Integral de $$$3 \cdot 2^{- x}$$$

Encontre $$$\int 3 \cdot 2^{- x}\, dx$$$.

Aplique a regra do múltiplo constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ usando $$$c=3$$$ e $$$f{\left(x \right)} = 2^{- x}$$$:

$${\color{red}{\int{3 \cdot 2^{- x} d x}}} = {\color{red}{\left(3 \int{2^{- x} d x}\right)}}$$

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

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

Assim,

$$3 {\color{red}{\int{2^{- x} d x}}} = 3 {\color{red}{\int{\left(- 2^{u}\right)d u}}}$$

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

$$3 {\color{red}{\int{\left(- 2^{u}\right)d u}}} = 3 {\color{red}{\left(- \int{2^{u} d u}\right)}}$$

Apply the exponential rule $$$\int{a^{u} d u} = \frac{a^{u}}{\ln{\left(a \right)}}$$$ with $$$a=2$$$:

$$- 3 {\color{red}{\int{2^{u} d u}}} = - 3 {\color{red}{\frac{2^{u}}{\ln{\left(2 \right)}}}}$$

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

$$- \frac{3 \cdot 2^{{\color{red}{u}}}}{\ln{\left(2 \right)}} = - \frac{3 \cdot 2^{{\color{red}{\left(- x\right)}}}}{\ln{\left(2 \right)}}$$

Portanto,

$$\int{3 \cdot 2^{- x} d x} = - \frac{3 \cdot 2^{- x}}{\ln{\left(2 \right)}}$$

Adicione a constante de integração:

$$\int{3 \cdot 2^{- x} d x} = - \frac{3 \cdot 2^{- x}}{\ln{\left(2 \right)}}+C$$

$$$\int 3 \cdot 2^{- x}\, dx = - \frac{3 \cdot 2^{- x}}{\ln\left(2\right)} + C$$$A