Integral de $$$\left(\frac{x}{8} - 2\right)^{3}$$$

Halla $$$\int \left(\frac{x}{8} - 2\right)^{3}\, dx$$$.

Sea $$$u=\frac{x}{8} - 2$$$.

Entonces $$$du=\left(\frac{x}{8} - 2\right)^{\prime }dx = \frac{dx}{8}$$$ (los pasos pueden verse »), y obtenemos que $$$dx = 8 du$$$.

La integral puede reescribirse como

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

Aplica la regla del factor constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ con $$$c=8$$$ y $$$f{\left(u \right)} = u^{3}$$$:

$${\color{red}{\int{8 u^{3} d u}}} = {\color{red}{\left(8 \int{u^{3} d u}\right)}}$$

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

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

Recordemos que $$$u=\frac{x}{8} - 2$$$:

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

Por lo tanto,

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

Simplificar:

$$\int{\left(\frac{x}{8} - 2\right)^{3} d x} = \frac{\left(x - 16\right)^{4}}{2048}$$

Añade la constante de integración:

$$\int{\left(\frac{x}{8} - 2\right)^{3} d x} = \frac{\left(x - 16\right)^{4}}{2048}+C$$

$$$\int \left(\frac{x}{8} - 2\right)^{3}\, dx = \frac{\left(x - 16\right)^{4}}{2048} + C$$$A