Integral of $$$2 - a^{2}$$$

Find $$$\int \left(2 - a^{2}\right)\, da$$$.

Integrate term by term:

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

Apply the constant rule $$$\int c\, da = a c$$$ with $$$c=2$$$:

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

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

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

Therefore,

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

Simplify:

$$\int{\left(2 - a^{2}\right)d a} = \frac{a \left(6 - a^{2}\right)}{3}$$

Add the constant of integration:

$$\int{\left(2 - a^{2}\right)d a} = \frac{a \left(6 - a^{2}\right)}{3}+C$$

$$$\int \left(2 - a^{2}\right)\, da = \frac{a \left(6 - a^{2}\right)}{3} + C$$$A