find the integral 0-1 of (xe)^-(x)^2 dx with derivatives and integrals of the exponential function, i am very lost please help, i know that i plugg in the 0 and 1 into u and dx but after that i am very confused and the answer is (e-1)/(2e)^2 help

Question

find the integral 0-1 of (xe)^-(x)^2 dx  with derivatives and integrals of the exponential function, i am very lost please help, i know that i plugg in the 0 and 1 into u and dx but after that i am very confused and the answer is (e-1)/(2e)^2 help

aum · Answer

Is it
$$\int\limits_{0}^{1}xe^{-x^2}dx$$

anonymous · Answer

yes,

anonymous · Answer

a nice u sub will do it in one step

aum · Answer

Let u = x^2
du = 2xdx
xdx = du/2

when x = 0, u = 0
when x = 1, u = 1
$ \large \int\limits_{0}^{1}xe^{-x^2}dx = \frac 12\int\limits_{0}^{1}e^{-u}du$

anonymous · Answer

or you could have made $u=-x^2$ etc

aum · Answer

yes.

anonymous · Answer

i can see how the demoninator is 2e^2, but how is the numerator 1-e, kinda lost there

anonymous · Answer

do you take the derivative of e^-u?

aum · Answer

If the problem I wrote in my first reply is correct, the answer I am getting is (e - 1) / (2e).

anonymous · Answer

oh yes, that is it, mixed the two