Tricky integration

**deezy** · November 30th 2011, 07:34 PM

$\int_0^2 r^3 \sqrt{4-r^2} dr$

Not sure how to integrate this... I tried a u substitution for $4-r^2$ but couldn't find anything to replace $r^3$ with.

**Amer** · November 30th 2011, 07:50 PM

try sub u=4-r^2
so r^2 = 4-u

**Prove It** · November 30th 2011, 07:56 PM

Originally Posted by deezy

$\int_0^2 r^3 \sqrt{4-r^2} dr$

Not sure how to integrate this... I tried a u substitution for $4-r^2$ but couldn't find anything to replace $r^3$ with.

$\displaystyle \begin{align*} \int{r^3\sqrt{4 - r^2}\,dr} = -\frac{1}{2}\int{-2r\cdot r^2\sqrt{4 - r^2}\,dr} \end{align*}$

Let $\displaystyle \begin{align*} u = 4 - r^2 \implies du = -2r\,dr \end{align*}$ and the integral becomes

$\displaystyle \begin{align*} -\frac{1}{2}\int{-2r\cdot r^2\sqrt{4 - r^2}\,dr} &= -\frac{1}{2}\int{\left(4 - u\right)\sqrt{u}\,du} \\ &= -\frac{1}{2}\int{4u^{\frac{1}{2}} - u^{\frac{3}{2}}\,du} \\ &= -\frac{1}{2}\left(\frac{4u^{\frac{3}{2}}}{\frac{3}{ 2}} - \frac{u^{\frac{5}{2}}}{\frac{5}{2}}\right) + C \end{align*}$

Now after you change the terminals to u terminals and insert them, you will have the answer for the definite integral.

**Intrusion** · November 30th 2011, 10:47 PM

You can also do

$u = r^2$
$du = 2rdr$
$dr = \frac{1}{2}du$

so $\frac{1}{2}\int (4-u)udu$

**Prove It** · November 30th 2011, 11:28 PM

Originally Posted by Intrusion

You can also do

$u = r^2$
$du = 2rdr$
$dr = \frac{1}{2}du$

so $\frac{1}{2}\int (4-u)udu$

Where did the square root go?

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