What is F(x) = int xe^(-x) dx if F(0) = 2 ? Calculus Techniques of Integration Evaluating the Constant of Integration 1 Answer Eddie Sep 13, 2016 F(x) = 3 - e^(-x) (x+1) Explanation: F(x) = int xe^(-x) dx = int xd/dx(-e^(-x)) dx = x(-e^(-x)) + int d/dx(x) e^(-x) dx + C = -x e^(-x) + int e^(-x) dx + C = -x e^(-x) - e^(-x) +C \implies F(x) = - e^(-x) (x+1) +C F(0) = 2 implies 2 = - 1 + C \implies F(x) = 3 - e^(-x) (x+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 1854 views around the world You can reuse this answer Creative Commons License