Integral de $$$e^{- \frac{t}{4}}$$$

Halla $$$\int e^{- \frac{t}{4}}\, dt$$$.

Sea $$$u=- \frac{t}{4}$$$.

Entonces $$$du=\left(- \frac{t}{4}\right)^{\prime }dt = - \frac{dt}{4}$$$ (los pasos pueden verse »), y obtenemos que $$$dt = - 4 du$$$.

Entonces,

$${\color{red}{\int{e^{- \frac{t}{4}} d t}}} = {\color{red}{\int{\left(- 4 e^{u}\right)d u}}}$$

Aplica la regla del factor constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ con $$$c=-4$$$ y $$$f{\left(u \right)} = e^{u}$$$:

$${\color{red}{\int{\left(- 4 e^{u}\right)d u}}} = {\color{red}{\left(- 4 \int{e^{u} d u}\right)}}$$

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

$$- 4 {\color{red}{\int{e^{u} d u}}} = - 4 {\color{red}{e^{u}}}$$

Recordemos que $$$u=- \frac{t}{4}$$$:

$$- 4 e^{{\color{red}{u}}} = - 4 e^{{\color{red}{\left(- \frac{t}{4}\right)}}}$$

Por lo tanto,

$$\int{e^{- \frac{t}{4}} d t} = - 4 e^{- \frac{t}{4}}$$

Añade la constante de integración:

$$\int{e^{- \frac{t}{4}} d t} = - 4 e^{- \frac{t}{4}}+C$$

$$$\int e^{- \frac{t}{4}}\, dt = - 4 e^{- \frac{t}{4}} + C$$$A