Integral de $$$e^{32 y}$$$

Halla $$$\int e^{32 y}\, dy$$$.

Sea $$$u=32 y$$$.

Entonces $$$du=\left(32 y\right)^{\prime }dy = 32 dy$$$ (los pasos pueden verse »), y obtenemos que $$$dy = \frac{du}{32}$$$.

Por lo tanto,

$${\color{red}{\int{e^{32 y} d y}}} = {\color{red}{\int{\frac{e^{u}}{32} d u}}}$$

Aplica la regla del factor constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ con $$$c=\frac{1}{32}$$$ y $$$f{\left(u \right)} = e^{u}$$$:

$${\color{red}{\int{\frac{e^{u}}{32} d u}}} = {\color{red}{\left(\frac{\int{e^{u} d u}}{32}\right)}}$$

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

$$\frac{{\color{red}{\int{e^{u} d u}}}}{32} = \frac{{\color{red}{e^{u}}}}{32}$$

Recordemos que $$$u=32 y$$$:

$$\frac{e^{{\color{red}{u}}}}{32} = \frac{e^{{\color{red}{\left(32 y\right)}}}}{32}$$

Por lo tanto,

$$\int{e^{32 y} d y} = \frac{e^{32 y}}{32}$$

Añade la constante de integración:

$$\int{e^{32 y} d y} = \frac{e^{32 y}}{32}+C$$

$$$\int e^{32 y}\, dy = \frac{e^{32 y}}{32} + C$$$A