# Thread: integral test.

1. ## integral test.

I believe this problem involves the integral test, its very new to me. Please help me with this one.

integral int 0 infinity

xe^5x dx

thankyou for any help.

2. Originally Posted by rcmango
I believe this problem involves the integral test, its very new to me. Please help me with this one.

integral int 0 infinity

xe^5x dx

thankyou for any help.
The integral diverges. Maybe you mean xe^(-5x)

3. Yes I do mean xe^(-5x) dx sorry about that. where do I start?

thanks alot.

4. More generally:

Let $m\ge0,\,n\ge1.$

$\int_0^\infty {x^m e^{ - nx} \,dx} = \frac{{m!}}
{{n^{m + 1} }}.$

Proof:

Set $u=nx,$ then the integral becomes

$\frac{1}
{{n^{m + 1} }}\int_0^\infty {u^m e^{ - u} \,du} = \frac{{\Gamma (m + 1)}}
{{n^{m + 1} }} = \frac{{m!}}
{{n^{m + 1} }}\quad\blacksquare$

Now plug $(m,n)=(1,5)$ and we're done.

5. Originally Posted by rcmango
Yes I do mean xe^(-5x) dx sorry about that. where do I start?

thanks alot.
Have you learned integration by parts? Yes - then use that technique. No - then give up until you have.

6. Originally Posted by mr fantastic
Have you learned integration by parts? Yes - then use that technique. No - then give up until you have.
A shortcut to integration by parts is tabular integration.

7. Originally Posted by colby2152
A shortcut to integration by parts is tabular integration.
I don't really see this as being a shortcut. However I will say that it does organize the process pretty well for a long chain of integration by parts.

-Dan

8. $\int_{0}^{\infty}xe^{5x}dx$

$\lim_{L\rightarrow{\infty}}\int_{0}^{L}xe^{5x}dx$

IBP: $\int{u}dv=uv-\int{v}du$

Now, use parts by letting $u=x, \;\ dv=e^{5x}dx, \;\ du=dx, \;\ v=\frac{1}{5}e^{5x}$

Putting it all together via parts gives us:

$\lim_{L\rightarrow{\infty}}\left[e^{5L}(\frac{L}{5}-\frac{1}{25})+\frac{1}{25}\right]$

Now, what's the limit as $L\rightarrow{\infty}$?.

For what it's worth, here's a graph showing the Riemann sum. As you can see, it's pretty large just from 0 to 10.

9. $\int_0^{\infty} xe^{-5x} dx = -\frac{1}{5}xe^{-5x}\bigg|_0^{\infty} + \frac{1}{5} \int_0^{\infty} e^{-5x} dx = - \frac{1}{25} e^{-5x} \bigg|_0^{\infty} = \frac{1}{25}$