Integral of $$$e^{- \frac{x}{40}}$$$

Find $$$\int e^{- \frac{x}{40}}\, dx$$$.

Let $$$u=- \frac{x}{40}$$$.

Then $$$du=\left(- \frac{x}{40}\right)^{\prime }dx = - \frac{dx}{40}$$$ (steps can be seen »), and we have that $$$dx = - 40 du$$$.

So,

$${\color{red}{\int{e^{- \frac{x}{40}} d x}}} = {\color{red}{\int{\left(- 40 e^{u}\right)d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=-40$$$ and $$$f{\left(u \right)} = e^{u}$$$:

$${\color{red}{\int{\left(- 40 e^{u}\right)d u}}} = {\color{red}{\left(- 40 \int{e^{u} d u}\right)}}$$

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

$$- 40 {\color{red}{\int{e^{u} d u}}} = - 40 {\color{red}{e^{u}}}$$

Recall that $$$u=- \frac{x}{40}$$$:

$$- 40 e^{{\color{red}{u}}} = - 40 e^{{\color{red}{\left(- \frac{x}{40}\right)}}}$$

Therefore,

$$\int{e^{- \frac{x}{40}} d x} = - 40 e^{- \frac{x}{40}}$$

Add the constant of integration:

$$\int{e^{- \frac{x}{40}} d x} = - 40 e^{- \frac{x}{40}}+C$$

$$$\int e^{- \frac{x}{40}}\, dx = - 40 e^{- \frac{x}{40}} + C$$$A