# How do you integrate int xsqrt(4-x) using substitution?

Nov 3, 2016

$\int \textcolor{w h i t e}{.} x \sqrt{4 - x} \textcolor{w h i t e}{.} \mathrm{dx} = - \frac{2}{15} \left(3 x + 8\right) {\left(4 - x\right)}^{\frac{3}{2}} + C$

#### Explanation:

Let $u = 4 - x$

Then $\mathrm{du} = - \mathrm{dx}$

$\int \textcolor{w h i t e}{.} x \sqrt{4 - x} \textcolor{w h i t e}{.} \mathrm{dx} = \int \textcolor{w h i t e}{.} \left(u - 4\right) \sqrt{u} \textcolor{w h i t e}{.} \mathrm{du}$

$\textcolor{w h i t e}{\int \textcolor{w h i t e}{.} x \sqrt{4 - x} \textcolor{w h i t e}{.} \mathrm{dx}} = \int \textcolor{w h i t e}{.} {u}^{\frac{3}{2}} - 4 {u}^{\frac{1}{2}} \textcolor{w h i t e}{.} \mathrm{du}$

$\textcolor{w h i t e}{\int \textcolor{w h i t e}{.} x \sqrt{4 - x} \textcolor{w h i t e}{.} \mathrm{dx}} = \frac{2}{5} {u}^{\frac{5}{2}} - \frac{8}{3} {u}^{\frac{3}{2}} + C$

$\textcolor{w h i t e}{\int \textcolor{w h i t e}{.} x \sqrt{4 - x} \textcolor{w h i t e}{.} \mathrm{dx}} = \frac{2}{5} {\left(4 - x\right)}^{\frac{5}{2}} - \frac{8}{3} {\left(4 - x\right)}^{\frac{3}{2}} + C$

$\textcolor{w h i t e}{\int \textcolor{w h i t e}{.} x \sqrt{4 - x} \textcolor{w h i t e}{.} \mathrm{dx}} = \left(\frac{2}{5} \left(4 - x\right) - \frac{8}{3}\right) {\left(4 - x\right)}^{\frac{3}{2}} + C$

$\textcolor{w h i t e}{\int \textcolor{w h i t e}{.} x \sqrt{4 - x} \textcolor{w h i t e}{.} \mathrm{dx}} = \frac{2}{15} \left(3 \left(4 - x\right) - 20\right) {\left(4 - x\right)}^{\frac{3}{2}} + C$

$\textcolor{w h i t e}{\int \textcolor{w h i t e}{.} x \sqrt{4 - x} \textcolor{w h i t e}{.} \mathrm{dx}} = - \frac{2}{15} \left(3 x + 8\right) {\left(4 - x\right)}^{\frac{3}{2}} + C$