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

Find $$$\int \left(- e^{- y}\right)\, dy$$$.

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

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

Let $$$u=- y$$$.

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

Thus,

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

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

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

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

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

Recall that $$$u=- y$$$:

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

Therefore,

$$\int{\left(- e^{- y}\right)d y} = e^{- y}$$

Add the constant of integration:

$$\int{\left(- e^{- y}\right)d y} = e^{- y}+C$$

$$$\int \left(- e^{- y}\right)\, dy = e^{- y} + C$$$A