Integral de $$$- \sqrt{3 - x}$$$

Encontre $$$\int \left(- \sqrt{3 - x}\right)\, dx$$$.

Aplique a regra do múltiplo constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ usando $$$c=-1$$$ e $$$f{\left(x \right)} = \sqrt{3 - x}$$$:

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

Seja $$$u=3 - x$$$.

Então $$$du=\left(3 - x\right)^{\prime }dx = - dx$$$ (veja os passos »), e obtemos $$$dx = - du$$$.

A integral torna-se

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

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

$$- {\color{red}{\int{\left(- \sqrt{u}\right)d u}}} = - {\color{red}{\left(- \int{\sqrt{u} d u}\right)}}$$

Aplique a regra da potência $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ com $$$n=\frac{1}{2}$$$:

$${\color{red}{\int{\sqrt{u} d u}}}={\color{red}{\int{u^{\frac{1}{2}} d u}}}={\color{red}{\frac{u^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}}}={\color{red}{\left(\frac{2 u^{\frac{3}{2}}}{3}\right)}}$$

Recorde que $$$u=3 - x$$$:

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

Portanto,

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

Adicione a constante de integração:

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

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