# Integrating exponential

1. Aug 24, 2008

### dj023102

if f(x) = pi*xe^(-x^2)
integrating this function if the lower boundary is 0 and the upper boundary is infinity is the answer pi*(2e-1). is this right?

Last edited: Aug 24, 2008
2. Aug 24, 2008

### d_leet

Can you think of an appropriate substitution that may help with this integral?

3. Aug 24, 2008

### dj023102

i think i worked out.

4. Aug 24, 2008

### d_leet

Could you show your work? I'm pretty sure that isn't the correct answer, but it is difficult to tell if you don't show how you arrived at it.

5. Aug 24, 2008

### dj023102

sure i used substitution method.
let u = -x^2
then du/dx=-2x, dx=du/-2x
this gives the integral of xe^u/-2xdu
=pi/2 integrate e^u lower boundary 0 upper boundary infinity
then separating the two integral, one integral with lower boundary 0 and upper boundary 1 and the second integral with lower boundary 1 and the upper boundary infinty.
the first integral gives e^1-e^0 and the second integral we get e^1.
then adding them together we get
(pi/2) * (e-1+e) is that right?

6. Aug 24, 2008

### Defennder

It shouldn't be e^u, it's slightly off. And what do you mean by "separating the integral"? You mean integration by parts? It's not needed here.

7. Aug 24, 2008

### HallsofIvy

Staff Emeritus
More simply, if u= -x2, then du= -2x dx or -(1/2)du= dx

$xe^{-x^2}dx$ becomes $e^{-x^2}(xdx)= -(1/2)e^u du$

When x= 0, u= 0 and when x="infinity", u is -"infinity" The integral becomes
$$\int_0^\infty xe^{-x^2} dx= -\frac{1}{2}\int_0^{-\infty} e^u du$$