Integral of $$$23 e^{- x}$$$

Find $$$\int 23 e^{- x}\, dx$$$.

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=23$$$ and $$$f{\left(x \right)} = e^{- x}$$$:

$${\color{red}{\int{23 e^{- x} d x}}} = {\color{red}{\left(23 \int{e^{- x} d x}\right)}}$$

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

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

Therefore,

$$23 {\color{red}{\int{e^{- x} d x}}} = 23 {\color{red}{\int{\left(- e^{u}\right)d u}}}$$

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

$$23 {\color{red}{\int{\left(- e^{u}\right)d u}}} = 23 {\color{red}{\left(- \int{e^{u} d u}\right)}}$$

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

$$- 23 {\color{red}{\int{e^{u} d u}}} = - 23 {\color{red}{e^{u}}}$$

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

$$- 23 e^{{\color{red}{u}}} = - 23 e^{{\color{red}{\left(- x\right)}}}$$

Therefore,

$$\int{23 e^{- x} d x} = - 23 e^{- x}$$

Add the constant of integration:

$$\int{23 e^{- x} d x} = - 23 e^{- x}+C$$

$$$\int 23 e^{- x}\, dx = - 23 e^{- x} + C$$$A