# What is  int_(1/3)^(2/3) \ x^3sqrt(4-9x^2) \ dx ?

Sep 24, 2017

${\int}_{\frac{1}{3}}^{\frac{2}{3}} \setminus {x}^{3} \sqrt{4 - 9 {x}^{2}} \setminus \mathrm{dx} = \frac{11}{405} \sqrt{3}$

#### Explanation:

We seek:

$I = {\int}_{\frac{1}{3}}^{\frac{2}{3}} \setminus {x}^{3} \sqrt{4 - 9 {x}^{2}} \setminus \mathrm{dx}$

We can perform a substitution of the form:

$u = 4 - 9 {x}^{2} \iff 9 {x}^{2} = 4 - u$, and $\frac{\mathrm{du}}{\mathrm{dx}} = - 18 x$

This substitution will require an associated change of limits from $x$ to $u$, and we have:

When $x = \left\{\begin{matrix}\frac{1}{3} \\ \frac{2}{3}\end{matrix}\right. \implies u = \left\{\begin{matrix}3 \\ 0\end{matrix}\right.$

And we manipulate the integral as follows:

$I = {\int}_{\frac{1}{3}}^{\frac{2}{3}} \setminus \frac{9 {x}^{2}}{9} \frac{- 18 x}{- 18} \sqrt{4 - 9 {x}^{2}} \setminus \mathrm{dx}$
$\setminus \setminus = - \frac{1}{162} \setminus {\int}_{\frac{1}{3}}^{\frac{2}{3}} \setminus \left(9 {x}^{2}\right) \left(- 18 x\right) \sqrt{4 - 9 {x}^{2}} \setminus \mathrm{dx}$
$\setminus \setminus = - \frac{1}{162} \setminus {\int}_{3}^{0} \setminus \left(4 - u\right) \sqrt{u} \setminus \mathrm{du}$
$\setminus \setminus = \frac{1}{162} \setminus {\int}_{0}^{3} \setminus \left(4 - u\right) {u}^{\frac{1}{2}} \setminus \mathrm{du}$
$\setminus \setminus = \frac{1}{162} \setminus {\int}_{0}^{3} \setminus 4 {u}^{\frac{1}{2}} - {u}^{\frac{3}{2}} \setminus \mathrm{du}$
$\setminus \setminus = \frac{1}{162} \setminus {\left[\frac{4 {u}^{\frac{3}{2}}}{\frac{3}{2}} - \frac{{u}^{\frac{5}{2}}}{\frac{5}{2}}\right]}_{0}^{3}$

$\setminus \setminus = \frac{2}{162} {\left[\frac{4 {u}^{\frac{3}{2}}}{3} - \frac{{u}^{\frac{5}{2}}}{5}\right]}_{0}^{3}$

$\setminus \setminus = \frac{1}{81} \left(\frac{4}{3} {\left(3\right)}^{\frac{3}{2}} - \frac{1}{5} {\left(3\right)}^{\frac{5}{2}}\right)$

$\setminus \setminus = \frac{1}{81} \left(\frac{4}{3} \left(3 \sqrt{3}\right) - \frac{1}{5} \left(9 \sqrt{3}\right)\right)$

$\setminus \setminus = \frac{1}{81} \left(\frac{12}{3} - \frac{9}{5}\right) \sqrt{3}$

$\setminus \setminus = \frac{1}{81} \left(\frac{11}{5}\right) \sqrt{3}$

$\setminus \setminus = \frac{11}{405} \sqrt{3}$

So the last answer is correct.