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 1082 views around the world You can reuse this answer Creative Commons License