Integral of $$$e^{- 5 t}$$$

Find $$$\int e^{- 5 t}\, dt$$$.

Let $$$u=- 5 t$$$.

Then $$$du=\left(- 5 t\right)^{\prime }dt = - 5 dt$$$ (steps can be seen »), and we have that $$$dt = - \frac{du}{5}$$$.

The integral becomes

$${\color{red}{\int{e^{- 5 t} d t}}} = {\color{red}{\int{\left(- \frac{e^{u}}{5}\right)d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=- \frac{1}{5}$$$ and $$$f{\left(u \right)} = e^{u}$$$:

$${\color{red}{\int{\left(- \frac{e^{u}}{5}\right)d u}}} = {\color{red}{\left(- \frac{\int{e^{u} d u}}{5}\right)}}$$

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

$$- \frac{{\color{red}{\int{e^{u} d u}}}}{5} = - \frac{{\color{red}{e^{u}}}}{5}$$

Recall that $$$u=- 5 t$$$:

$$- \frac{e^{{\color{red}{u}}}}{5} = - \frac{e^{{\color{red}{\left(- 5 t\right)}}}}{5}$$

Therefore,

$$\int{e^{- 5 t} d t} = - \frac{e^{- 5 t}}{5}$$

Add the constant of integration:

$$\int{e^{- 5 t} d t} = - \frac{e^{- 5 t}}{5}+C$$

$$$\int e^{- 5 t}\, dt = - \frac{e^{- 5 t}}{5} + C$$$A