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

Halla $$$\int e^{- 2 y}\, dy$$$.

Sea $$$u=- 2 y$$$.

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

Entonces,

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

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 $$$u=- 2 y$$$:

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

Por lo tanto,

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

Añade la constante de integración:

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

$$$\int e^{- 2 y}\, dy = - \frac{e^{- 2 y}}{2} + C$$$A