Integral of $$$s^{2} \left(s - 1\right)$$$

Find $$$\int s^{2} \left(s - 1\right)\, ds$$$.

Expand the expression:

$${\color{red}{\int{s^{2} \left(s - 1\right) d s}}} = {\color{red}{\int{\left(s^{3} - s^{2}\right)d s}}}$$

Integrate term by term:

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

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

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

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

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

Therefore,

$$\int{s^{2} \left(s - 1\right) d s} = \frac{s^{4}}{4} - \frac{s^{3}}{3}$$

Simplify:

$$\int{s^{2} \left(s - 1\right) d s} = \frac{s^{3} \left(3 s - 4\right)}{12}$$

Add the constant of integration:

$$\int{s^{2} \left(s - 1\right) d s} = \frac{s^{3} \left(3 s - 4\right)}{12}+C$$

$$$\int s^{2} \left(s - 1\right)\, ds = \frac{s^{3} \left(3 s - 4\right)}{12} + C$$$A