I need help with this integral: $ int_0 ^ 2 (x-1) e ^ {(x-1) ^ 2} ; mathrm {d} x $

Okey. I choose $ 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 could use one of the properties of the symmetric functions integral there:

- If $ f $ is then just $ int _ {- a} ^ f (x) ; mathrm {d} x = 2 int_ {0} ^ a f (x) ; mathrm {d} x $
- If $ f $ is so strange $ int _ {- a} ^ a f (x) ; mathrm {d} x = 0 $

To let $ h (u) = ue ^ {u ^ 2} $, then $ h (-u) = – ue ^ {(- u) ^ 2} = – ue ^ {u ^ 2} $, That's why it's strange. So I can use the second property and close it $ int_0 ^ 2 (x-1) e ^ {(x-1) ^ 2} ; mathrm {d} x = int _ {- 1} ^ 1 ue ^ {u ^ 2} ; mathrm {d} u = 0 $???

Or do I have a second substitution? $ int _ {- 1} ^ 1 ue ^ {u ^ 2} ; mathrm {d} u $?

$ v = u ^ 2 $, so $ dv = 2u you $. $ frac {1} {2} dv = u you $

$ 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 analyzes okay? If so, which: integral of the properties of symmetric functions or second substitution, or both? If not, how can I work with it? Thank you in advance.