# Integral sqrt(x)*e^-x

Tags:
1. Nov 5, 2016

### ooohffff

1. The problem statement, all variables and given/known data
Evaluate the following integral:
0 √(x)* e-x dx

2. Relevant equations
0 e-x2 dx = (√π)/2

3. The attempt at a solution
So far this is what I've done:
u = x1/2
du = 1/2 x-1/2

2 ∫ e-u2 u2 du

Now, I'm not really sure what to do? Or if what I've done so far is leading me down the right path

2. Nov 5, 2016

### andrewkirk

Try writing it as $2\int_0^\infty (e^{-u^2}u)\cdot u\,du$ and doing integration by parts. If you can get to where the only remaining integral to do is like $\int_0^\infty e^{-u^2}\,du$ you can use the fact that that integrand is a constant multiplied by the pdf of a standard normal distribution, for which the integral from 0 to $\infty$ is known.

3. Nov 5, 2016

### ooohffff

Okay, so I did (I'm don't remember how to do this very well so bear with me):

w = u dw = du
dv = e-u2 u du v = -1/2 e-u2

(-u/2) e-u2 - ∫ (-1/2) e-u2 du
-(u/2) e-u2 + (1/2)∫ e-u2 du

For the left side of the equation, would I plug u back in? If so then the limit as x approaches infinity would make that part 0. For the right side, I'm unsure as to whether I just replace the integral with √π /2. If this is all true, then 2( 0 + (1/2) (√π/2)) and would the final answer be √π/2 ?

4. Nov 5, 2016

### andrewkirk

Why have your integration limits disappeared?

The integration by parts rule for definite integrals is:

$$\int_a^b f'(t)g(t)\,dt=\left[f(t)g(t)\right]_a^b-\int_a^b f(t)g'(t)\,dt$$

5. Nov 5, 2016

### ooohffff

Okay, so:

2[[(-u/2) e-u2]0 - ∫0 (-1/2) e-u2 du]
2[[(-u/2) e-u2]0 + (1/2)∫0 e-u2 du]
2[(0) + (1/2) (√π/2)] = √π/2

6. Nov 5, 2016

### PeroK

Is that $\frac{\sqrt{\pi}}{2}$ or $\sqrt{\frac{\pi}{2}}$?

7. Nov 5, 2016

### ooohffff

The first one, (√π)/2

8. Nov 5, 2016

Looks right.