What is f(x) = int e^x-x dx if f(-1) = 1 ? Calculus Techniques of Integration Evaluating the Constant of Integration 1 Answer Bill K. Feb 24, 2016 f(x)=e^(x)-x^2/2+3/2-e^{-1} Explanation: The general antiderivative is \int (e^{x}-x)\ dx=e^{x}-x^{2}/2+C Therefore, f(x)=e^{x}-x^{2}/2+C for some constant C. Since f(-1)=1, we get 1=e^{-1}-(-1)^{2}/2+C=e^{-1}-1/2+C, which implies that C=3/2-e^{-1}. Hence, f(x)=e^(x)-x^2/2+3/2-e^{-1}. 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 1411 views around the world You can reuse this answer Creative Commons License