Answer
$$\int_0^2(x-1)e^{(x-1)^2}dx=0$$
Work Step by Step
To evaluate the integral
$$\int_0^2(x-1)e^{(x-1)^2}dx$$
we will use substitution $(x-1)^2=t$ which gives us $2(x-1)dx=dt\Rightarrow (x-1)dx=\frac{dt}{2}$. The integration bounds would be: for $x=0$ we have $t=1$ and for $x=2$ we have $t=1$. Putting this into the integral we get:
$$\int_0^2(x-1)e^{(x-1)^2}dx=\int_{1}^1e^t\frac{dt}{2}=0$$
