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Unformatted text preview: MATH 118, LECTURES 9: NUMERICAL INTEGRATION 1 Numerical Integration In evaluating definite integrals, we required that we were able to find the antiderivative of the function, but consider being asked to evaluate integraldisplay b a e x 2 dx. There is no closedform function whose derivative is e x 2 , so we have no manner in which to evaluate the antiderivative at the endpoints! We cannot evaluate the definite integral in the same manner as we have been. The solution to this apparent problem is to recall our alternative in tepretation of the definite integral as the area under the curve. For many functions, we can find the exact value of the definite integral by performing Riemann sums (i.e. dividing the region to be integrated into small steps, computing rectangles, and then taking the limit as the step size goes to zero). The important thing to realize here is that all we needed to know in order to perform Riemann sums was the value of the function at various points in the region to be integratedat no point do we require knowledge of the antiderivative of the function. This suggests an alternative method by which to evaluate the area under the curve, and therefore the definite integral. Rather than worrying about the antiderivative, we can approximate the area under the curve by dividing the region into small pieces and adding up the approximate area of the curve of all these small regions. We will not be able to take the limit as the region size goes to zero, but we will be able to make the value as close to the actual value as we like by taking successively small region sizes. In this lecture, we will consider three different ways to approximate the area under a curve: the rectangular rule, trapezoidal rule, and Simpsons rule. Each of these regions has an associated formula (corresponding to summing over all of the subregions) and an error bound which goes to zero as the region size is taken to zero. 1 Rule Formula & Error Bound Rectangular Rule integraldisplay b a f ( x ) dx = h n summationdisplay i =1 f ( x i ) Trapezoidal Rule integraldisplay b a f ( x ) dx = h 2 bracketleftBigg f ( a ) + 2 n 1 summationdisplay i =1 f ( x i ) + f ( b ) bracketrightBigg Error:  T n  M ( b a ) 3 12 n 2 M = max  f primeprime ( x )  on a x b Simpsons Rule integraldisplay b a f ( x ) dx = h 3 f ( a ) + 4 n/ 2 summationdisplay i =1 f ( x 2 i 1 ) + 2 n/ 2 1 summationdisplay i =1 f ( x 2 i ) + f ( b ) Error:  S n  M ( b a ) 5 180 n 4 M = max  f primeprimeprimeprime ( x )  on a x b Note that in each of these definitions we take n to be the number of intervals and let h = x i x i 1 be the same for all i = 1 , . . . , n so that h = ( b a ) /n . We let x = a and x n = b ....
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 Spring '09
 Soares

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