Answer
$2k$
Work Step by Step
Use the substitution $t=x-2$; it follows that the limits of integration of the integral with respect to $t$ are
when $x=0$ so $t=0-2=-2$
when $x=4$ so $t=4-2=2$
$$dt=(x-2)'dx=dx$$
$$\int_0^4 xe^{(x-2)^4}dx=\int_{-2}^2(t+2)e^{t^4}dt=\int_{-2}^2te^{t^4}dt+2\int_{-2}^2e^{t^4}dt$$
Similary by doing the same method for the given integral it follows that:
$$\int_0^4 e^{(x-2)^4}dx=\int_{-2}^2 e^{t^4}dt=k$$
so:
$$\int_0^4 xe^{(x-2)^4}dx=\int_{-2}^2(t+2)e^{t^4}dt=\int_{-2}^2te^{t^4}dt+2k$$
Since $f(-t)=-te^{(-t)^4}=-te^{t^4}=-f(t)$ it follows that $f$ is odd and $ \int_{-2}^2te^{t^4}dt=0$ so:
$$\int_0^4 xe^{(x-2)^4}dx=\int_{-2}^2(t+2)e^{t^4}dt=0+2k=2k$$
