Integral of $$$e^{- \frac{x}{a}}$$$ with respect to $$$x$$$

Find $$$\int e^{- \frac{x}{a}}\, dx$$$.

Let $$$u=- \frac{x}{a}$$$.

Then $$$du=\left(- \frac{x}{a}\right)^{\prime }dx = - \frac{1}{a} dx$$$ (steps can be seen »), and we have that $$$dx = - a du$$$.

The integral becomes

$${\color{red}{\int{e^{- \frac{x}{a}} d x}}} = {\color{red}{\int{\left(- a e^{u}\right)d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=- a$$$ and $$$f{\left(u \right)} = e^{u}$$$:

$${\color{red}{\int{\left(- a e^{u}\right)d u}}} = {\color{red}{\left(- a \int{e^{u} d u}\right)}}$$

The integral of the exponential function is $$$\int{e^{u} d u} = e^{u}$$$:

$$- a {\color{red}{\int{e^{u} d u}}} = - a {\color{red}{e^{u}}}$$

Recall that $$$u=- \frac{x}{a}$$$:

$$- a e^{{\color{red}{u}}} = - a e^{{\color{red}{\left(- \frac{x}{a}\right)}}}$$

Therefore,

$$\int{e^{- \frac{x}{a}} d x} = - a e^{- \frac{x}{a}}$$

Add the constant of integration:

$$\int{e^{- \frac{x}{a}} d x} = - a e^{- \frac{x}{a}}+C$$

$$$\int e^{- \frac{x}{a}}\, dx = - a e^{- \frac{x}{a}} + C$$$A