ch. 5 pp the definite integral

7/29/2019 Ch. 5 Pp the Definite Integral

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The Definite Integral


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Riemann Sums

Sigma notation enables us to express alarge sum in compact form

1 2

1

.....n

k n

k

a a a a


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Riemann Sums

LRAM, MRAM,and RRAM are examples ofRiemann sums

Sn =

This sum, which depends on the partition Pand the choice of the numbers ck,is a

Riemann sum for f on the interval [a,b]

1

( )n

k k

k

f c x


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Definite Integral as a Limit of

Riemann SumsLet f be a function defined on a closed interval [a,b]. For

any partition P of [a,b], let the numbers ck be chosenarbitrarily in the subintervals [xk-1,xk].

If there exists a numberIsuch that

no matter how P and the cks are chosen, then f isintegrable on [a,b] and Iis the definite integral of f over[a,b].

01

lim ( )n

k kP

k

f c x I


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Definite Integral of a continuous

function on [a,b]

Let f be continuous on [a,b], and let [a,b] bepartitioned into n subintervals of equal length x= (b-a)/n. Then the definite integral of f over[a,b] is given by

where each ck is chosen arbitrarily in the kthsubinterval.

1

lim ( )n

kn

k

f c x


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Definite integral

This is read as the integral from ato boffofxdee x or sometimes as the integral

from ato bof f ofxwith respect to x.

( )b

af x dx


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Using Definite integral notation

2

1

32

1

lim (3( ) 2 5)

(3 2 5)

n

k kn

k

m m x

x x dx

The function being integrated is f(x) = 3x2 2x + 5over the interval [-1,3]


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Definition: Area under a curve

If y = f(x) is nonnegative and integrable over

a closed interval [a,b], then the area

under the curve of y = f(x) from a to b is

the integral of f from a to b,

( )b

aA f x dx


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Nonpositive regions

If the graph is nonpositive from a to b then

( )

b

aA f x dx


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Area of any integrable function

= (area above the x-axis)

(area below x-axis)

( )b

af x dx


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Integral of a Constant

If f(x) = c, where c is a constant, on the

interval [a,b], then

( ) ( )b b

a af x dx cdx c b a


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Evaluating Integrals using areas

We can use integrals to calculate areas andwecan use areas to calculate integrals.

Using areas, evaluate the integrals:

1)

2)

3

2( 1)x dx

22

24 x dx


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Evaluating Integrals using areas

Evaluate using areas:

3)

4) (a


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Evaluating integrals using areas

Evaluate the discontinuous function:

Since the function is discontinuous at x = 0, wemust divide the areas into two pieces and find thesum of the areas

= -1 + 2 = 1

2

1

xdx

x


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Integrals on a Calculator

You can evaluate integrals numerically using

the calculator. The book denotes this by

using NINT. The calculator function fnInt

is what you will use.

= fnInt(xsinx,x,-1,2) is approx.

2.04

2

1

sinx xdx


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Evaluate Integrals on calculator

Evaluate the following integralsnumerically:

1) = approx. 3.14

2) = approx. .89

1

20

4

1dx

x

25

0

xe dx


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Rules for Definite Integrals

1) Order of Integration:

( ) ( )a b

b a

f x dx f x dx


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2) Zero: ( ) 0a

af x dx


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3) Constant Multiple:

( ) ( )

( ) ( )

b b

a a

b b

a a

kf x dx k f x dx

f x dx f x dx


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4) Sum and Difference:

( ( ) ( )) ( ) ( )b b b

a a a

f x g x dx f x dx g x dx


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5) Additivity:

( ) ( ) ( )

b c c

a b af x dx f x dx f x dx


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6) Max-Min Inequality: If max f and min f

are the maximum and minimum values of

f on [a,b] then:

min f (b a) max f (b a)( )b

af x dx


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7) Domination: f(x) g(x) on [a,b]

f(x) 0 on [a,b] 0

( ) ( )b b

a af x dx g x dx

( )b

af x dx


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Using the rules for integration

Suppose:

Find each of the following integrals, if possible:a) b) c)

d) e) f)

1

1( ) 5f x dx

4

1( ) 2f x dx

1

1( ) 7h x dx

1

4( )f x dx

4

1( )f x dx

1

12 ( ) 3 ( )f x h x dx

1

0( )f x dx

2

2( )h x dx

4

1( ) ( )f x h x dx


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Using the rules for definite integrals

Show that the value of

is less than 3/2

The Max-Min Inequality rule says the

max f. (b a) is an upper bound.The maximum value of(1+cosx) on [0,1] is 2 so

the upper bound is:

2(1 0) = 2 , which is less than 3/2

1

01 cosxdx


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Average (Mean) Value

If f is integrable on [a,b], its average (mean)value on [a,b] is:

av(f) =

Find the average value of f(x) = 4 x2 on[0,3] . Does f actually take on this value atsome point in the given interval?

1 ( )b

af x dx

b a


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Applying the Mean Value

Av(f) =

= 1/3(3) = 1

4 x2 = 1 when x = 3 but only 3 falls inthe interval from [0,3], so x = 3 is theplace where the function assumes theaverage.

32

0

1(4 )

3 0x dx


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Mean Value Theorem for Definite

Integrals

If f is continuous on [a,b], then at some point

c in [a,b],

1( ) ( )b

af c f x dx

b a


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The Fundamental Theorem of

Calculus, Part I

If f is continuous on [a,b], then the function

F(x) =

has a derivative at every point x in [a,b],

and

( )x

a

f t dt

( ) ( )x

a

dfdx f t dt f x

dx


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Applications of The Fundamental

Theorem of Calculus, Part I

1.

2.

3.

cos cosxd

tdt xdx

2 20

1 1

1 1

xddt

dx t x

2

2 2

1cos cos (2 ) 2 cos

xdtdt x x x x

dx


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Find dy/dx.

y =

Since this has an x on both ends of theintegral, it must be separated.

2

2

1

2

x

tx

dte


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=

2 20

2 2 0

1 1 1

2 2 2

x x

t t tx xdt dt dt

e e e

22

0 0

1 1

2 2

x x

t tdt dt

e e


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=

=

22

1 1(2) (2 )

2 2x x

xe e

2 2

2 2

22xx

x

ee


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The Fundamental Theorem of

Calculus, Part 2

If f is continuous at every point of [a,b], and if F is

any antiderivative of f on [a,b], then

This part of the Fundamental Theorem is alsocalled the Integral Evaluation Theorem.

( ) ( ) ( )b

af x dx F b F a


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Trapezoidal Rule

To approximate , use

T = (y0

+ 2y1

+ 2y2

+ . 2yn-1

+ yn

)

where [a,b] is partitioned into n

subintervals of equal length h = (b-a)/n.

( )b

af x dx

2

h


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Using the trapezoidal rule

Use the trapezoidal rule with n = 4 toestimate

h = (2-1)/4 or , so

T = 1/8( 1+2(25/16)+2(36/16)+2(49/16)+4)= 75/32 or about 2.344

22

1

x dx


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Simpson Rule

To approximate , use

S = (y0 + 4y1 + 2y2 + 4y3. 2yn-2 +4yn-1 + yn)

where [a,b] is partitioned into an even

numbernsubintervals of equal length h =(ba)/n.

( )ba

f x dx

3

h


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Using Simpsons Rule

Use Simpsons rule with n = 4 to estimate

h = (2 1)/4 = , so

S = 1/12 (1 + 4(25/16) + 2(36/16) + 4(49/16) + 4)

= 7/3

22

1

x dx

ch. 5 pp the definite integral

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