Evaluate $\int_0^2 x^2 e^{-x^3} \, dx$ using substitution The integral in question is:
$$\int_0^2 x^2 e^{-x^3} \, dx.$$
I'm not sure if I should put $u=x^2$ and $du=2x\,dx$ which would lead to $(1/2)\,du=x\,dx$ or put $u=e^{-x^3}$ which I'm not sure would change anything since the derivative of $e^x$ is $e^x$.
 A: Better method:
$$\begin{align} \int^2_0 x^2e^{-x^3}dx &= - \frac{1}{3} \int^2_0 \frac{d}{dx}e^{-x^3}dx
\\
&= - \frac{1}{3} e^{-x^3}\bigg|_0^2 \\
\end{align}$$ 
Noticing this is somewhat similar to choosing the substitution $u=x^3$ (and therefore using $du=3x^2 dx$)
A: I guess someone will answer using the substitution $\;u=x^3\;$ , but you don't really need this. Observe that using the Chain Rule, we have
$$\int f'(x)\,e^{f(x)}\;dx=e^{f(x)}+C$$
and in our case 
$$x^2=-\frac13(-3x^2)=-\frac13(-x^3)'$$
so that
$$\int x^2e^{-x^3}dx=-\frac13\int(-x^3)'e^{-x^3}dx=-\frac13e^{-x^3}+C$$
A: By Inspection: the term in front of the exponential ($x^2$) is the differential of the exponent ($x^3$). So what we can say is 'assume the solution to the integral is $I = e^{-x^3}$'. Does this work? Lets check by differentiating this, which should give back the integrand that you have:
$\frac{d}{dx}e^{-x^3} = -3x^2\cdot e^{-x^3}$, by the chain rule and so we can see that our original solution was just off by a factor of -3. Hence our true solution is $\,-\frac{1}{3}e^{-x^3}$. This would be the ideal way of solving this since you can just see the answer.
By substitution: method requires that you take $u = x^3 \implies du = 3u^{\frac{2}{3}}dx$. Hence you get (omitting the limits):
$\int dx\,x^2e^{-x^3} \equiv \frac{1}{3}\int du \, e^{-u}$, which is easy to solve.
A: Here's what I consider a good hint:
$$
\int_0^2 e^{-x^3}\Big(x^2 \, dx\Big)
$$
If anyone doesn't understand this hint, then that immediately tells me specifically what they need to work on understanding.
