Integral of $$$e^{- \frac{y}{t}}$$$ with respect to $$$y$$$

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

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

Then $$$du=\left(- \frac{y}{t}\right)^{\prime }dy = - \frac{1}{t} dy$$$ (steps can be seen »), and we have that $$$dy = - t du$$$.

So,

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

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

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

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

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

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

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

Therefore,

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

Add the constant of integration:

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

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