Question

Use series to approximate the definite integral to within 3 decimal places.
\[\int_0^1 xcos(x^3)dx\]

Answer 1

\[=\int_0^1\sum_{n=0}^{\infty}(-1)^n\frac{x^{6n+1}}{(2n)!}\]

Answer 2

\[\left[\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}\frac{x^{6n+2}}{6n+2}\right]_0^1\]

Answer 3

It's similar to the previous problem I just realized...

Answer 4

I think I got it.