# How do you find the integral int_0^1x*sqrt(1-x^2)dx ?

Sep 22, 2014

Begin by making a $u$-substitution.

Let $u = 1 - {x}^{2}$

$\int x \sqrt{u} \mathrm{dx} = \int {u}^{\frac{1}{2}} x \mathrm{dx}$

$\mathrm{du} = - 2 x \mathrm{dx}$

$\frac{\mathrm{du}}{- 2} = \frac{- 2 x \mathrm{dx}}{- 2}$

$\frac{- 1}{2} \cdot \mathrm{du} = x \mathrm{dx}$

$\int {u}^{\frac{1}{2}} x \mathrm{dx}$, Note that $x \mathrm{dx}$ can be replaced with $\frac{- 1}{2} \cdot \mathrm{du}$

$\int {u}^{\frac{1}{2}} \cdot \frac{- 1}{2} \cdot \mathrm{du}$

$= \frac{- 1}{2} \int {u}^{\frac{1}{2}} \mathrm{du}$

$= \frac{- 1}{2} \left[{u}^{\frac{3}{2}} / \left(\frac{3}{2}\right)\right]$

$= \frac{- 1}{2} \left[{u}^{\frac{3}{2}} \cdot \left(\frac{2}{3}\right)\right]$

$= \frac{- 1}{2} \left[\frac{2 {u}^{\frac{3}{2}}}{3}\right]$

$= \left(- 1\right) \left[\frac{{u}^{\frac{3}{2}}}{3}\right]$

$= - \left[\frac{{u}^{\frac{3}{2}}}{3}\right]$, now switch from $u$ back to $1 - {x}^{2}$

$= - {\left[\frac{{\left(1 - {x}^{2}\right)}^{\frac{3}{2}}}{3}\right]}_{0}^{1}$, remember to put the original boundaries

$= - \left[\frac{{\left(1 - {\left(1\right)}^{2}\right)}^{\frac{3}{2}}}{3} - \frac{{\left(1 - {\left(0\right)}^{2}\right)}^{\frac{3}{2}}}{3}\right]$

$= - \left[\frac{{\left(1 - 1\right)}^{\frac{3}{2}}}{3} - \frac{{\left(1 - 0\right)}^{\frac{3}{2}}}{3}\right]$

$= - \left[\frac{{\left(0\right)}^{\frac{3}{2}}}{3} - \frac{{\left(1\right)}^{\frac{3}{2}}}{3}\right]$

$= - \left[\frac{0}{3} - \frac{1}{3}\right]$

$= - \left[- \frac{1}{3}\right]$

$= \frac{1}{3} \to S o l u t i o n$