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
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Use the indicated entry in the Table of Integrals to evaluate the integral.
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2 Answers
See the answer below:
# 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
# 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 #