What is f(x) = int (x+3)^2-2x dx if f(1) = 0 ? Calculus Techniques of Integration Evaluating the Constant of Integration 1 Answer Øko Dec 19, 2017 f(x)=1/3(x+3)^3-x^2-61/3 Explanation: f(x)=int(x+3)^2-2xdx Use inth(x)+g(x)dx=inth(x)dx+intg(x)dx f(x)=int(x+3)^2dx-int2xdx f(x)=1/3(x+3)^3-x^2+C Evaluate the constant of integration 0=1/3(1+3)^3-1^2+C 0=1/3(4)^3-1+C 0=64/3-3/3+C 0=61/3+C C=-61/3 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 1386 views around the world You can reuse this answer Creative Commons License