Integral de $$$- 2 f x + 4$$$ em relação a $$$x$$$

Encontre $$$\int \left(- 2 f x + 4\right)\, dx$$$.

Integre termo a termo:

$${\color{red}{\int{\left(- 2 f x + 4\right)d x}}} = {\color{red}{\left(\int{4 d x} - \int{2 f x d x}\right)}}$$

Aplique a regra da constante $$$\int c\, dx = c x$$$ usando $$$c=4$$$:

$$- \int{2 f x d x} + {\color{red}{\int{4 d x}}} = - \int{2 f x d x} + {\color{red}{\left(4 x\right)}}$$

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

$$4 x - {\color{red}{\int{2 f x d x}}} = 4 x - {\color{red}{\left(2 f \int{x d x}\right)}}$$

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

$$- 2 f {\color{red}{\int{x d x}}} + 4 x=- 2 f {\color{red}{\frac{x^{1 + 1}}{1 + 1}}} + 4 x=- 2 f {\color{red}{\left(\frac{x^{2}}{2}\right)}} + 4 x$$

Portanto,

$$\int{\left(- 2 f x + 4\right)d x} = - f x^{2} + 4 x$$

Simplifique:

$$\int{\left(- 2 f x + 4\right)d x} = x \left(- f x + 4\right)$$

Adicione a constante de integração:

$$\int{\left(- 2 f x + 4\right)d x} = x \left(- f x + 4\right)+C$$

$$$\int \left(- 2 f x + 4\right)\, dx = x \left(- f x + 4\right) + C$$$A