Integral of $$$e^{- \frac{x}{5}}$$$

Find $$$\int e^{- \frac{x}{5}}\, dx$$$.

Let $$$u=- \frac{x}{5}$$$.

Then $$$du=\left(- \frac{x}{5}\right)^{\prime }dx = - \frac{dx}{5}$$$ (steps can be seen »), and we have that $$$dx = - 5 du$$$.

Thus,

$${\color{red}{\int{e^{- \frac{x}{5}} d x}}} = {\color{red}{\int{\left(- 5 e^{u}\right)d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=-5$$$ and $$$f{\left(u \right)} = e^{u}$$$:

$${\color{red}{\int{\left(- 5 e^{u}\right)d u}}} = {\color{red}{\left(- 5 \int{e^{u} d u}\right)}}$$

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

$$- 5 {\color{red}{\int{e^{u} d u}}} = - 5 {\color{red}{e^{u}}}$$

Recall that $$$u=- \frac{x}{5}$$$:

$$- 5 e^{{\color{red}{u}}} = - 5 e^{{\color{red}{\left(- \frac{x}{5}\right)}}}$$

Therefore,

$$\int{e^{- \frac{x}{5}} d x} = - 5 e^{- \frac{x}{5}}$$

Add the constant of integration:

$$\int{e^{- \frac{x}{5}} d x} = - 5 e^{- \frac{x}{5}}+C$$

$$$\int e^{- \frac{x}{5}}\, dx = - 5 e^{- \frac{x}{5}} + C$$$A