Long time no post, revising some calculus from Stewart's book and got stuck on a problem in ch5 Problems Plus.
Problem Statement:
$$
\text{If} \int_0^4 e^{(x-2)^4}\mathrm{d}x = k, \;\; \text{find} \;\; \int_0^4 xe^{(x-2)^4}\mathrm{d}x
$$
My attempt:
First thought was that this should be something similar to integrating $e^x \sin(x)$, i.e. integrate by parts twice and solve for the unknown integral. I thought it could be something similar here if I could rearrange the second integral with some clever substitution. My attempt was roughly this:
$$
\begin{aligned}
\int_0^4 xe^{(x-2)^4}\mathrm{d}x & = \int_{-2}^2(u+2)e^{u^4}\mathrm{d}u \\
&= \int_{-2}^{2}ue^{u^4}\mathrm{d}u + 2\int_{-2}^{2}e^{u^4}\mathrm{d}u \\
&= \int_0^4(x-2)e^{(x-2)^4}\mathrm{d}x + 2\int_0^4 e^{(x-2)^4}\mathrm{d}x \\
&= \frac{1}{2}\int_0^4 e^{w^2}\mathrm{d}w + 2k
\end{aligned}
$$
Now it looks like I've simplified the integral but I don't think the $e^{w^2}$ integral has a closed form and I'm not sure where to take it from here. I tried writing it out as a limit of a Riemann sum just to see if I can maybe compute the limit, i.e. get the integral from first principles, but that left me with a nasty limit:
$$
\lim_{n \to \infty} \sum_{i=0}^{n} \frac{4}{n} e^{\frac{16i^2}{n}}
$$
so this seems like a bit of a dead end.
Any pointers for an approach? Not looking for a full solution. Mind, this is at the end of a chapter on definite/indefinite integrals, fundamental theorem of calculus and substitution rule. So it shouldn't be anything too complicated. I feel like I'm missing a simple trick somewhere.
Best Answer
The first integral in your penultimate line is actually $0$. By the substitution $u=x-2$, $$\int_0^4 (x-2)e^{(x-2)^4} dx = \int_{-2}^2 ue^{u^4} du = 0 $$ due to the oddness of the integrand. Your substitution, $w=(x-2)^2$, doesn’t work because $(x-2)^2$ is not a one-one function in $(0,4)$.