What is #f(x) = int xsqrt(x^2-1) dx# if #f(3) = 0 #?

1 Answer
Aug 2, 2016

#f(x) = 1/3(x^2-1)^(3/2) - (16 sqrt 2)/(3) #

Explanation:

# int xsqrt(x^2-1) dx #

#= int \ d/dx ( 2/3* 1/2(x^2-1)^(3/2) ) dx #

#= int \ d/dx ( 1/3(x^2-1)^(3/2) ) dx #

#= 1/3(x^2-1)^(3/2) + C #

using the IV:

#0 = 1/3(3^2-1)^(3/2) + C implies C = - (16 sqrt 2)/(3)#

So #f(x) = 1/3(x^2-1)^(3/2) - (16 sqrt 2)/(3) #