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 $b=2$
$u=g(x)=x-1$
$g(a)=g(0)=(0-1)^2=(-1)^2=1$
$g(b)=g(2)=(2-1)^2=(1)^2=1$
$$\int_0^2 (x-1)e^{(x-1)^2}\;\mathrm{d}x$$
$$=\int_0^2 e^{(x-1)^2}(x-1)\;\mathrm{d}x$$
$$=\int_1^1 e^{u}\frac{1}{2}\;\mathrm{d}u$$
$$=\frac{1}{2}\int_1^1 e^{u}\;\mathrm{d}u$$
$$=0$$
 A: It looks like you've done it two ways and got the same answer. That's good.  I think they are both correct, except for your second substitution.  
The second integral does yield zero, as it should.   This is because $ue^{u^2}$ has a primitive, $\dfrac{e^{u^2}}2$, which has the same value at $1$ and $-1$.  
But the second substitution you did won't always work.  Say you had $\int_{-1}^1u^2\operatorname du$.  If you substitute $y=u^2$ and change the limits to $1$ and $1$, you get zero, which is incorrect.  
To avoid this sort of thing,  it's a good idea to draw a picture.
I guess the moral of the story is that you can't always make any substitution you want,  change the limits accordingly,  and get the right answer. You will have noticed this, or will eventually notice, fairly often. 
More accurately,  if you look at integration by substitution, it appears our $\varphi $ would have to be $\varphi (x)=\sqrt x$.  This isn't even defined on $ [-1,1]$, much less differentiable.  
A: 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:
$$\int_{-1}^0f(u)du+\int_0^1f(u)du=\int_0^1f(u)du+\int_0^1f(-u)du=\int_0^1f(u)du-\int_0^1f(u)du=0$$
