Taylor/Maclaurin series problem

[ThirdEyeBlind]

Homework Statement

Use series to show that the integral from 0 to 1/10 of arctan (x^2)dx is approx. 1/3000. What bound can you place on the error in this approximation?

Homework Equations

The Attempt at a Solution

Manipulated Maclaurin series to get the sum from n=0 to infinity of (-1)^n ((x^2)^(2n+1)) / 2n+1

Simplify to get

. Now I don't know to take the integral of

, assuming that is the correct way to proceed.

 

[praharmitra]

\int dx \left(\sum\limits_k f(x,k) \right) = \sum\limits_k \int dx f(x,k)

Does that help?

 

[ThirdEyeBlind]

[STRIKE]I couldn't get Latex to correctly show what I have so I did it in paint. Is this correct? If it is, how do I evaluate this? Any help would be appreciated.
[/STRIKE]

EDIT: I think I figured it out. tan^-1 (x^2) = x^(2) - x^(6)/3 + x^(10)/5 -... and if you integrate the first term from 0 to 1/10, you get x^3 / 3 which is .1^(3) / 3 which equals 1/3000
EDIT2: For the error bound, is the |error| less than or equal x^7 / 21 -> (1/10)^7 / 21 since the alternating series test applies?

 

[praharmitra]

[STRIKE]I couldn't get Latex to correctly show what I have so I did it in paint. Is this correct? If it is, how do I evaluate this? Any help would be appreciated.
[/STRIKE]

EDIT: I think I figured it out. tan^-1 (x^2) = x^(2) - x^(6)/3 + x^(10)/5 -... and if you integrate the first term from 0 to 1/10, you get x^3 / 3 which is .1^(3) / 3 which equals 1/3000
EDIT2: For the error bound, is the |error| less than or equal x^7 / 21 -> (1/10)^7 / 21 since the alternating series test applies?

I'm no expert on errors and stuff, but i think it's correct