1. ## 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.

2. $\displaystyle \int udv = uv - \int vdu$

3. Hello, cinder!

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

4. I really don't remember (or get) what you do to the $\displaystyle u$ and $\displaystyle dv$ to change them like that.

5. 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