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

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

Aplica la regla del factor constante $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ con $$$c=-1$$$ y $$$f{\left(t \right)} = e^{- t}$$$:

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

Sea $$$u=- t$$$.

Entonces $$$du=\left(- t\right)^{\prime }dt = - dt$$$ (los pasos pueden verse »), y obtenemos que $$$dt = - du$$$.

Entonces,

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

Aplica la regla del factor constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ con $$$c=-1$$$ y $$$f{\left(u \right)} = e^{u}$$$:

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

La integral de la función exponencial es $$$\int{e^{u} d u} = e^{u}$$$:

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

Recordemos que $$$u=- t$$$:

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

Por lo tanto,

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

Añade la constante de integración:

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

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