What is F(x) = int x^2+e^(4-2x) dx if F(0) = 1 ? Calculus Techniques of Integration Evaluating the Constant of Integration 1 Answer Ratnaker Mehta Sep 21, 2016 F(x)=x^3/3-e^4/2(e^(-2x)+1)+1. Explanation: F(x)=int(x^2+e^(4-2x))dx=intx^2dx+inte^(4-2x)dx rArr F(x)=x^3/3+e^(4-2x)/-2+C To determine C, we use the cond. F(0)=1 rArr 0^3/3+e^(4-0)/-2+C=1. rArr C=1-e^4/2. Sub.img, in F(x), we get, F(x)=x^3/3-e^(4-2x)/2+1-e^4/2, i.e., x^3/3-e^4/2(e^(-2x)+1)+1. Enjoy Maths.! 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 1356 views around the world You can reuse this answer Creative Commons License