(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:


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    $\begingroup$ Can you post what you have got so far? $\endgroup$ – Nate Eldredge Jul 5 '11 at 1:57
  • $\begingroup$ @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. $\endgroup$ – André Nicolas Jul 5 '11 at 4:26

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

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

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

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


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 $x^{2}=t$.

  • $\begingroup$ 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. $\endgroup$ – tzs Jul 6 '11 at 3:59

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