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

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

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

Then $$$du=\left(\frac{x}{8} - 2\right)^{\prime }dx = \frac{dx}{8}$$$ (steps can be seen »), and we have that $$$dx = 8 du$$$.

Thus,

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

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=8$$$ and $$$f{\left(u \right)} = u^{3}$$$:

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

Apply the power rule $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=3$$$:

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

Recall that $$$u=\frac{x}{8} - 2$$$:

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

Therefore,

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

Simplify:

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

Add the constant of integration:

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

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