Question

Evaluate the following integral:

`int_0^1 sqrt(x(x - 1))  "d"x`

Answer 1

`int_0^1 sqrt(x(x - 1))  "d"x = int_0^1 sqrt(x^2 - x)  "d"x`

= `int_0^1 sqrt((x - 1/2)^2 - (1/2)^2)`

This is of the form `int_0^2 sqrt(x^2 - "a"^2)`

= `[((x - 1/2))/2 sqrt((x - 1/2)^2 - (1/2)^2) = 1/((4)(2)) log|(x - 1/2) + sqrt(x(x - 1))|]_0^1`

= `((1 - 1/2))/2 sqrt((1 - 1/2)^2 - (1/2)^2) - 1/8 log|(1 - 1/2) + sqrt(1(1 - 1))|`

= `(-((-1)/2))/2 sqrt(((-1)/2)^2 - (1/2)^2) + 1/8 log|(-1)/2 + sqrt(0(0 - 1))|`

= `1/4 (0) - 1/8 log 1/2 + 1/4 (0) - 1/8 log 1/2`

= 0