Question

What is `int_(0)^(1) (1+x)(1-x)^(1/2)dx`?

Answer 1

The answer is `=14/15`

Explanation:

Calculate the indefinite integral first.

Let `u=1-x`, `=>`, `du=-dx`

`1+x=1+1-u=2-u`

Therefore, the integral is

`I=int(1+x)sqrt(1-x)dx=int(u-2)sqrtudu`

`=int(u^(3/2-2u^(1/2)))du`

`=2/5u^(5/2)-4/3u^(3/2)`

`=2/5(1-x)^(5/2)-4/3(1-x)^(3/2)+C`

Calculate the definite integral

`int_0^1(1+x)sqrt(1-x)dx=[2/5(1-x)^(5/2)-4/3(1-x)^(3/2)]_0^1`

`=(0)-(2/5-4/3)`

`=14/15`

Answer 2

`14/15`

Explanation:

Using product rule treating (1+x) as first function and `(1-x)^(1/2` as 2nd function, the given integral would be

`(1+x) int (1-x)^(1/2) dx- int d/dx (1+x) int (1-x)^(1/2) dx`

=`(1+x) (-2/3) (1-x)^(3/2)-int (-2/3) (1-x)^(3/2) dx`

=`- 2/3 (1+x)(1-x)^(3/2) + 2/3 int (1-x)^(3/2) dx`

=`-2/3 (1+x)(1-x)^(3/2) + 2/3 (-2/5) (1-x)^(5/2)`

=`-2/3 (1+x) (1-x)^(3/2) - 4/15 (1-x)^(5/2) ] _0^1`

=`2/3 +4/15`

=`14/15`