What is #f(x) = int 5x-2xe^(x)dx# if #f(0)=-2 #? Calculus Techniques of Integration Evaluating the Constant of Integration 1 Answer Cem Sentin Jul 11, 2017 #f(x)=int5x*dx-int2x*e^x*dx# #=5/2*x^2-(2x*e^x-int2e^x*dx)# #=5/2*x^2-(2x-2)*e^x+C# Let #x=0#, #C+2=-2#, hence #C=-4#. Thus, #f(x)=5/2*x^2-(2x-2)*e^x-4# Explanation: 1) Take integral. 2) Impose #f(0)=-2# condition. 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 1284 views around the world You can reuse this answer Creative Commons License