Integral de $$$2 x^{2} \left(2 x - 4\right)$$$

Halla $$$\int 2 x^{2} \left(2 x - 4\right)\, dx$$$.

La entrada se reescribe: $$$\int{2 x^{2} \left(2 x - 4\right) d x}=\int{x^{2} \left(4 x - 8\right) d x}$$$.

Simplificar el integrando:

$${\color{red}{\int{x^{2} \left(4 x - 8\right) d x}}} = {\color{red}{\int{4 x^{2} \left(x - 2\right) d x}}}$$

Aplica la regla del factor constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ con $$$c=4$$$ y $$$f{\left(x \right)} = x^{2} \left(x - 2\right)$$$:

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

Expand the expression:

$$4 {\color{red}{\int{x^{2} \left(x - 2\right) d x}}} = 4 {\color{red}{\int{\left(x^{3} - 2 x^{2}\right)d x}}}$$

Integra término a término:

$$4 {\color{red}{\int{\left(x^{3} - 2 x^{2}\right)d x}}} = 4 {\color{red}{\left(- \int{2 x^{2} d x} + \int{x^{3} d x}\right)}}$$

Aplica la regla de la potencia $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ con $$$n=3$$$:

$$- 4 \int{2 x^{2} d x} + 4 {\color{red}{\int{x^{3} d x}}}=- 4 \int{2 x^{2} d x} + 4 {\color{red}{\frac{x^{1 + 3}}{1 + 3}}}=- 4 \int{2 x^{2} d x} + 4 {\color{red}{\left(\frac{x^{4}}{4}\right)}}$$

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

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

Aplica la regla de la potencia $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ con $$$n=2$$$:

$$x^{4} - 8 {\color{red}{\int{x^{2} d x}}}=x^{4} - 8 {\color{red}{\frac{x^{1 + 2}}{1 + 2}}}=x^{4} - 8 {\color{red}{\left(\frac{x^{3}}{3}\right)}}$$

Por lo tanto,

$$\int{x^{2} \left(4 x - 8\right) d x} = x^{4} - \frac{8 x^{3}}{3}$$

Simplificar:

$$\int{x^{2} \left(4 x - 8\right) d x} = x^{3} \left(x - \frac{8}{3}\right)$$

Añade la constante de integración:

$$\int{x^{2} \left(4 x - 8\right) d x} = x^{3} \left(x - \frac{8}{3}\right)+C$$

$$$\int 2 x^{2} \left(2 x - 4\right)\, dx = x^{3} \left(x - \frac{8}{3}\right) + C$$$A