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I was asked to evaluate the following using Simpson's rule (by using 2 strips). Below is the function and my answer to it. What am I doing wrong?

$$\begin{align} \color{red}{\int_1^{1.6}\dfrac{\sin2t}{t}\,\mathrm dt} & \color{blue}{=\int_1^{1.6}\dfrac{\cos2t^2-\sin2t}{t^2}=\dfrac{0.6}2\big[f(1)+4(1.3)+f(3)\big]} \\\,\\ &\color{blue}{=\big[-1.32-5.1+0.6\big]} \end{align}$$

When I plugged in the values I put my calculator in Radian mode but the answer I get is negative. The correct answer is 0.2460

Thank you.

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Your formula is wrong. With $f(x) = \sin(2x)/x\;$ you get using the simple Simpson rule $$\frac{0.6}{6}\left(f(1.0) + 4f(1.3)+f(1.6)\right)\approx 0.245897.$$ and if you take two strips, you get $$\frac{0.6}{6}\left(f(1.0) + 4 f(1.15)+ 2 f(1.3)+ 4 f(1.45)+f(1.6)\right) \approx 0.245982.$$ What is purpose of the $\cos\;$ expression?

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  • $\begingroup$ Oh I see. I guess I forgot that the whole point of using this formula is for functions that can not be integrated! thanks $\endgroup$ – Cash Vai Nov 25 '13 at 15:20
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Formula for Simpson's rule is h/2 ( f(a) + 4*f(a+h) + f(a+2h) ) Here h = 0.6 and you have chosen a = 1, then why is a+h = 1.3 ?

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  • $\begingroup$ I used h= b-a/2 and the simpson formula is h/n [f(1) + 4f(x1) + 2f(x2) + ... f(xn)]. So is the integration part correct? $\endgroup$ – Cash Vai Nov 25 '13 at 15:12

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