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

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

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

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

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

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

Por lo tanto,

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

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

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

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 $$$u=- x$$$:

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

Por lo tanto,

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

Añade la constante de integración:

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

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