Can someone please show me how to do this? Thank you so much!

$\displaystyle a \int_0^{+\infty} x \, e^{-ax} \, dx$

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- May 30th 2008, 09:41 AMbloomsgal8Integration by parts help!
Can someone please show me how to do this? Thank you so much!

$\displaystyle a \int_0^{+\infty} x \, e^{-ax} \, dx$ - May 30th 2008, 09:50 AMMoo
Hello,

First of all, remember the formula :

$\displaystyle \int_m^n u(x) \cdot v'(x) dx=\left[u(x) \cdot v(x)\right]_m^n-\int_m^n u'(x) \cdot v(x) dx$

To sum up, you will have to integrate one function and derivate another one.

When you have a polynomial (here, x) and an exponential, integrate the exponential and derivate the polynomial.

---> $\displaystyle u(x)=x$ and $\displaystyle v'(x)=e^{ax}$

~ - May 30th 2008, 10:12 AMbloomsgal8
Thank you, that helps! The only problems is I dont know how to integrate e^ax...(Worried)

- May 30th 2008, 10:16 AMMoo
Hmm...

By a substitution for example :

$\displaystyle \int e^{ax} dx$

Let $\displaystyle t=ax$

therefore, $\displaystyle \frac{dt}{dx}=a$

---> $\displaystyle dx=\frac 1a \cdot dt$

We substitute : $\displaystyle \int e^{ax} dx=\int \frac 1a \cdot e^t dt=\frac 1a \int e^t dt$

And the integral of $\displaystyle e^t$ is known .. :) - May 30th 2008, 03:41 PMKrizalid
This is just a special case related to Gamma function. Have you seen what Gamma function is? Look at wiki, it's pretty clear. (Or some other place you want.)

As for the problem, make the substitution $\displaystyle z=ax$ and the integral becomes $\displaystyle \frac{1}{a^{2}}\int_{0}^{\infty }{ze^{-z}\,dz}=\frac{\Gamma (1)}{a^{2}}=\frac{1}{a^{2}}.$ (Full answer is actually $\displaystyle \frac1a.$)