Approximate the integral $\int_0^{0.5}{x^2e^{x^2}}dx$ correct to four decimal places using a Maclaurin series.

Question

I got $$\int_0^{0.5}{\sum_0^\infty}\frac{x^{2n+2}}{n!}dx$$ for the taylor series representation, but I'm not sure what to do next.

Do I use 0 and 0.5 as bounds for z for the Lagrange Error Bound? And what does the remainder have to be less than for it to be correct to 4 decimal places?

Answer 1

I took the integral and did trial and error with x = 0.5 (the whole thing = 0 when x = 0) until I got a term that was less than 0.0001. It requires 3 terms, or n = 3.

Answer 2

Integrating gives $$\sum_{n=0}^\infty\frac{2^{-n-3}}{n!(2n+3)}$$ If we drop all the terms before $n=k$ the remainder is $$2^{-3}\sum_{n=k}^\infty\frac{4^{-n}}{n!(2n+3)}<\frac{2^{-3}}{k!(2k+3)}\frac{4^{1-k}}{3}$$

No doubt one can do something more sophisticated, but this ought to be good enough.