I need help with this integral: $\int\_0^2 (x-1)e^{(x-1)^2}\;\mathrm{d}x$

Okey. I'm choosing $u=x-1$, so $du=dx$.

$a=0$ and $b=2$

$u=g(x)=x-1$

$g(a)=g(0)=0-1=-1$

$g(b)=g(2)=2-1=1$

$$\int\_0^2 (x-1)e^{(x-1)^2}\;\mathrm{d}x$$
$$=\int\_{-1}^1 ue^{u^2}\;\mathrm{d}u$$

I'm stuck here. I was wondering if I can use one of the integral of symmetric functions properties there:

* If $f$ is even, then $\int\_{-a}^a f(x)\;\mathrm{d}x=2\int\_{0}^a f(x)\;\mathrm{d}x$
* If $f$ is odd, then $\int\_{-a}^a f(x)\;\mathrm{d}x=0$

Let be $h(u)=ue^{u^2}$, then $h(-u)=-ue^{(-u)^2}=-ue^{u^2}$. Therefore is odd. So I can use the second property and conclude that $\int\_0^2 (x-1)e^{(x-1)^2}\;\mathrm{d}x=\int\_{-1}^1 ue^{u^2}\;\mathrm{d}u=0$???

Or shall I make a second substitution with $\int\_{-1}^1 ue^{u^2}\;\mathrm{d}u$?

$v=u^2$, so $dv=2u du$, $\frac{1}{2}dv=u du$ 

$v=(1)^2=1$ and $v=(-1)^2=1$

Then,

$$\int\_0^2 (x-1)e^{(x-1)^2}\;\mathrm{d}x$$
$$=\int\_{-1}^1 ue^{u^2}\;\mathrm{d}u$$
$$=\int\_{1}^1 e^{v}\cdot\frac{1}{2}\mathrm{d}v$$
$$=\frac{1}{2}\int\_{1}^1 e^{v}\mathrm{d}v$$
$$= \tfrac{1}{2} [e^{v}] \Big|\_{1}^1$$
$$= \tfrac{1}{2} [e^{1}-e^{1}]$$
$$= \tfrac{1}{2} [e-e]$$
$$= \tfrac{1}{2} [0]$$
$$=0$$

Are these analysis all right? If so, which one: integral of symmetric functions properties or second substitution or both? If not how can I work with this? Thanks in advance.

UPDATE! I found another way to do it. Faster. 

$\int\_0^2 (x-1)e^{(x-1)^2}\;\mathrm{d}x$

Is choosing $u=(x-1)^2$, so $du=2(x-1)dx$ $\Rightarrow$ $\frac{1}{2}du=(x-1)dx$.

$a=0$ and $b=2$

$u=g(x)=x-1$

$g(a)=g(0)=(0-1)^2=(-1)^2=1$

$g(b)=g(2)=(2-1)^2=(1)^2=1$

$$\int\_0^2 (x-1)e^{(x-1)^2}\;\mathrm{d}x$$

$$=\int\_0^2 e^{(x-1)^2}(x-1)\;\mathrm{d}x$$

$$=\int\_1^1 e^{u}\frac{1}{2}\;\mathrm{d}u$$

$$=\frac{1}{2}\int\_1^1 e^{u}\;\mathrm{d}u$$

$$=0$$