Question

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

Answer 1

`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`