Originally Posted by

**Bowlbase** I am not getting the correct answer for this so I'm sure I'm messing up some arithmetic or missing a step. Any help would be appreciated.

Use Simpson's Approx. for: $\displaystyle \int^1_0 \frac{tan(x)}{x}dx\;n=10$

$\displaystyle \frac{b-a}{n}= \frac {1-0}{10}=.1=\Delta x$

$\displaystyle \frac{\Delta x}{3}=\frac{1}{30}$

Plugging in 0,.1,.2,...,.9,1 into the function I get

$\displaystyle S_{10}=\frac{1}{30}(0+4(1.003)+2(1.0136)+4(1.0311) +2(1.0570)+4(1.0930)+2(1.1402)+4(1.2033)+2(1.2870) +4(1.4002)+1.5574)$

This equals roughly 1.11585

However, when I evaluate the integral

$\displaystyle \int^1_0 \frac{tan(x)}{x}dx$

I end up with 1.14915. Where is the discrepancy coming from? Thanks for the help!