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.
More generally:
Let $\displaystyle m\ge0,\,n\ge1.$
$\displaystyle \int_0^\infty {x^m e^{ - nx} \,dx} = \frac{{m!}}
{{n^{m + 1} }}.$
Proof:
Set $\displaystyle u=nx,$ then the integral becomes
$\displaystyle \frac{1}
{{n^{m + 1} }}\int_0^\infty {u^m e^{ - u} \,du} = \frac{{\Gamma (m + 1)}}
{{n^{m + 1} }} = \frac{{m!}}
{{n^{m + 1} }}\quad\blacksquare$
Now plug $\displaystyle (m,n)=(1,5)$ and we're done.
A shortcut to integration by parts is tabular integration.
$\displaystyle \int_{0}^{\infty}xe^{5x}dx$
$\displaystyle \lim_{L\rightarrow{\infty}}\int_{0}^{L}xe^{5x}dx$
IBP: $\displaystyle \int{u}dv=uv-\int{v}du$
Now, use parts by letting $\displaystyle u=x, \;\ dv=e^{5x}dx, \;\ du=dx, \;\ v=\frac{1}{5}e^{5x}$
Putting it all together via parts gives us:
$\displaystyle \lim_{L\rightarrow{\infty}}\left[e^{5L}(\frac{L}{5}-\frac{1}{25})+\frac{1}{25}\right]$
Now, what's the limit as $\displaystyle L\rightarrow{\infty}$?.
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.