Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

(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:

share|cite|improve this question
Can you post what you have got so far? – Nate Eldredge Jul 5 '11 at 1:57
@Miles: This is not just a practice "exercise." When one looks at the normal distribution, its mean, and its variance, a relationship like the one in the problem shows up. – André Nicolas Jul 5 '11 at 4:26
up vote 10 down vote accepted

Hint: $x^2 e^{-x^2} = x ( x e^{-x^2})$ and the second factor is a derivative.

share|cite|improve this answer
Thanks, this was the hint I needed - so obvious in hindsight. – Mark Jul 5 '11 at 2:19
Half a derivative to be accurate :-) – Asaf Karagila Jul 5 '11 at 5:35
@Asaf: $[2,\frac{d}{dx}] = 0$. So lhf is perfectly okay in saying what he said. – Willie Wong Jul 5 '11 at 9:59
@Willie: I was hoping that by now most people would know that 83% of my comments are meant as tongue in cheek :-) – Asaf Karagila Jul 5 '11 at 10:06
I'd bet Willie's comment was, too... – PseudoNeo Jul 5 '11 at 17:48

Hint: Consider integration by parts of $\int_0^1 {e^{ - x^2 } 1 \, dx}.$

Edit: $$ \int_0^1 {e^{ - x^2 } 1 \,dx} = e^{ - x^2 } x|_0^1 - \int_0^1 {e^{ - x^2 } ( - 2x)x \,dx} = e^{ - 1} + 2\int_0^1 {x^2 e^{ - x^2 } \,dx} . $$

share|cite|improve this answer

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 $x^{2}=t$.

share|cite|improve this answer
I don't see how you use the result in the first line to evaluate anything here. After you substitute $x^2=t$ in the middle integral, I don't see what functions as $f$ to allow use of the first identity. – tzs Jul 6 '11 at 3:59

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.