Express $\int^1_0x^2 e^{-x^2} dx$ in terms of $\int^1_0e^{-x^2} dx$ (Apologies, this was initially incorrectly posted on mathoveflow)
In the MIT 18.01 practice questions for Exam 4 problem 3b (link below), we are asked to express $\int^1_0x^2 e^{-x^2} dx$ in terms of $\int^1_0e^{-x^2} dx$
I understand that this should involve using integration by parts but the given solution doesn't show working and I'm not able to obtain the same answer regardless of how I set up the integration.
Link to the practice exam:
http://ocw.mit.edu/courses/mathematics/18-01-single-variable-calculus-fall-2006/exams/prexam4a.pdf

You can use this result as well: $$\int e^{x} \bigl[ f(x) + f'(x)\bigr] \ dx = e^{x} f(x) +C$$
So your integral can be rewritten as 
\begin{align*}
\int\limits_{0}^{1} x^{2}e^{-x^{2}} \ dx & = -\int\limits_{0}^{1} \Bigl[-x^{2} -2x\Bigr] \cdot e^{-x^{2}} -\int\limits_{0}^{1} 2x \cdot e^{-x^{2}}\ dx
\end{align*}
The second part of the integral can be $\text{easily evaluated}$ by putting $x^{2}=t$.
