What is #f(x) = int (x-1)^3 dx# if #f(-1) = 1 #? Calculus Techniques of Integration Evaluating the Constant of Integration 1 Answer Noah G Jan 3, 2017 #f(x) = 1/4(x- 1)^4 - 3# Explanation: Let #u = x - 1#. Then #du = dx#. #f(x) = int u^3 du# #f(x) = 1/4u^4 + C# #f(x) = 1/4(x- 1)^4 + C# Now, solve for #C# knowing that when #x= -1#, #y = 1#. #1 = 1/4(-1 - 1)^4 + C# #1 = 1/4(16) + C# #1 = 4 + C# #C = -3# Hopefully this helps! 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 1258 views around the world You can reuse this answer Creative Commons License