Trapezoid and Simpsons Rule to approximate e^10 by using....

[krisaldine]

Hi I was trying to figure out this problem for a while now, can someone help me on it? I'm confused on how to approach it.

1) Given a computer program to find approximations to integrals using either the Trapezoid Rule or Simpson's Rule, explain how to approximate e^10 by using the integral of 1/x.

Much appreciated!

There's also one problem with integration by parts and integration using trig functions, anyone can help me on this:

2) Integral of x tan^-1x dx

thanks again! :-D

 

[krisaldine]

I know that for the integral of tan -1 x dx you can use the formula, designate u as tan ^-1 and v' as dx, thus making u' dx/(1+x^2) and v as x
dx= xtan^-1x - integral of x/(1+x^2)dx, and thus xtan ^-1 x - (1/2)ln (1+x^2)+C but I forgot if there is an x added before the tangent...what do I do??

 

[skeeter]

\(\displaystyle \L \int x \arctan{x} dx\)

\(\displaystyle \L u = \arctan{x}\), \(\displaystyle \L dv = x dx\)

\(\displaystyle \L du = \frac{1}{1+x^2} dx\), \(\displaystyle \L v = \frac{x^2}{2}\)


\(\displaystyle \L \int x \arctan{x} dx = \frac{x^2\arctan{x}}{2} - \int \frac{x^2}{2(1+x^2)} dx\)

\(\displaystyle \L \int x \arctan{x} dx = \frac{x^2\arctan{x}}{2} - \frac{1}{2}\int \frac{x^2+ 1 - 1}{1+x^2} dx\)

\(\displaystyle \L \int x \arctan{x} dx = \frac{x^2\arctan{x}}{2} - \frac{1}{2}\int 1 - \frac{1}{1+x^2} dx\)

\(\displaystyle \L \int x \arctan{x} dx = \frac{x^2\arctan{x}}{2} - \frac{x}{2} + \frac{\arctan{x}}{2} + C\)

\(\displaystyle \L \int x \arctan{x} dx = \frac{1}{2} \left[(x^2+1)\arctan{x} - x\right] + C\)