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

Feb 20, 2017

The function $f$ is:

$f \left(x\right) = \frac{1}{2} \left({e}^{2 x - 1} + {e}^{1 - 2 x}\right) + {e}^{x} + 3 - {e}^{2} - \frac{1}{2} \left({e}^{3} + {e}^{- 3}\right)$.

#### Explanation:

By integrating the value of $f$:

$f \left(x\right) = \int \left({e}^{2 x - 1} - {e}^{1 - 2 x} + {e}^{x}\right) \mathrm{dx} =$

$= \int {e}^{2 x - 1} \mathrm{dx} - \int {e}^{1 - 2 x} \mathrm{dx} + \int {e}^{x} \mathrm{dx} =$

$= \frac{1}{2} {e}^{2 x - 1} - \left(- \frac{1}{2}\right) {e}^{1 - 2 x} + {e}^{x} + C =$

$= \frac{1}{2} \left({e}^{2 x - 1} + {e}^{1 - 2 x}\right) + {e}^{x} + C$

Then, substituting the known point of $f$:

$f \left(2\right) = \frac{1}{2} \left({e}^{3} + {e}^{- 3}\right) + {e}^{2} + C = 3$,

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

$C = 3 - {e}^{2} - \frac{1}{2} \left({e}^{3} + {e}^{- 3}\right)$.

The function $f$ will have the following formula:

$f \left(x\right) = \frac{1}{2} \left({e}^{2 x - 1} + {e}^{1 - 2 x}\right) + {e}^{x} + 3 - {e}^{2} - \frac{1}{2} \left({e}^{3} + {e}^{- 3}\right)$.