Evaluate $\displaystyle\int\limits_0^{\infty}\frac x{20}e^{-x/20} dx$ I tried doing $u$-substitution and got $-20e$ as my final answer, but I think the correct answer is just $20$. I'm not sure what I did wrong, but probably had to do with plugging in infinity... could someone explain the process of solving this integral?
 A: Hint
$$\frac{x}{20} = y$$
$$20\int_0^{+\infty}ye^{-y}\ \text{d}y$$
Many ways to evaluate it. By parts once, or just knowing it's the gamma function:
$$20\int_0^{+\infty}ye^{-y}\ \text{d}y = 20\cdot\Gamma(2) = 20\cdot 1 = 20$$
What Gamma Function is
Euler Gamma Function is defined as
$$\Gamma(x) = \int_0^{+\infty}t^{x-1}e^{-t}\ \text{d}t$$
more here
https://en.wikipedia.org/wiki/Gamma_function
By parts
Simply call $f = x$ and $g' = e^{-x}$ and proceed, it easy!
In this case you applythe integration by parts use, obtaining
$$-xe^{-x}\bigg|_0^{+\infty} - \left(\int_0^{+\infty} -e^{-x}\  \text{d}x\right)$$
The first term is zero because at infinity the exponential dominates, and in zero the $x$ function dominates.
The second term is simply
$$-e^{-x}\bigg|_0^{+\infty} = -e^{-\infty} - (-e^0) = 0 - (-1) = 1$$
Remember the $20$ factor above and the answer is 
$$\boxed{20}$$
A: I think integration by parts is the way to go here, not $u$-sub. Recall, we have $$\int^\infty_0 u(x) v'(x) dx = \left[u(x)v(x) \right]_0^\infty - \int^\infty_0 u'(x) v(x) dx.$$I'm abusing notation here; of course we shouldn't just "plug in" $\infty$, we should take a limit but of course this amounts to the same thing. Putting $u(x) = x/20$ and $v(x) =-20e^{-x/20}$, we see $$\int^\infty_0 \frac x {20} e^{-x/20} dx = \left[ - xe^{-x/20}  \right]^\infty_0 + \int^\infty_0  e^{-x/20} dx.$$ The bracketed term is zero so $$\int^\infty_0 \frac x {20} e^{-x/20} dx = \left[-20e^{-x/20}\right]^\infty_0 = 20.$$
A: \begin{align}
\int_0^\infty \frac x{20}e^{-x/20} \, dx & = 20 \int_0^\infty \left(\frac x{20}\right) e^{-x/20} \, \left(\frac{dx}{20}\right) \\[6pt]
& = 2\underbrace{0 \int_0^\infty u e^{-u}\,du}_\text{substitution} = \underbrace{20 \int u\,dv = 20\left( uv - \int v\,du \right)}_\text{integration by parts} \\[10pt]
& = \left[-20 u e^{-u} \vphantom{\frac 11} \right]_0^\infty - \int_0^\infty (-e^{-u})\,du = \text{etc.}
\end{align}
To evaluate the part in brackets, you need $\lim\limits_{u\to\infty} ue^{-u} = \lim\limits_{u\to\infty} \dfrac u {e^u}$. L'Hopital's rule handles that quickly.
