 70 Exercise 1A and 1B

TJ exercise 1


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Standard forms

From the differentiation exercise we know:

2

( )

1(ln )

(tan ) sec

x xde e

dx

d

xdx x

dx x

dx

This gives us three new antiderivatives.

2

1

ln

sec tan

x xe dx e c

dx x c x

x dx x c

Note: when 0 but when 0x x x x x x


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1 23 3

0 11. Find x xe dx e dx

1 2 23 3 3

0 1 0

x x xe dx e dx e dx 2

3

0

13

xe

61 13 3

e

52. Find dxx

5 15dx dx

x x 5lnx c

5lnx c


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23. Find tan x dx(We need to use a little trig here and our knowledge of integrals.)

From a few pages ago we know 2sec tanx dx x c

2 2sin cos 1x x 2 2

2 2 2

sin cos 1

cos cos cos

x x

x x x 2 2tan 1 secx x

2 2tan sec 1x dx x dx

2 2tan sec 1 tanx dx x dx x x c

Page 72 Exercise 2A and if time some of 2B

TJ Exercise 2 and some of 3

`


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Integration by Substitution

When differentiating a composite function we made use of the chain rule.

3(2 3)y x Let 2 3u x 3y u

23dy

u

du

and 2du

dx

dy dy du

dx du dx 23 2u

26u 26(2 3)x

When integrating, we must reduce the function to a standard form one for whichwe know the antiderivative.

This can be awkward, but under certain conditions, we can use the chain rule in reverse.


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2 3 52. Find 3x ( 4)x dx

3

Let 4u x

2

then 3du x dx 2 3 5 53x ( 4)x dx u du

61

6 u c

3 6( 4)

6

xc


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33. Find 8cosxsin x dx

Let sinu x

then cosdu x dx

3 38cosxsin 8x dx u du

4

2u c

42sin x c

Page 74 Exercise 3 Odd Numbers

TJ Exercise 4.


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For many questions the choice of substitution will not always be obvious.

You may even be given the substitution and in that case you must use it.

4ln1. Find . Let lnxdx u x x

1du dx

x

4ln4

xdx u du

x

22u c

22(ln )x c


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2 12. Find (1 ) . Let sinx dx u x

cosdx u du

2 2(1 ) 1 sin cosx dx u u du 2

cos u du

sinx u Substituting gives,

We can not integrate this yet. Let us use trig.

2 1cos (1 cos2 )2

u u

1 1cos2

2 2

u du 1

sin2

2 4

uu c

1.2sin cos

2 4

uu u c

1sin cos

2u u u c


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1 21 sin 1

2x x x c

Now for some trig play..

sin , but what does cos equal?x u u

1

sin cos2

u u u c

2 2sin cos 1u u 2 2cos 1 sinu

2

cos 1 sinu u

21

(1 ) sin cos2

x dx u u u c


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2sinx

sin2

x

1sin

2

x

2

22 2sin cos

4

xdx c

x

2 24sinx

24 4cos 2 24cos 4 x

22cos 4 x

21cos 42

x

1 212sin 2. . 4

2 2 2

x xx c

1 22sin 42 2

x xx c


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Page 75 Exercise 4A Odd numbersTJ Exercise 5 and 6


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Now for some not very obvious substitutions.

51. Find sin x dx5 4sin sin sinx x x 2 2sin (sin )x x 2 2sin (1 cos )x x

5 2 2sin sin (1 cos )x dx x x dx Let cosu x sindu x dx

5 2 2sin (1 )x dx u du 2 4(1 2 )u u du

2 41 2u u du 3 52 1

3 5u u u c

3 52 1cos cos cos3 5

x x x c


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12. Find

1dx

x Let 1u x

1

212

du x dx

1

22dx x du 2 1dx u du

1 2( 1)

1

udx du

ux

2

2 duu

2 2lnu u c

2 2 2ln 1x x c

Page 76 Exercise 4B Questions 1 to 5


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Substitution and definite integralsAssuming the function is continuous over the interval, then exchanging the limits for

xby the corresponding limits for uwill save you having to substitute back after theintegration process.

2

2 3

1

1. Evaluate (2x+4)(x 4 )x dx2Let 4u x x 2 4du x dx When 2, 12; 1, 5x u x u

2 122 3 3

1 5(2x+4)(x 4 )x dx u du

12

4

5

1

4u

5027.75


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Special (common) forms

Some substitutions are so common that they can be treated as standards and, when

their form is established, their integrals can be written down without further ado.

1( ) ( )f ax b dx F ax b c

a

`( )ln ( )

( )

f xdx f x c

f x

21`( ) ( ) ( ( ))

2

f x f x dx f x c

Page 80 Exercise 6A Questions 1(c), (d) 10 plus a few more

Do some of exercise 6B if time.


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Area under a curve

a b

y = f(x)

( )b

aA f x dx

a b

y = f(x)

( )b

aA f x dx


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Area between the curve and y - axis

b

a

y = f(x)

( )b

aA f y dy


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1. Calculate the area shown in the diagram below.

5

2

y = x2+ 1

2 1y x 2 1x y

1x y

15 2

2( 1)A y dy

53

2

2

2( 1)

3y

14units squared.

3

TJ Exercise 7, 8 and 9


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Volumes of revolution

Volumes of revolution are formed when a curve is rotated about the x or y axis.

2b

aV y dx

2d

cV x dy


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1. Find the volume of revolution obtained betweenx= 1 and x = 2 when the curve

y=x2+ 2 is rotated about

(i) the xaxis (ii) the yaxis.

22

1)i V y dx

24 2

1( 4 1)x x dx

25 3

1

4

5 3

x xx

32 32 1 12 15 3 5 3

3263 units15

( ) when 1, 3 and when 2, 6ii x y x y

6

3( 2)y dy

62

3

22

yy

9

18 12 62

315

units2

62

3V x dy


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Page 91 Exercise 10B Questions 4, 5 and 10.

TJ Exercise 10.


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Displacement, velocity, acceleration for

rectilinear motion

We have already seen in chapter 2 on differential calculus that where distance is

given as a function of time, S = f (t), then

velocityds

v dt

2

2accelerationd s dv

a dt dt

Hence,

( ) ( )v t a t dt ( ) ( )s t v t dt


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1. A particle starts from rest and, at time t seconds, the velocity is given by

v = 3t2+ 4t1. Determine the distance, velocity and acceleration at t = 4 seconds.

23 4 1v t t

dva

dt 6 4t

( )s v t dt 3 22t t t c When t = 0, s = 0

3 22t t t

When t = 4,34 2.16 4s

92 units

3.16 16 1v

63 units/ s

24 4a 228 units / s


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chapter 3 integral calculus

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