Why the integral of $e^{-x}\;$ is $\;-e^{-x}$, and not $e^{-x}$? I thought that the integral of $e^{x}$ is always $e^{x}$. Why does it change its sign to a negative when there is a negative exponent?
 A: If you differentiate each of $e^{−x}$ and $−e^{−x}$, which of the two gives you $e^{−x}$ as its derivative?
The phenomenon you are seeing is due to "inverting" the chain rule, so to speak. 
If we let $\;u = -x,\;$ then $\;du = -dx \;\implies \;dx = -du.\;$ So 
$$\int e^{-x} dx \quad = \quad\int e^u (- du) \quad = \quad - \int e^u \, du\quad = \quad -e^u +C \quad = \quad-e^{-x}+C$$
A: If $g(x)$ is the anti-derivative of $f(x)$, then $g(-x)$ is the antiderivative of $-f(-x)$.
This follows from Chain rule, since if $$\dfrac{dg(x)}{dx} = f(x),$$ we then have $$\dfrac{dg(-x)}{dx} = \left. \dfrac{dg(y)}{dy} \right\vert_{y=-x} \cdot \left. \dfrac{dy}{dx} \right \vert_{y=-x} = \left. f(y) \right \vert_{y=-x} \times \dfrac{d(-x)}{dx} = f(-x) \times (-1) = -f(-x)$$
A: $\int e^x dx$ is always $e^x$. In this case, you are really looking at $\int e^u du$, where $u=-x$ (and therefore, $du=-dx$). Thus $\int e^{-x} dx = -\int e^u du = -e^u = -e^{-x}$.
A: Also, worth mentioning is that the integral of $ e^x $ is $e^x$ when the integral is taken with respect to x.  Similarly, one could say the integral of $e^{-x}$ with respect to $-x$ is $e^{-x}$; in other words,
$$\int e^{-x}d(-x) = e^{-x} $$
while,
$$ \int e^{-x} dx = -e^{-x} $$
