Integral of $$$e^{- \frac{x}{3}}$$$

Find $$$\int e^{- \frac{x}{3}}\, dx$$$.

Let $$$u=- \frac{x}{3}$$$.

Then $$$du=\left(- \frac{x}{3}\right)^{\prime }dx = - \frac{dx}{3}$$$ (steps can be seen »), and we have that $$$dx = - 3 du$$$.

The integral can be rewritten as

$${\color{red}{\int{e^{- \frac{x}{3}} d x}}} = {\color{red}{\int{\left(- 3 e^{u}\right)d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=-3$$$ and $$$f{\left(u \right)} = e^{u}$$$:

$${\color{red}{\int{\left(- 3 e^{u}\right)d u}}} = {\color{red}{\left(- 3 \int{e^{u} d u}\right)}}$$

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

$$- 3 {\color{red}{\int{e^{u} d u}}} = - 3 {\color{red}{e^{u}}}$$

Recall that $$$u=- \frac{x}{3}$$$:

$$- 3 e^{{\color{red}{u}}} = - 3 e^{{\color{red}{\left(- \frac{x}{3}\right)}}}$$

Therefore,

$$\int{e^{- \frac{x}{3}} d x} = - 3 e^{- \frac{x}{3}}$$

Add the constant of integration:

$$\int{e^{- \frac{x}{3}} d x} = - 3 e^{- \frac{x}{3}}+C$$

$$$\int e^{- \frac{x}{3}}\, dx = - 3 e^{- \frac{x}{3}} + C$$$A