Evaluate $\displaystyle\int\limits_0^{\infty}\frac x{20}e^{-x/20} dx$

Question

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?

Answer 1

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

$$\boxed{20}$$

Answer 2

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 $$\int^\infty_0 \frac x {20} e^{-x/20} dx = \left[-20e^{-x/20}\right]^\infty_0 = 20.$$

Answer 3

\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.