Answer
$\int_0^4xe^{(x-2)^4}dx=2k$
Work Step by Step
Let $y=x-2$.
Then, $\int_0^4e^{(x-2)^4}dx=\int_{-2}^2e^{y^4}dy$.
$\int_0^4xe^{(x-2)^4}dx=\int_{-2}^2(y+2)e^{y^4}dy$
$=\int_{-2}^2ye^{y^4}+2e^{y^4}dy$ (Apply the properties for integrals)
$=\int_{-2}^2ye^{y^4}dy+2\int_{-2}^2e^{y^4}dy$ (See that $f(y)=ye^{4}$ is an odd function)
$=0+2\int_{-2}^2e^{y^4}dy$ (Since if $f$ is an odd function, $\int_{-a}^af =0$)
$=2\int_{-2}^2e^{y^4}dy$
$=2\int_0^4e^{(x-2)^4}dx$
$=2k$
Thus,
$\int_0^4xe^{(x-2)^4}dx=2k$