Integral of $$$\sqrt{q}$$$

Find $$$\int \sqrt{q}\, dq$$$.

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

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

Therefore,

$$\int{\sqrt{q} d q} = \frac{2 q^{\frac{3}{2}}}{3}$$

Add the constant of integration:

$$\int{\sqrt{q} d q} = \frac{2 q^{\frac{3}{2}}}{3}+C$$

$$$\int \sqrt{q}\, dq = \frac{2 q^{\frac{3}{2}}}{3} + C$$$A