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I'm stuck on this one. I've tried integration by parts but to no avail:
$$\int_{0}^{4} x e^{ (x-2)^4 } dx = \left[ x \int e^{ (x-2)^4 } dx \right]_{0}^{4} - \underbrace{\int_{0}^{4} e^{ (x-2)^4 } dx}_{=k} $$
I suspect this is a dead end. I think there's some simple trick here I'm missing. Any hints are appreciated.
$$ I = \int_{0}^{4} xe^{(x-2)^4} \quad dx $$ substitute $t = 4-x$
$$ \begin{align} I &= \int_{4}^{0} (4-t)e^{(4-t-2)^4} \quad (-dt) &= \int_{0}^{4} (4-t)e^{(t-2)^4} \quad dt \end{align}$$ Adding the 2 results $$\begin{align} 2I = 4 \int_{0}^{4} e^{(x-2)^4} \quad dx & = 4k \end{align} $$
Note that
$$k = \int_{-2}^2 dx \, e^{x^4} $$
Then
$$\int_0^4 dx \, x \, e^{(x-2)^4} = \int_{-2}^2 dx \, (x+2) e^{x^4} = 2 \int_{-2}^2 dx \, e^{x^4}= 2 k$$
Note that the result desired comes out of symmetry, as clearly
$$\int_{-2}^2 dx \, x \, e^{x^4} = 0$$
Well, the function $(x-2)\,e^{-(x-2)^4}$ is continuous and symmetric with respect to $x=2$, hence its integral over $(0,4)$ is zero. The claim easily follows through $x=(x-2)+2$.
