# Simpson's Approximation

• Nov 14th 2011, 07:43 PM
Bowlbase
Simpson's Approximation
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: $\int^1_0 \frac{tan(x)}{x}dx\;n=10$
$\frac{b-a}{n}= \frac {1-0}{10}=.1=\Delta x$
$\frac{\Delta x}{3}=\frac{1}{30}$

Plugging in 0,.1,.2,...,.9,1 into the function I get
$S_{10}=\frac{1}{30}(0+4(1.003)+2(1.0136)+4(1.0311) +2(1.0570)+4(1.0930)+2(1.1402)+4(1.2033)+2(1.2870) +4(1.4002)+1.5574)$

This equals roughly 1.11585

However, when I evaluate the integral

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

I end up with 1.14915. Where is the discrepancy coming from? Thanks for the help!
• Nov 14th 2011, 10:59 PM
CaptainBlack
Re: Simpson's Approximation
Quote:

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: $\int^1_0 \frac{tan(x)}{x}dx\;n=10$
$\frac{b-a}{n}= \frac {1-0}{10}=.1=\Delta x$
$\frac{\Delta x}{3}=\frac{1}{30}$

Plugging in 0,.1,.2,...,.9,1 into the function I get
$S_{10}=\frac{1}{30}(0+4(1.003)+2(1.0136)+4(1.0311) +2(1.0570)+4(1.0930)+2(1.1402)+4(1.2033)+2(1.2870) +4(1.4002)+1.5574)$

This equals roughly 1.11585

However, when I evaluate the integral

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

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

Reconsider the value of the integrand at x=0.

CB
• Nov 15th 2011, 06:46 AM
Bowlbase
Re: Simpson's Approximation
Tan(0)/0 should be 0 no? Or, at the very least, undefined.

I should add that I'm using a calculator and Wolframalpha alpha to evaluate. Is this just a reasonable amount of error from the approximation?
• Nov 15th 2011, 07:39 AM
Bowlbase
Re: Simpson's Approximation
Okay, I got it.

I needed to use limit as x approaches 0 = 1. That gets me a lot closer to the answer.