Can someone please show me how to do this? Thank you so much!
$\displaystyle a \int_0^{+\infty} x \, e^{-ax} \, dx$
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 $\displaystyle v'(x)=e^{ax}$
~
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 ..
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 $\displaystyle \frac1a.$)