Integration using substitution - applying integral of symmetric functions properties or second substitution? 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 $b=2$
$u=g(x)=x-1$
$g(a)=g(0)=(0-1)^2=(-1)^2=1$
$g(b)=g(2)=(2-1)^2=(1)^2=1$
$$\int_0^2 (x-1)e^{(x-1)^2}\;\mathrm{d}x$$
$$=\int_0^2 e^{(x-1)^2}(x-1)\;\mathrm{d}x$$
$$=\int_1^1 e^{u}\frac{1}{2}\;\mathrm{d}u$$
$$=\frac{1}{2}\int_1^1 e^{u}\;\mathrm{d}u$$
$$=0$$

note:
$$I=\int_0^2(x-1)e^{(x-1)^2}dx$$
by letting $u=x-1$ like you suggested this can be turned into:
$$I=\int_{-1}^1ue^{u^2}du$$
now notice that splitting this up gives:
$$\int_{-1}^0f(u)du+\int_0^1f(u)du=\int_0^1f(u)du+\int_0^1f(-u)du=\int_0^1f(u)du-\int_0^1f(u)du=0$$
