Integral of $$$e^{- u}$$$

Find $$$\int e^{- u}\, du$$$.

Let $$$v=- u$$$.

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

So,

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

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

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

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

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

Recall that $$$v=- u$$$:

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

Therefore,

$$\int{e^{- u} d u} = - e^{- u}$$

Add the constant of integration:

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

$$$\int e^{- u}\, du = - e^{- u} + C$$$A