Integral de $$$e^{- 3 x}$$$

Halla $$$\int e^{- 3 x}\, dx$$$.

Sea $$$u=- 3 x$$$.

Entonces $$$du=\left(- 3 x\right)^{\prime }dx = - 3 dx$$$ (los pasos pueden verse »), y obtenemos que $$$dx = - \frac{du}{3}$$$.

Entonces,

$${\color{red}{\int{e^{- 3 x} d x}}} = {\color{red}{\int{\left(- \frac{e^{u}}{3}\right)d u}}}$$

Aplica la regla del factor constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ con $$$c=- \frac{1}{3}$$$ y $$$f{\left(u \right)} = e^{u}$$$:

$${\color{red}{\int{\left(- \frac{e^{u}}{3}\right)d u}}} = {\color{red}{\left(- \frac{\int{e^{u} d u}}{3}\right)}}$$

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

$$- \frac{{\color{red}{\int{e^{u} d u}}}}{3} = - \frac{{\color{red}{e^{u}}}}{3}$$

Recordemos que $$$u=- 3 x$$$:

$$- \frac{e^{{\color{red}{u}}}}{3} = - \frac{e^{{\color{red}{\left(- 3 x\right)}}}}{3}$$

Por lo tanto,

$$\int{e^{- 3 x} d x} = - \frac{e^{- 3 x}}{3}$$

Añade la constante de integración:

$$\int{e^{- 3 x} d x} = - \frac{e^{- 3 x}}{3}+C$$

$$$\int e^{- 3 x}\, dx = - \frac{e^{- 3 x}}{3} + C$$$A