Approximate Integration (trapezoids, midpoint, and simpson's rule)

• Mar 8th 2008, 01:47 PM
FalconPUNCH!
Approximate Integration (trapezoids, midpoint, and simpson's rule)
http://xs225.xs.to/xs225/08106/untitled-1132.jpg

I couldn't get the answer for the trapezoid correct, so I haven't tried the other two rules.

Here's what I tried:

$\displaystyle \triangle{x} = \frac{\pi}{10}$

I used the trapezoid rule

$\displaystyle T_n = \frac{\triangle{x}}{2} [ f(x_0) + 2f(x_1) + 2f(x_2) + ... + 2f(x_{n-1}) + f(x_n)]$

Here's my work:

$\displaystyle T_n = \frac{\frac{\pi}{10}}{2} [ sin(0) + 2sin(\frac{\pi}{10}) + 2sin(\frac{\pi}{5}) + 2sin(\frac{3\pi}{10}) + 2sin(\frac{\pi}{2.5})$$\displaystyle + 2sin(\frac{\pi}{2}) + 2sin(\frac{3\pi}{5}) + 2sin(\frac{7\pi}{10}) + 2sin(\frac{4\pi}{5}) + 2sin(\frac{9\pi}{10}) + sin(\pi) I get approximately .086106 as my answer, but the book has the answer to almost 1.9. Any idea what I did wrong? • Mar 9th 2008, 11:48 AM CaptainBlack Quote: Originally Posted by FalconPUNCH! http://xs225.xs.to/xs225/08106/untitled-1132.jpg I couldn't get the answer for the trapezoid correct, so I haven't tried the other two rules. Here's what I tried: \displaystyle \triangle{x} = \frac{\pi}{10} I used the trapezoid rule \displaystyle T_n = \frac{\triangle{x}}{2} [ f(x_0) + 2f(x_1) + 2f(x_2) + ... + 2f(x_{n-1}) + f(x_n)] Here's my work: \displaystyle T_n = \frac{\frac{\pi}{10}}{2} [ sin(0) + 2sin(\frac{\pi}{10}) + 2sin(\frac{\pi}{5}) + 2sin(\frac{3\pi}{10}) + 2sin(\frac{\pi}{2.5})$$\displaystyle + 2sin(\frac{\pi}{2}) + 2sin(\frac{3\pi}{5}) + 2sin(\frac{7\pi}{10}) + 2sin(\frac{4\pi}{5}) + 2sin(\frac{9\pi}{10}) + sin(\pi)$

I get approximately .086106 as my answer, but the book has the answer to almost 1.9.

Any idea what I did wrong?