Question

Question #73dd0

Answer 1

`1/3`

Explanation:

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

`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`: `u=1-0^2=1`

For `x=1`: `u=1-1^2=0`

So we now have

`-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)` as `u^(1/2)` so it's easier to integrate:

`1/2 int_0^1 u^(1/2) du`

Now integrate:

`1/2 (u^(3/2))/(3/2)`

Take out the `1/3/2` and then plug in the integrands:

`(1/2)(1/(3/2)) (1^(3/2)-0^(3/2))`

Simplify:

`(1/3)(1)=1/3`