5.5 numerical integration
DESCRIPTION
Photo by Vickie Kelly, 1998. Greg Kelly, Hanford High School, Richland, Washington. 5.5 Numerical Integration. Mt. Shasta, California. Using integrals to find area works extremely well as long as we can find the antiderivative of the function. - PowerPoint PPT PresentationTRANSCRIPT
![Page 1: 5.5 Numerical Integration](https://reader030.vdocuments.site/reader030/viewer/2022013004/56813460550346895d9b46a5/html5/thumbnails/1.jpg)
5.5 Numerical Integration
Mt. Shasta, CaliforniaGreg Kelly, Hanford High School, Richland, WashingtonPhoto by Vickie Kelly, 1998
![Page 2: 5.5 Numerical Integration](https://reader030.vdocuments.site/reader030/viewer/2022013004/56813460550346895d9b46a5/html5/thumbnails/2.jpg)
Using integrals to find area works extremely well as long as we can find the antiderivative of the function.
Sometimes, the function is too complicated to find the antiderivative.
At other times, we don’t even have a function, but only measurements taken from real life.
What we need is an efficient method to estimate area when we can not find the antiderivative.
![Page 3: 5.5 Numerical Integration](https://reader030.vdocuments.site/reader030/viewer/2022013004/56813460550346895d9b46a5/html5/thumbnails/3.jpg)
211
8y x
43
0
1
24A x x
4 2
0
11
8A x dx
0 4x
Actual area under curve:
20
3A 6.6
![Page 4: 5.5 Numerical Integration](https://reader030.vdocuments.site/reader030/viewer/2022013004/56813460550346895d9b46a5/html5/thumbnails/4.jpg)
211
8y x 0 4x
Left-hand rectangular approximation:
Approximate area: 1 1 1 31 1 1 2 5 5.75
8 2 8 4
(too low)
![Page 5: 5.5 Numerical Integration](https://reader030.vdocuments.site/reader030/viewer/2022013004/56813460550346895d9b46a5/html5/thumbnails/5.jpg)
Approximate area: 1 1 1 31 1 2 3 7 7.75
8 2 8 4
211
8y x 0 4x
Right-hand rectangular approximation:
(too high)
![Page 6: 5.5 Numerical Integration](https://reader030.vdocuments.site/reader030/viewer/2022013004/56813460550346895d9b46a5/html5/thumbnails/6.jpg)
Averaging the two:
7.75 5.756.75
2
1.25% error (too high)
![Page 7: 5.5 Numerical Integration](https://reader030.vdocuments.site/reader030/viewer/2022013004/56813460550346895d9b46a5/html5/thumbnails/7.jpg)
Averaging right and left rectangles gives us trapezoids:
1 9 1 9 3 1 3 17 1 171 3
2 8 2 8 2 2 2 8 2 8T
![Page 8: 5.5 Numerical Integration](https://reader030.vdocuments.site/reader030/viewer/2022013004/56813460550346895d9b46a5/html5/thumbnails/8.jpg)
1 9 1 9 3 1 3 17 1 171 3
2 8 2 8 2 2 2 8 2 8T
1 9 9 3 3 17 171 3
2 8 8 2 2 8 8T
1 27
2 2T
27
4 6.75 (still too high)
![Page 9: 5.5 Numerical Integration](https://reader030.vdocuments.site/reader030/viewer/2022013004/56813460550346895d9b46a5/html5/thumbnails/9.jpg)
Trapezoidal Rule:
0 1 2 12 2 ... 22 n n
hT y y y y y
( h = width of subinterval )
This gives us a better approximation than either left or right rectangles.
![Page 10: 5.5 Numerical Integration](https://reader030.vdocuments.site/reader030/viewer/2022013004/56813460550346895d9b46a5/html5/thumbnails/10.jpg)
1.031251.28125
1.781252.53125
211
8y x 0 4x
Compare this with the Midpoint Rule:
Approximate area: 6.625 (too low)0.625% error
The midpoint rule gives a closer approximation than the trapezoidal rule, but in the opposite direction.
![Page 11: 5.5 Numerical Integration](https://reader030.vdocuments.site/reader030/viewer/2022013004/56813460550346895d9b46a5/html5/thumbnails/11.jpg)
Midpoint Rule: 6.625 (too low)0.625% error
Trapezoidal Rule: 6.750 1.25% error (too high)
Notice that the trapezoidal rule gives us an answer that has twice as much error as the midpoint rule, but in the opposite direction.
If we use a weighted average:
2 6.625 6.7506.6
3
This is the exact answer!
Oooh!
Ahhh!
Wow!
![Page 12: 5.5 Numerical Integration](https://reader030.vdocuments.site/reader030/viewer/2022013004/56813460550346895d9b46a5/html5/thumbnails/12.jpg)
1x 2x 3x 4xh h h h
This weighted approximation gives us a closer approximation than the midpoint or trapezoidal rules.
Midpoint:
1 3 1 32 2 2M h y h y h y y
Trapezoidal:
0 2 2 4
1 12 2
2 2T y y h y y h
0 2 2 4T h y y h y y
0 2 42T h y y y
1 3 0 2 4
14 2
3h y y h y y y
twice midpoint trapezoidal
1 3 0 2 44 4 23
hy y y y y
0 1 2 3 44 2 43
hy y y y y
2
3
M T
![Page 13: 5.5 Numerical Integration](https://reader030.vdocuments.site/reader030/viewer/2022013004/56813460550346895d9b46a5/html5/thumbnails/13.jpg)
Simpson’s Rule:
0 1 2 3 2 14 2 4 ... 2 43 n n n
hS y y y y y y y
( h = width of subinterval, n must be even )
Example: 211
8y x 1 9 3 17
1 4 2 4 33 8 2 8
S
1 9 171 3 3
3 2 2
120
3 6.6
![Page 14: 5.5 Numerical Integration](https://reader030.vdocuments.site/reader030/viewer/2022013004/56813460550346895d9b46a5/html5/thumbnails/14.jpg)
Simpson’s rule can also be interpreted as fitting parabolas to sections of the curve, which is why this example came out exactly.
Simpson’s rule will usually give a very good approximation with relatively few subintervals.
It is especially useful when we have no equation and the data points are determined experimentally.
p
![Page 15: 5.5 Numerical Integration](https://reader030.vdocuments.site/reader030/viewer/2022013004/56813460550346895d9b46a5/html5/thumbnails/15.jpg)
Trapezoidal Rule:
0 1 2 12 2 ... 22 n n
hT y y y y y
( h = width of subinterval )
Simpson’s Rule: 0 1 2 3 2 14 2 4 ... 2 4
3 n n n
hS y y y y y y y
( h = width of subinterval, n must be even )
Your Turn!Use the 2 rules aboveto estimate the integralto the right:
Then determine the error of each method by comparing to the exact integral value calculated by the antiderivative.