Integral de $$$e^{- 2 x}$$$

Halla $$$\int e^{- 2 x}\, dx$$$.

Sea $$$u=- 2 x$$$.

Entonces $$$du=\left(- 2 x\right)^{\prime }dx = - 2 dx$$$ (los pasos pueden verse »), y obtenemos que $$$dx = - \frac{du}{2}$$$.

Entonces,

$${\color{red}{\int{e^{- 2 x} d x}}} = {\color{red}{\int{\left(- \frac{e^{u}}{2}\right)d u}}}$$

Aplica la regla del factor constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ con $$$c=- \frac{1}{2}$$$ y $$$f{\left(u \right)} = e^{u}$$$:

$${\color{red}{\int{\left(- \frac{e^{u}}{2}\right)d u}}} = {\color{red}{\left(- \frac{\int{e^{u} d u}}{2}\right)}}$$

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

$$- \frac{{\color{red}{\int{e^{u} d u}}}}{2} = - \frac{{\color{red}{e^{u}}}}{2}$$

Recordemos que $$$u=- 2 x$$$:

$$- \frac{e^{{\color{red}{u}}}}{2} = - \frac{e^{{\color{red}{\left(- 2 x\right)}}}}{2}$$

Por lo tanto,

$$\int{e^{- 2 x} d x} = - \frac{e^{- 2 x}}{2}$$

Añade la constante de integración:

$$\int{e^{- 2 x} d x} = - \frac{e^{- 2 x}}{2}+C$$

$$$\int e^{- 2 x}\, dx = - \frac{e^{- 2 x}}{2} + C$$$A