Integral of $$$\left(y - 4\right)^{2}$$$

Find $$$\int \left(y - 4\right)^{2}\, dy$$$.

Let $$$u=y - 4$$$.

Then $$$du=\left(y - 4\right)^{\prime }dy = 1 dy$$$ (steps can be seen »), and we have that $$$dy = du$$$.

Therefore,

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

Apply the power rule $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=2$$$:

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

Recall that $$$u=y - 4$$$:

$$\frac{{\color{red}{u}}^{3}}{3} = \frac{{\color{red}{\left(y - 4\right)}}^{3}}{3}$$

Therefore,

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

Add the constant of integration:

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

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