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In this problem we will approximate the integral of $\cos(x^3)$ over the interval $[0, 2]$.
(a) Write an expression for MN, TN and SN with $N = 4$.
(b) For each of the approximations determine an N so that the error is guaranteed to be less than $10^{-4}.$ You will find it useful to know that on the interval $[0, 2], |\frac {d^4\cos(x^3)}{dx^4} | \le 850$.

Please help me with the question. Thanks a lot!

Ross Millikan
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user77117
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  • dont know why after d/dx cos(x^3) got cut off. It should be fourth derivative of d/dx cos(x^3)< and = 850. – user77117 May 10 '13 at 02:55
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    Welcome to MSE! I edited some of your post using MathJax (see FAQ). It helps to share your thoughts and approach so we can provide better answers. Regards – Amzoti May 10 '13 at 02:59
  • Surely your textbook has formulas for $M_N$, $T_N$, and $S_N$. Have you tried applying them to this situation? – Adam Saltz May 10 '13 at 03:09
  • You could define MN, TN, and SN and be consistent with upper and lower case between the title and the body. – Ross Millikan May 10 '13 at 03:09
  • @user77117: Why did you repost this as a new question? You have the same question here: http://math.stackexchange.com/questions/386688/calculus-help-mn-sn-tn – Amzoti May 10 '13 at 04:11

1 Answers1

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Hints:

For Simpson's rule, we write:

$$\displaystyle \int_0^2 \cos x^3 dx = \frac{1}{12} \left(1 + 2 \sum_{n=1}^{4-1} \cos n^3 + 4 \sum_{n=1}^{4} \cos (\frac{1}{8} (-1+2 n)^3)+ \cos 8\right) = 0.878445$$

Can you now follow the approach and derive the trapezoidal and midpoint rule?

The results should be:

  • Midpoint Rule = $1.07095$
  • Trapezoidal Rule = $0.493433$
  • Exact result = $0.8554746507540773$
Amzoti
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