1. ## integrate

$\displaystyle \frac{1}{2}xe^{\frac{-x}{2}}$

i used parts and wondering wether im correct.thats to the power of -x/2 (its a bit small)!

2. Originally Posted by skystar
$\displaystyle \frac{1}{2}xe^{\frac{-x}{2}}$

i used parts and wondering wether im correct.thats to the power of -x/2 (its a bit small)!
That would probably be the easiest way to integrate it. Sub in u = x/2 first, though.

-Dan

3. I get

$\displaystyle \int \frac{1}{2} x e^{- \frac{x}{2}} dx = -e^{-\frac{x}{2}}(x+2)$

4. i got different nooooo!

5. can someone show me how im such an idiot

6. Originally Posted by skystar
can someone show me how im such an idiot

7. ok first of all substutute $\displaystyle u=\frac{x}{2}$

therefore $\displaystyle x=2u$ and $\displaystyle du=\frac{1}{2}dx$

so we have
$\displaystyle 2 \int u e^{-u} du$

using parts once we have

$\displaystyle 2(-ue^{-u}+\int e^{-u} du)$

$\displaystyle 2(-ue^{-u}-e^{-u})+c$

$\displaystyle -2e^{-u}(u+1)+c$

$\displaystyle -2e^{-\frac{x}{2}}(\frac{x}{2}+1)+c$

$\displaystyle -2e^{-\frac{x}{2}}(\frac{x+2}{2})+c$

$\displaystyle -e^{-\frac{x}{2}}(x+2)+c$