What is f(x) = int xsqrt(x^2-1) dx if f(3) = 0 ? Calculus Techniques of Integration Evaluating the Constant of Integration 1 Answer Eddie 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) Answer link Related questions How do you find the constant of integration for intf'(x)dx if f(2)=1? What is a line integral? What is f(x) = int x^3-x if f(2)=4 ? What is f(x) = int x^2+x-3 if f(2)=3 ? What is f(x) = int xe^x if f(2)=3 ? What is f(x) = int x - 3 if f(2)=3 ? What is f(x) = int x^2 - 3x if f(2)=1 ? What is f(x) = int 1/x if f(2)=1 ? What is f(x) = int 1/(x+3) if f(2)=1 ? What is f(x) = int 1/(x^2+3) if f(2)=1 ? See all questions in Evaluating the Constant of Integration Impact of this question 1175 views around the world You can reuse this answer Creative Commons License