Question

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

Answer 1

The function `f` is:

`f (x) = 1/2 (e^{2x-1} + e^{1-2x}) + e^x + 3 - e^2 - 1/2 (e^3 + e^{- 3})`.

Explanation:

By integrating the value of `f`:

`f (x) = int (e^{2 x - 1} - e^{1 - 2x} + e^x) dx =`

`= int e^{2x-1} dx - int e^{1-2x} dx + int e^x dx =`

`= 1/2 e^{2x-1} - (- 1/2) e^{1 - 2x} + e^x + C =`

`= 1/2 (e^{2x-1} + e^{1-2x}) + e^x + C`

Then, substituting the known point of `f`:

`f (2) = 1/2 (e^3 + e^{- 3}) + e^2 + C = 3`,

we can obtain the value of the constant `C`:

`C = 3 - e^2 - 1/2 (e^3 + e^{- 3})`.

The function `f` will have the following formula:

`f (x) = 1/2 (e^{2x-1} + e^{1-2x}) + e^x + 3 - e^2 - 1/2 (e^3 + e^{- 3})`.