Question #73dd0

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Question #73dd0

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Answer 1 · 2017-04-15T18:02:08

$\frac{1}{3}$ $1/3$

Explanation:

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

$\frac{1}{- 2} \int_{0}^{1} - 2 x \sqrt{1 - x^{2}} d x$ $1/-2 int_0^1 -2xsqrt(1-x^2)dx$

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

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

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

So we now have

$- \frac{1}{2} \int_{1}^{0} \sqrt{u} d u$ $-1/2 int_1^0 sqrt(u) du$

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

$\frac{1}{2} \int_{0}^{1} u^{\frac{1}{2}} d u$ $1/2 int_0^1 u^(1/2) du$

Now integrate:

$\frac{1}{2} \frac{u^{\frac{3}{2}}}{\frac{3}{2}}$ $1/2 (u^(3/2))/(3/2)$

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

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

Simplify:

$(\frac{1}{3}) (1) = \frac{1}{3}$ $(1/3)(1)=1/3$

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Question #73dd0

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