What is f(x) = int e^(2x-1)-e^(3-x)+e^x dxf(x)=∫e2x−1−e3−x+exdx if f(2) = 3 f(2)=3?

1 Answer
1s2s2p
Dec 9, 2017

f(x)=e^(2x-1)/2+e^(3-x)+e^x+3-e^3/2+e+e^2f(x)=e2x−12+e3−x+ex+3−e32+e+e2
f(x)=e^(2x-1)/2+e^(3-x)+e^x+3.06456947f(x)=e2x−12+e3−x+ex+3.06456947

Explanation:

Using inte^(ax+b)dx=e^(ax+b)/a∫eax+bdx=eax+ba

Our integration gives us:
f(x)=e^(2x-1)/2+e^(3-x)+e^x+Cf(x)=e2x−12+e3−x+ex+C

We also know that e^(2(2)-1)/2+e^(3-2)+e^2+C=3e2(2)−12+e3−2+e2+C=3

e^3/2+e+e^2+C=3e32+e+e2+C=3

C=3-e^3/2+e+e^2~~3.06456947C=3−e32+e+e2≈3.06456947

f(x)=e^(2x-1)/2+e^(3-x)+e^x+3-e^3/2+e+e^2f(x)=e2x−12+e3−x+ex+3−e32+e+e2
f(x)=e^(2x-1)/2+e^(3-x)+e^x+3.06456947f(x)=e2x−12+e3−x+ex+3.06456947

Related questions
See all questions in Evaluating the Constant of Integration
Impact of this question
1404 views around the world
You can reuse this answer
Creative Commons License