Integral of $$$e^{- \frac{y}{2}}$$$

Find $$$\int e^{- \frac{y}{2}}\, dy$$$.

Let $$$u=- \frac{y}{2}$$$.

Then $$$du=\left(- \frac{y}{2}\right)^{\prime }dy = - \frac{dy}{2}$$$ (steps can be seen »), and we have that $$$dy = - 2 du$$$.

The integral becomes

$${\color{red}{\int{e^{- \frac{y}{2}} d y}}} = {\color{red}{\int{\left(- 2 e^{u}\right)d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=-2$$$ and $$$f{\left(u \right)} = e^{u}$$$:

$${\color{red}{\int{\left(- 2 e^{u}\right)d u}}} = {\color{red}{\left(- 2 \int{e^{u} d u}\right)}}$$

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

$$- 2 {\color{red}{\int{e^{u} d u}}} = - 2 {\color{red}{e^{u}}}$$

Recall that $$$u=- \frac{y}{2}$$$:

$$- 2 e^{{\color{red}{u}}} = - 2 e^{{\color{red}{\left(- \frac{y}{2}\right)}}}$$

Therefore,

$$\int{e^{- \frac{y}{2}} d y} = - 2 e^{- \frac{y}{2}}$$

Add the constant of integration:

$$\int{e^{- \frac{y}{2}} d y} = - 2 e^{- \frac{y}{2}}+C$$

$$$\int e^{- \frac{y}{2}}\, dy = - 2 e^{- \frac{y}{2}} + C$$$A