# What is f(x) = int e^x-e^(-2x) dx if f(0)=-2 ?

Mar 8, 2018

the antiderivative of $f \left(x\right) = {e}^{x} - {e}^{-} \left(2 x\right)$

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

that is to say that the derivative of $F \left(x\right)$ is equal to $f \left(x\right)$
$= \frac{d}{\mathrm{dx}} \left({e}^{x} - {e}^{2 x} / 2 + C\right) = {e}^{x} - {e}^{- 2 x}$

now we just need to find C to get the total antiderivative of $f \left(x\right)$
and we can use the fact that at $x = 0$ the value comes out as -2
$\therefore f \left(0\right) = - 2$

therefore
${e}^{0} - {e}^{2 \cdot 0} / 2 + C = - 2$
$= 1 - \frac{1}{2} + C = - 2$

therefore,
$C = - 2 \frac{1}{2}$
=
therefore, the entire function becomes
${e}^{x} - {e}^{2 x} / 2 - 2 \frac{1}{2}$