# Math Help - Integration Question

1. ## Integration Question

ok so maybe im just being stupid here...

$\int_0^\infty xe^{-x} dx$

so by using integration by parts....

= $\left[ -xe^{-x} - \int -e^{-x} dx \right]_0^\infty$

= $\left[ -xe^{-x} - e^{-x} \right]_0^\infty$

= $\left[ (-x-1)e^{-x}\right]_0^\infty$

so is that right? if so, pluging in $\infty$ doesn't really make sense or give me an answer that looks right!

any help will be much appreciated!

2. First, you have to take the limit, so

$\int_0^\infty xe^{-x}\,dx=\lim_{a\to\infty}\int_0^a xe^{-x}\,dx$

Now integrate by parts and take the limit.

3. Originally Posted by UbikPkd
ok so maybe im just being stupid here...

$\int_0^\infty xe^{-x} dx$

so by using integration by parts....

= $\left[ -xe^{-x} - \int -e^{-x} dx \right]_0^\infty$

= $\left[ -xe^{-x} - e^{-x} \right]_0^\infty$

= $\left[ (-x-1)e^{-x}\right]_0^\infty$

so is that right? if so, pluging in $\infty$ doesn't really make sense or give me an answer that looks right!

any help will be much appreciated!
To continue:
$= \left[ (-x-1)e^{-x}\right]_0^\infty$

$= \lim_{x \to \infty}[(-x - 1)e^{-x}] - (-0 - 1)e^{-0}$

$= 0 + 1 = 1$

-Dan

4. thanks! could you also tell me why the first bit = 0 ?

is that: $(-\infty -1) e ^ {-\infty} = 0$ if so how does that work?

5. Try putting ever larger numbers in place of infinity,
you will find that it quickly gives you smaller and smaller answers.

In the LIMIT it TENDS toward zero.

Infinity is of course a mathematical construct that has NO "real" existance
(I'm sure some people would argue with that!)
But it is useful...

Mark.

6. Originally Posted by UbikPkd
thanks! could you also tell me why the first bit = 0 ?

is that: $(-\infty -1) e ^ {-\infty} = 0$ if so how does that work?
It is not $(-\infty -1) e ^ {-\infty}$. $\infty$ is not a number so you can't substitute it in directly.

But we can do this:
$\lim_{x \to \infty} (-x - 1)e^{-x} = - \lim_{x \to \infty} \frac{x + 1}{e^x}$

Now, as x gets very large, what happens to this fraction? It gets smaller and smaller because $e^x$ grows much faster than x + 1 does. So
$\lim_{x \to \infty} (-x - 1)e^{-x} = - \lim_{x \to \infty} \frac{x + 1}{e^x} = 0$

-Dan

7. yep, i get it, thanks!