Integral of $$$\frac{e^{- x}}{25}$$$

Find $$$\int \frac{e^{- x}}{25}\, dx$$$.

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{1}{25}$$$ and $$$f{\left(x \right)} = e^{- x}$$$:

$${\color{red}{\int{\frac{e^{- x}}{25} d x}}} = {\color{red}{\left(\frac{\int{e^{- x} d x}}{25}\right)}}$$

Let $$$u=- x$$$.

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

The integral becomes

$$\frac{{\color{red}{\int{e^{- x} d x}}}}{25} = \frac{{\color{red}{\int{\left(- e^{u}\right)d u}}}}{25}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=-1$$$ and $$$f{\left(u \right)} = e^{u}$$$:

$$\frac{{\color{red}{\int{\left(- e^{u}\right)d u}}}}{25} = \frac{{\color{red}{\left(- \int{e^{u} d u}\right)}}}{25}$$

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

$$- \frac{{\color{red}{\int{e^{u} d u}}}}{25} = - \frac{{\color{red}{e^{u}}}}{25}$$

Recall that $$$u=- x$$$:

$$- \frac{e^{{\color{red}{u}}}}{25} = - \frac{e^{{\color{red}{\left(- x\right)}}}}{25}$$

Therefore,

$$\int{\frac{e^{- x}}{25} d x} = - \frac{e^{- x}}{25}$$

Add the constant of integration:

$$\int{\frac{e^{- x}}{25} d x} = - \frac{e^{- x}}{25}+C$$

$$$\int \frac{e^{- x}}{25}\, dx = - \frac{e^{- x}}{25} + C$$$A