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

enter image source here

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

enter image source here

2 Answers
Frederico Guizini S.
Aug 2, 2017

See the answer below:
enter image source here

Steve M
Aug 2, 2017

# 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 #

Related questions
Impact of this question
13622 views around the world
You can reuse this answer
Creative Commons License