What is f(x) = int e^(4x-1)-e^(3x-2)+e^x dx if f(2) = 3 ?

1 Answer
Feb 12, 2018

f(x)=(e^(4x-1)-e^7)/4+(e^(3x-2)-e^4)/3+e^x-e^2+3

Explanation:

We can write f(x)=inte^(4x-1)-e^(3x-2)+e^xdx as

f(x)=inte^(4x-1)dx-inte^(3x-2)dx+inte^xdx

Now considering u=4x-1, we have du=4dx and

inte^(4x-1)dx=inte^u(du)/4=e^u/4=e^(4x-1)/4

and similarly using v=3x-2, we can get inte^(3x-2)dx=inte^v(dv)/3=e^v/3=e^(3x-2)/3

and hence f(x)=e^(4x-1)/4+e^(3x-2)/3+e^x+c

As f(2)=3, we have e^(4xx2-1)/4+e^(3xx2-2)/3+e^2+c=3

or e^7/4+e^4/3+e^2+c=3

Hence c=3-e^7/4-e^4/3-e^2 and

f(x)=e^(4x-1)/4+e^(3x-2)/3+e^x+3-e^7/4-e^4/3-e^2

or f(x)=(e^(4x-1)-e^7)/4+(e^(3x-2)-e^4)/3+e^x-e^2+3