Integration by parts help!

**bloomsgal8** · May 30th 2008, 09:41 AM

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

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

**Moo** · May 30th 2008, 09:50 AM

Hello,

Originally Posted by bloomsgal8

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

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

First of all, remember the formula :

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

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.

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

~

**bloomsgal8** · May 30th 2008, 10:12 AM

Thank you, that helps! The only problems is I dont know how to integrate e^ax...

**Moo** · May 30th 2008, 10:16 AM

Originally Posted by bloomsgal8

Thank you, that helps! The only problems is I dont know how to integrate e^ax...

Hmm...

By a substitution for example :

$\int e^{ax} dx$

Let $t=ax$

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

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

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

And the integral of $e^t$ is known ..

**Krizalid** · May 30th 2008, 03:41 PM

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

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