Question

How do you evaluate `int_0^1` `sqrt(x - x^2)` dx?

Use the indicated entry in the Table of Integrals to evaluate the integral.

`int_0^1` `sqrt(x - x^2)` dx; entry 113

~~~~~~~~~~~~~~~~~~~~~~~

Answer 1

See the answer below:

Answer 2

`int_0^1 \ sqrt(x-x^2) \ dx = pi/8`

Explanation:

Your own question provides the solution to the integral so it simply a case of using the information that you have been provided with :

We seek:

`I = int_0^1 \ sqrt(x-x^2) \ dx`

You are given that:

`int \ sqrt(2au-u^2) \ du = (u-1)/2sqrt(2au-u^2) + a^2/2cos^(-1)((a-u)/a) + C`

Compare `2au-u^2` with `x-x^2` and we see that `a=1/2` provides the solution that we seek; thus, with `a=1/2` we get using the formula provided , that:

`I = [(u-1)/2sqrt(2(1/2)u-u^2) + (1/2)^2/2cos^(-1)(((1/2)-u)/(1/2))]_0^1`

`\ \ = [(u-1)/2sqrt(u-u^2) + 1/8cos^(-1)(1-2u)]_0^1`

`\ \ = (0 + 1/8cos^(-1)(-1)) - (0 + 1/8cos^(-1)(1) )`

`\ \ = (0 + 1/8 pi) - (0 + 0 )`

`\ \ = pi/8`