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

Find $$$\int \left(- e^{- t}\right)\, dt$$$.

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

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

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

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

The integral can be rewritten as

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

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,

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

Add the constant of integration:

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

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