# integration question (again)

• Jan 21st 2007, 07:46 PM
cinder
integration question (again)
Sorry, but it's been a while since I've done calculus.

Evaluate the integral:

$\displaystyle \int re^{r/2}dr$

Do you let $\displaystyle u = r$ and $\displaystyle dv = e^{r/2}dr$?

From there, I'm not sure what to do.
• Jan 21st 2007, 08:00 PM
Jameson
$\displaystyle \int udv = uv - \int vdu$

• Jan 21st 2007, 08:01 PM
Soroban
Hello, cinder!

Quote:

Evaluate the integral: .$\displaystyle \int re^{\frac{r}{2}}dr$

Do you let: $\displaystyle u = r$ and $\displaystyle dv = e^{\frac{r}{2}}dr$? . . . . Yes!

Integration by parts . . .

. . $\displaystyle \begin{array}{ccc}u = r & dv = e^{\frac{r}{2}}dr \\ du = dr & v = 2e^{\frac{r}{2}} \end{array}$

And we have: .$\displaystyle 2re^{\frac{r}{2}} - 2\int e^{\frac{r}{2}}\,dr \;=\;2re^{\frac{r}{2}} - 4e^{\frac{r}{2}} + C \;=\;2e^{\frac{r}{2}}(r - 2) + C$

• Jan 21st 2007, 08:06 PM
cinder
I really don't remember (or get) what you do to the $\displaystyle u$ and $\displaystyle dv$ to change them like that. :confused:
• Jan 21st 2007, 08:23 PM
CaptainBlack
Quote:

Originally Posted by cinder
Sorry, but it's been a while since I've done calculus.

Evaluate the integral:

$\displaystyle \int re^{r/2}dr$

Do you let $\displaystyle u = r$ and $\displaystyle dv = e^{r/2}dr$?

From there, I'm not sure what to do.

set $\displaystyle u=r$, $\displaystyle dv=e^{r/2}$, then:

$\displaystyle \int re^{r/2}dr=uv-\int v\,du=r\,2e^{r/2}-2\int e^{r/2}dr=r\,2e^{r/2}-4 e^{r/2}=2e^{r/2}(r-2)$

RonL