# Question #73dd0

Apr 15, 2017

$\frac{1}{3}$

#### Explanation:

Set $u = 1 - {x}^{2}$. Then $\mathrm{du} = - 2 x$ We don't have a $- 2$ though. So we'll multiply by $- \frac{2}{-} 2$. Put the $- 2$ on the top of the numerator inside the integral and leave the $\frac{1}{-} 2$ outside the integral.

$\frac{1}{-} 2 {\int}_{0}^{1} - 2 x \sqrt{1 - {x}^{2}} \mathrm{dx}$

Make the substitution. But note that if you substitute, you have to change the integrand:

For $x = 0$: $u = 1 - {0}^{2} = 1$

For $x = 1$: $u = 1 - {1}^{2} = 0$

So we now have

$- \frac{1}{2} {\int}_{1}^{0} \sqrt{u} \mathrm{du}$

We can flip the integrand by using the negative on the outside. We can also write $\sqrt{u}$ as ${u}^{\frac{1}{2}}$ so it's easier to integrate:

$\frac{1}{2} {\int}_{0}^{1} {u}^{\frac{1}{2}} \mathrm{du}$

Now integrate:

$\frac{1}{2} \frac{{u}^{\frac{3}{2}}}{\frac{3}{2}}$

Take out the $\frac{1}{3} / 2$ and then plug in the integrands:

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

Simplify:

$\left(\frac{1}{3}\right) \left(1\right) = \frac{1}{3}$