Originally Posted by

**lllll** the function is:

$\displaystyle \int^{1}_{0} x cos(x^3) \ dx $ find the error to 3 decimal places.

so I expressed cos as a Taylor Series which gives me:

$\displaystyle \int^{1}_{0} x \sum^{\infty}_{n=0} -1^n \frac{x^{6n}}{(2n)!} \ dx $

$\displaystyle \int^{1}_{0} \sum^{\infty}_{n=0} -1^n \frac{x^{6n+1}}{(2n)!} \ dx $

$\displaystyle \sum^{\infty}_{n=0} -1^n \frac{x^{6n+2}}{(6n+2)(2n)!} \bigg{|}^{1}_{0} = \sum^{\infty}_{n=0} \frac{-1^n}{(6n+2)(2n)!} $

but to find the error to whatever decimal place I get stuck, if someone could show me how to find it, and explain the process behind their approach that would really help me out.