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

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Answer 1 · 2016-08-02T12:36:06

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

# 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) #