Integral of $$$\sqrt{z}$$$

Find $$$\int \sqrt{z}\, dz$$$.

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

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

Therefore,

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

Add the constant of integration:

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

$$$\int \sqrt{z}\, dz = \frac{2 z^{\frac{3}{2}}}{3} + C$$$A