finding the area under a curve: riemann, trapezoidal, and simpsons rule adguary calwile laura rogers...
TRANSCRIPT
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
Right Riemann- Example
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
Right Riemann- Example
)]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
Within the brackets!
)(0xf )(
1xnf
Left Riemann- Example
Remember: Left Only
Given this problem below, what all do we need to know in order to find the area under
the curve using Left Riemann?
dxxf x4
0
3)( 4 partitions
)](...)()()()([13210 xxxxx n
fffffn
abA
Left Riemann- Example
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
Left Riemann- Example
)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
Within the brackets!
)(2/1xf )(
2/1xnf
Midpoint Riemann- Example
Remember: Midpoint Only
Given this problem below, what all do we need to know in order to find the area under
the curve using Midpoint Riemann?
dxxf x4
0
3)( 4 partitions
)](...)()()()([2/12/72/52/32/1 xxxxx n
fffffn
abA
Midpoint Riemann- Example
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
2/7
2/5
2/3
2/1
0
4
3
2
1
0
xxxxx
Midpoint Riemann- Example
)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
Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #27
Applications of Approximating Areas
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.
Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #28
Applications of Approximating Areas
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
Given this problem below, what all do we need to know in order to find the area under
the curve using Trapezoidal Rule?
dxxf x4
0
3)( 4 partitions
)](...)(2)(2)(2)([2 3210 xxxxx n
fffffn
abA
Trapezoidal Example
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
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
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
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
Sources
• http://www.intmath.com/Integration
© Laura Rogers, Adguary Calwile; 2011