# Tricky integration

• Nov 30th 2011, 07:34 PM
deezy
Tricky integration
$\displaystyle \int_0^2 r^3 \sqrt{4-r^2} dr$

Not sure how to integrate this... I tried a u substitution for $\displaystyle 4-r^2$ but couldn't find anything to replace $\displaystyle r^3$ with.
• Nov 30th 2011, 07:50 PM
Amer
Re: Tricky integration
try sub u=4-r^2
so r^2 = 4-u
• Nov 30th 2011, 07:56 PM
Prove It
Re: Tricky integration
Quote:

Originally Posted by deezy
$\displaystyle \int_0^2 r^3 \sqrt{4-r^2} dr$

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

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

\displaystyle \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.
• Nov 30th 2011, 10:47 PM
Intrusion
Re: Tricky integration
You can also do

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

so $\displaystyle \frac{1}{2}\int (4-u)udu$
• Nov 30th 2011, 11:28 PM
Prove It
Re: Tricky integration
Quote:

Originally Posted by Intrusion
You can also do

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

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

Where did the square root go?