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 October 2nd, 2011, 08:49 PM #1 Senior Member   Joined: Sep 2011 From: New York, NY Posts: 333 Thanks: 0 Simpson's Rule approximation How large should n be to guarantee that the Simpson's Rule approximation to is accurate to within 0.0001? $\int_0^115e{^x}^2\, dx$
 October 2nd, 2011, 10:16 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Simpson's Rule approximation My old calculus text states concerning the error bound for Simpson's Rule: If there exists a number M > 0 such that $\|f^{\small{(4)}}(x)\|\le M$ for all x in [a,b] then $E_n\le\frac{M(b-a)}{180n^4}$ So, for $f(x)=15e^{x^2}$ we compute: $f^{\small{(4)}}(x)=60e^{x^2}$$4x^4+12x^2+3$$$ Analysis of $f^{\small{(5)}}(x)$ shows that $f^{\small{(4)}}(x)$ is increasing on (0,1] thus we find: $M=f^{\small{(4)}}(1)=60e$$4+12+3$$=1140e$ Thus: $\frac{1140e(1-0)}{180n^4}\le0.00005$ $\frac{19e}{3n^4}\le0.00005$ $\frac{380000}{3}e\le n^4$ $24.22\le n$ Since n must be even we find the minimum value of n is 26.
 October 2nd, 2011, 10:55 PM #3 Senior Member   Joined: Sep 2011 From: New York, NY Posts: 333 Thanks: 0 Re: Simpson's Rule approximation some how it worked out the be 22
 October 2nd, 2011, 11:08 PM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Simpson's Rule approximation Maybe I should have used: $\frac{19e}{3n^4}\le0.0001$ $\frac{190000}{3}e\le n^4$ $20.37\le n$ Then n would be 22.

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