Hi does anyone knows how to solve this question.
Use power series to approximate the definite integral to within the given accuracy
$\int_{0}^{1}x^{2}\sin(x^{4})dx$
Error $<0.001$
I managed to integreate the function but do not know how to proceed from here.
We have $$ x^2\sin(x^4)=\sum_{k=0}^\infty \frac{(-1)^kx^{8k+6}}{(2k+1)!}, $$ and hence $$ \int_0^1 x^2\sin(x^4)\,dx=\sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)!(8k+7)}. $$ Also, $$ \sin x=x-\frac{x^3}{3!}+\cdots+\frac{(-1)^nx^{2n+1}}{(2n+1)!}\sin^{(2n+1)}(\xi), $$ for some $\xi\in (0,x)$. Hence $$ \left|\,\sin x-x+\frac{x^3}{3!}+\cdots+\frac{(-1)^{k}x^{2k-1}}{(2k-1)!}\,\right|<\frac{1}{(2k+1)!}, $$ when $\lvert x\rvert<1$, and thus $$ \left|\,x^2\sin x^4-x^6+\frac{x^{10}}{3!}+\cdots+\frac{(-1)^{k}x^{8k+6}}{(2k-1)!}\,\right|<\frac{1}{(2k+1)!}, $$ as well. So $$ \left|\,\int_0^1 x^2\sin x^4\,dx-\sum_{k=0}^{n-1} \frac{(-1)^k}{(2k+1)!(8k+7)}\right|<\frac{1}{(2n+1)!}. $$ Thus, you need to pick $n$, so that $\frac{1}{(2n+1)!}<10^{-3}$. Note that $n=3$ does your approximation.
