Answer
$\int_{0}^{2}(x-1)~e^{(x-1)^2}~dx = 0$
Work Step by Step
$\int_{0}^{2}(x-1)~e^{(x-1)^2}~dx=\int_{0}^{1}(x-1)~e^{(x-1)^2}~dx+\int_{1}^{2}(x-1)~e^{(x-1)^2}~dx$
Let $u = (x-1)^2$
$\frac{du}{dx} = 2(x-1)$
$dx = \frac{du}{2(x-1)}$
When $x = 0$, then $u = 1$
When $x = 1$, then $u = 0$
When $x = 2$, then $u = 1$
$\int_{1}^{0} (x-1)e^u~\frac{du}{2(x-1)}+\int_{0}^{1} (x-1)e^u~\frac{du}{2(x-1)}$
$=\int_{1}^{0} \frac{1}{2}e^u~du+\int_{0}^{1} \frac{1}{2}e^u~du$
$=-\int_{0}^{1} \frac{1}{2}e^u~du+\int_{0}^{1} \frac{1}{2}e^u~du$
$= 0$
