What is #f(x) = int xe^(2x) + 3x^3 dx# if #f(-1 ) = 1 #?

1 Answer
bp
Dec 11, 2016

#f(x)=(xe^(2x))/2 -e^(2x)/4 +(3x^4)/4 +3/(4e^2) -3/4#

Explanation:

#int xe^(2x) +3x^3 dx= int xe^(2x) dx +int 3x^3 dx# +C

=#(xe^(2x)) /2 -int e^(2x) /2dx + (3x^4) /4#+C

f(x)=#(xe^(2x))/2 -e^(2x)/4 +(3x^4)/4 +C#

Now letting x= -1, #f(-1)=1= -1/(2e^2) -1/(4e^2) +3/4 +C#

This gives C= #3/(4e^2)-3/4#

Therefore #f(x)=(xe^(2x))/2 -e^(2x)/4 +(3x^4)/4 +3/(4e^2) -3/4#

Related questions
See all questions in Evaluating the Constant of Integration
Impact of this question
1241 views around the world
You can reuse this answer
Creative Commons License