CHAPTER 4SECTION 4.3

RIEMANN SUMS AND DEFINITE INTEGRALS

Riemann Sum

1. Partition the interval [a,b] into n subintervalsa = x0 < x1 … < xn-1< xn = b

• Call this partition P

• The kth subinterval is xk = xk-1 – xk

• Largest xk is called the norm, called || ||

• If all subintervals are of equal length, the norm is called regular.

2. Choose an arbitrary value from each subinterval, call it ic

Riemann Sum3. Form the sum

This is the Riemann sum associated with• the function f• the given partition P• the chosen subinterval representatives

• We will express a variety of quantities in terms of the Riemann sum

1 1 2 21

( ) ( ) ... ( ) ( )n

n n n i ii

R f c x f c x f c x f c x

1 1 2 21

( ) ( ) ... ( ) ( )n

n n n i ii

R f c x f c x f c x f c x

ic

This illustrates that the size of ∆x is allowed to vary

a x1 x2 x3 x4 x5

x1* x2* x3* x4* x5*

Then a < x1 < x2 < x3 < x4 ….etc. is a partition of [ a, b ] Notice the partition ∆x does not have to be the same size for each rectangle.

y = f (x)

And x1* , x2* , x3* , etc… are x coordinates such that a < x1* < x1, x1 < x2* < x2 , x2 < x3* < x3 , … and are used to construct the height of the rectangles.

Etc…

The graph of a typical continuous function y = ƒ(x) over [a, b]. Partition [a, b] into n subintervals a < x1 < x2 <…xn < b. Select any number in each subinterval cck.k. Form the product f(ck)xk. Then take the sum of these products.

1

( )n

k kk

f c x

This is called the Riemann SumRiemann Sum of the partition of x.

The width of the largest subinterval of a partition is the normnorm of the partition, written ||x||.

As the number of partitions, n, gets larger and larger, the norm gets smaller and smaller.

As n, ||x|| 0 only if ||x|| are the same width!!!!

The Riemann SumCalculated

• Consider the function2x2 – 7x + 5

• Use x = 0.1

• Let the = left edgeof each subinterval

• Note the sum

x 2x 2̂-7x+5 dx * f(x)4 9 0.9

4.1 9.92 0.9924.2 10.88 1.0884.3 11.88 1.1884.4 12.92 1.2924.5 14 1.44.6 15.12 1.5124.7 16.28 1.6284.8 17.48 1.7484.9 18.72 1.872

5 20 25.1 21.32 2.1325.2 22.68 2.2685.3 24.08 2.4085.4 25.52 2.5525.5 27 2.75.6 28.52 2.8525.7 30.08 3.0085.8 31.68 3.1685.9 33.32 3.332

Riemann sum = 40.04

x 2x 2̂-7x+5 dx * f(x)4 9 0.9

4.1 9.92 0.9924.2 10.88 1.0884.3 11.88 1.1884.4 12.92 1.2924.5 14 1.44.6 15.12 1.5124.7 16.28 1.6284.8 17.48 1.7484.9 18.72 1.872

5 20 25.1 21.32 2.1325.2 22.68 2.2685.3 24.08 2.4085.4 25.52 2.5525.5 27 2.75.6 28.52 2.8525.7 30.08 3.0085.8 31.68 3.1685.9 33.32 3.332

Riemann sum = 40.04

ic

The Riemann Sum

• We have summed a series of boxes

• If the x were smaller, we would have gotten a better approximation

f(x) = 2x2 – 7x + 5

1

( ) 40.04n

i ii

f c x

Finer partitions of [a, b] create more rectangles with shorter bases.

0 1

lim ( )n

i ii

f c x L

1

( )n

i ii

f c x

The Definite Integral

• The definite integral is the limit of the Riemann sum

• We say that f is integrable when– the number I can be approximated as accurate

as needed by making || || sufficiently small– f must exist on [a,b] and the Riemann sum must

exist– is the same as saying n 0

0

1

lim( )n

k

b

a i if f c xxI dx

b

af x dx

IntegrationSymbol

lower limit of integration

upper limit of integration

integrandvariable of integration

Notation for the definite integral

Important for AP test [ and mine too !! ]

Recognizing a Riemann Sum as a Definite integral

lim ( )

.

.

.

. ( )

. ( )

ni

n i

n n

xn

b a

nx dx

i

na i x so a

b a so b

i

nx

Thus x dx

2 13

13

13

2 13

1

3 3 4

4 2 13

2

5 2 1

1

1

4

Recognizing a Riemann Sum as a Definite integral

lim

.

.

.

. ( )

ni

n i

n n

xn

b a

nx dx

a= b s a i xi

n

xi

n

Thus x dx

ince =

=

35 5

15

2 0 55

35

4 3

2

1

2

0

5

Recognizing a Riemann Sum as a Definite integral

From our textbook

lim ]

: (5 )

[

0

2

1

2

1

4

5 3

3

ii

n

i ic c x

answer x x dx

over [ 1,4 ]

Notice the text uses ∆ instead of ∆x, but it is basically the same as our ∆x , and ci is our xi *

Try the reverse : write the integral as a Riemann Sum … also on AP and my test

4

1 3 1010 3 7

2 37

3 4 37 7

2

3

10

2

1

37

x x dx

a b x=-

n n

a i xi

n

Thusi

n nni

n i

n

.

.

. lim ( )( )

so

Theorem 4.4 Continuity Implies Integrability

D CI

Relationship between Differentiability, Continuity, and Integrability

D – differentiable functions, strongest condition … all Diff ’ble functions are continuous and integrable.

C – continuous functions , all cont functions are integrable, but not all are diff ’ble.

I – integrable functions, weakest condition … it is possible they are not con‘ t, and not diff ‘ble.

Evaluate the following Definite Integral 32

4xdx

c cnn

1

k1

n

n(n 1)

2

First … remember these sums and definitions:

ci = a + i x xb a

n

lim ( )

lim [ (( )

)]

lim[ ( )]

ni

n

n

n

ni

n

nn

n

n n

n

182

6

182

6 1

2

36 54 11

36 54 18

1

3

3 26 6

2

4

1

1

xdx f c x

in n

n ii

n

i

ni

n

lim ( )

lim ( )

c cnn

1

k1

n

n(n 1)

2

ci = a + i x

xb a

n

EXAMPLE Evaluate the definite integral by the limit Evaluate the definite integral by the limit definitiondefinition

0 1

lim ( )n

i ii

f c x L

6

1x dx

1

5 51

n

i

if

n n

1

5 51

n

i

i

n n

21

5 25n

i

i

n n

21 1

1 255

n n

i i

in n

5x

n

51i

ic

n

Evaluate the definite integral by the Evaluate the definite integral by the limit definition, continuedlimit definition, continued

( )b

a

L f x dx0 1

lim ( )n

ii

f c xi L

6

1x dx

2

1 25 ( 1)5

2

n nn

n n

255 1

2n

n

6

1

25 25lim 5

2 2nx dx

n

25 255

2 2n

35

2

25 ( 1)5

2

n

n

21 1

1 255

n n

i i

in n

The Definite integral above represents the Area of the region under the curve y = f ( x) , bounded by the x-axis, and the vertical lines x = a, and x = b

y = f ( x)

a b

y

x

b

adxxf )(

Theorem 4.4 Continuity Implies Integrability

D CI

Relationship between Differentiability, Continuity, and Integrability

D – differentiable functions, strongest condition … all Diff ’ble functions are continuous and integrable.

C – continuous functions , all cont functions are integrable, but not all are diff ’ble.

I – integrable functions, weakest condition … it is possible they are not con‘ t, and not diff ‘ble.

3

Y = x

x

y

xdx0

3

Areas of common geometric shapes

Ax

A 2

0

3

2

9

2

1

23 3

9

2

Sol’n to definite integral A = ½ base * height

0

A Sight Integral ... An integral you should know on sight

2 2 21

2a x dx a

a

a

-

This is the Area of a semi-circle of radius a

a-a

1 0

2

. ( )

. ( ) ( )

f x dx

f x dx f x dx

a

a

a

b

b

a

Special Definite Integrals

for f (x ) integrable from a to b

1 5 0

29

2

2

2

3

0

0

3

. ( )

.

x dx

xdx xdx

EXAMPLE

Additive property of integrals

If is integrable over interval [ ],

where , then:

f a , b

a < c < b

f x dx f x dx f x dxa

b

a

c

c

b( ) ( ) ( )

a bc

y

x

More Properties of Integrals

For integrable on [ ], and is a

constant ... , then since and

are integrable on [ ], we have :

1.

f, g a, b k

kf f g

a, b

kf x dx k f x dx

f x g x dx f x dx g x dx

a

b

a

b

a

b

a

b

a

b

( ) ( )

. [ ( ) ( )] ( ) ( )2

EXAMPLE

Given x dx xdx

Solve x x dx

x dx xdx

+ 3

) + 3(3

2

2

2

2

1 1

2

2

1

2

2

1 1

2

7

3

3

2

3

3

37

37

9

2

23

2

: ( )

( )

Even – Odd Property of Integrals

For an even function:

For an odd function:

f ( x )

f x dx f x dx

f ( x )

f x dx

a

a

a

a

a

( ) ( )

( )

2

0

0

Even function: f ( x ) = f ( - x ) … symmetric about y - axis

Finally …. Inequality Properties

If is integrable and nonnegative on [ ] :

0

f a, b

f(x)dxa

b

If , are integrable on [ ] , and :

f g a, b f(x) g(x)

f(x)dx g x dxa

b

a

b

( )

END

Rules for definite integralsEvaluate the using the following values:

Example 2:

4

3

2

2x dx

4 4 4

3 3

2 2 2

2 2x dx x dx dx

4 4 4

3 3

2 2 2

2 2x dx x dx dx = 60 + 2(2) = 64

Using the TI 83/84 to checkcheck your answers

Find the area under on [1,5]

• Graph f(x)

• Press 2nd CALC 7• Enter lower limit 1 • Press ENTER• Enter upper limit 5• Press ENTER.

3y x

Set up a Definite Integral for finding the area of the shaded region. Then use geometry to find the area.

1. 4f x 2. 1f x x 6

4

2

5

6

4

2

5

Use the limit definition to find

3 2

13x dx

Set up a Definite Integral for finding the area of the shaded region. Then use geometry to find the area.

1. 4f x 2. 1f x x 6

4

2

5

6

4

2

5

5

1

4A dx 6

2

1x dA x 4 4

216 un

123 4 4 4

rectangle triangle

220 un