finding the area under a curve: riemann, trapezoidal, and simpsons rule adguary calwile laura rogers...

Finding the area under a curve: Riemann, Trapezoidal, and Simpson’s Rule

Adguary CalwileLaura Rogers

Autrey~ 2nd Per.

3/14/11

Introduction to area under a curve

Before integration was developed, people found the area under curves by dividing the space beneath into rectangles, adding the area, and approximating the answer.

As the number of rectangles, n, increases, so does the accuracy of the area approximation.

Introduction to area under a curve (cont.)

There are three methods we can use to find the area under a curve: Riemann sums, the trapezoidal rule, and Simpson’s rule.

For each method we must know:• f(x)- the function of the curve• n- the number of partitions or rectangles• (a, b)- the boundaries on the x-axis

between which we are finding the areadxxf

b

a )(

Finding Area with Riemann SumsFinding Area with Riemann Sums

• For convenienceFor convenience, the , the area of a partition is area of a partition is often divided into often divided into subintervals with equal subintervals with equal width – in other words, width – in other words, the rectangles all have the rectangles all have the same width. (see the same width. (see the diagram to the the diagram to the right for an example of right for an example of a Right Riemann a Right Riemann approximation)approximation)

4

3

2

1

2

f x = x2

Subintervals with equal width

Finding Area with Riemann SumsFinding Area with Riemann Sums

• It is possible to divide a region into It is possible to divide a region into different sized rectangles based on different sized rectangles based on an algorithm or rule (see graph an algorithm or rule (see graph above)above)

6

4

2

5 10 15

Unequal Subintervals

Riemann Sums

There are three types of Riemann SumsRight Riemann:

Left Riemann:

Midpoint Riemann:

)]()...()()([321 xxxx nffff

n

abA

)](...)()()()([13210 xxxxx n

fffffn

abA

)](...)()()()([2/12/72/52/32/1 xxxxx n

fffffn

abA

Right Riemann- Overview

Right Riemann places the right point of the rectangles along the curve to find the area.

The equation that is used for the RIGHT RIEMANN ALWAYS begins with:

And ends with

Within the brackets!

)(1xf )(xnf

Right Riemann- Example

)]()...()()([321 xxxx nffff

n

abA

Remember: Right Only

Given this problem below, what all do we need to know in order to find the area under

the curve using Right Riemann?

dxxf x4

0

3)( 4 partitions


For each method we must know:f(x)- the function of the curven- the number of partitions or rectangles(a, b)- the boundaries on the x-axis between which we are finding the area

4

3

2

1

0

4

3

2

1

0

xxxxx


)]4()3()2()1([4

04)( ffffxf

2100

)100(1

)642781(1

uA

Right Riemann TRY ME!

dxx

xf 2

1

1)(

Volunteer:___________________

4 Partitions

!Show All Your Work!

dxx

xf 2

1

1)( n=4

Did You Get It Right?

dxx

xf 2

1

1)( n=4

22 76.420

319

)105

319(

4

1

)]7/4()3/2()5/4()1[(4

1

uu

Left Riemann- Overview

Left Riemann uses the left corners of rectangles and places them along the curve to find the area. The equation that is used for the LEFT

RIEMANN ALWAYS begins with:

And ends with


)(0xf )(

1xnf

Left Riemann- Example

Remember: Left Only


the curve using Left Riemann?

dxxf x4

0

3)( 4 partitions

)](...)()()()([13210 xxxxx n

fffffn

abA



4

3

2

1

0

4

3

2

1

0

xxxxx


)3()2()1()0([4

04)( ffffxf

236

)36(1

)27810(1

uA

Left Riemann- TRY ME!

Volunteer:___________

dxxf x1

0

3)(

3 Partitions

!Show All Your Work!

dxxf x1

0

3)( n=3

Did You Get My Answer?

22 111.9

13

1)

27

9(

3

1

)27

8

27

10(

3

1

uu

dxxf x1

0

3)( n=3

Midpoint Riemann- Overview

Midpoint Riemann uses the midpoint of the rectangles and places them along the curve to

find the area. The equation that is used for MIDPOINT RIEMANN ALWAYS begins with:

And ends with


)(2/1xf )(

2/1xnf

Midpoint Riemann- Example

Remember: Midpoint Only


the curve using Midpoint Riemann?

dxxf x4

0

3)( 4 partitions

)](...)()()()([2/12/72/52/32/1 xxxxx n

fffffn

abA



2/7

2/5

2/3

2/1

0

4

3

2

1

0

xxxxx


)2/7()2/5()2/3()2/1([4

04)( ffffxf

)]8/343()8/125()8/27()8/1[(4

04

262

]62[1

uA

Midpoint Riemann- TRY ME

dxxf x )1)(3

0

3

(

6 partitions

Volunteer:_________

!Show Your Work!

dxxf x )1)(3

0

3

( n=6

Correct???

dxxf x )1)(3

0

3

( n=6

22 3125.2516

405

)8

405(

2

1

)64

1395

64

793

64

407

64

189

64

91

64

65(

2

1

uu

Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #26

Applications of Approximating Areas

EXAMPLEEXAMPLE

SOLUTIONSOLUTION

The velocity of a car (in feet per second) is recorded from the speedometer every 10 seconds, beginning 5 seconds after the car starts to move. See Table 2. Use a Riemann sum to estimate the distance the car travels during the first 60 seconds. (Note: Each velocity is given at the middle of a 10-second interval. The first interval extends from 0 to 10, and so on.)

Since measurements of the car’s velocity were taken every ten seconds, we will use . Now, upon seeing the graph of the car’s velocity, we can construct a Riemann sum to estimate how far the car traveled.

10x



5, 20

15, 44

25, 3235, 39

45, 65

55, 80

0

10

20

30

40

50

60

70

80

90

0 10 20 30 40 50 60

Time

Velocity

This is an example of using a midpoint Riemann sum to This is an example of using a midpoint Riemann sum to approximate an integral.approximate an integral.



Therefore, we estimate that the distance the car traveled is 2800 feet.

CONTINUECONTINUEDD

tvtvtvtvtvtv 55453525155

tvvvvvv 55453525155

10806539324420

2800

Trapezoidal Rule Overview

Trapezoidal Rule is a little more accurate than Riemann Sums because it uses trapezoids instead of

rectangles. You have to know the same 3 things as Riemann but the equation that is used for TRAPEZOIDAL RULE ALWAYS begins with:

and ends with

Within the brackets with every“ f ” being multiplied by 2

EXCEPT for the first and last terms

)(0xf )( nxf

Trapezoidal Rule- Example

Remember: Trapezoidal Rule Only


the curve using Trapezoidal Rule?

dxxf x4

0

3)( 4 partitions

)](...)(2)(2)(2)([2 3210 xxxxx n

fffffn

abA

Trapezoidal Example


4

3

2

1

0

4

3

2

1

0

xxxxx

Trapezoidal Rule- Example

)()(2)(2)(2)0([4

044321 xfxfxfxff

2136

)136(1

)64541620(1

)]64()27(2)8(2)1(2)0[(1

uA

Trapezoidal Rule- TRY Me

Volunteer:_____________

dxxf x4

0

2)(

4 Partitions

Trapezoidal Rule- TRY ME!!

dxxf x4

0

2)( n=4

Was this your answer?

dxxf x4

0

2)( n=4

222

)44(2

1

)1616840(2

1

u

Simpson’s Rule- Overview

Simpson’s rule is the most accurate method of finding the area under a curve. It is better than the

trapezoidal rule because instead of using straight lines to model the curve, it uses parabolic arches to approximate each part of the curve. The equation

that is used for Simpson’s Rule ALWAYS begins with: And ends with

Within the brackets with every “f” being multiplied by alternating coefficients of 4 and 2 EXCEPT the first and last terms.

In Simpson’s Rule, n MUST be even.

Simpson’s Rule- Example

Remember: Simpson’s Rule Only

)](...)(4)(2)(4)([3 3210 xxxxx n

fffffn

abA

Given this problem below, what all do we need to know in order to find the

area under the curve using Simpson’s Rule?

4 Partitions4

0

3)( dxxf x

Simpson’s Example


4

3

2

1

0

4

3

2

1

0

xxxxx

Simpson’s Rule- Example

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

044321 xfxfxfxff

264

)192)(12/4(

)641081640)(12/4(

)]64()27(4)8(2)1(4)0)[(12/4(

uA

Simpson’s Rule TRY ME!

dxxxf )1()(2

0

4 partitions

Volunteer:____________

!Show Your Work!

dxxxf )1()(2

0 n=4

Check Your Answer!

)]()(4)(2)(4)([12

02)( 43210 xfxfxfxfxfxf

22 33.46

26

]310661[6

1

)]2()2

3(4)2(2)

2

1(4)0([

6

1

uu

fffff

finding the area under a curve: riemann, trapezoidal, and simpsons rule adguary calwile laura rogers...

Documents

area slide

right riemann example

partitions slide

midpoint riemann example

left riemann example

right riemann try

midpoint riemann try

midpoint riemann sum