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

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

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

Therefore,

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

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 $$$u=- t$$$:

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

Therefore,

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

Add the constant of integration:

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

Answer: $$$\int{e^{- t} d t}=- e^{- t}+C$$$