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

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

Aplique a regra do múltiplo constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ usando $$$c=\frac{1}{2}$$$ e $$$f{\left(x \right)} = e^{- x}$$$:

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

Seja $$$u=- x$$$.

Então $$$du=\left(- x\right)^{\prime }dx = - dx$$$ (veja os passos »), e obtemos $$$dx = - du$$$.

Logo,

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

Aplique a regra do múltiplo constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ usando $$$c=-1$$$ e $$$f{\left(u \right)} = e^{u}$$$:

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

A integral da função exponencial é $$$\int{e^{u} d u} = e^{u}$$$:

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

Recorde que $$$u=- x$$$:

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

Portanto,

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

Adicione a constante de integração:

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

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