Integral of $$$\frac{\sqrt{x}}{2}$$$

Find $$$\int \frac{\sqrt{x}}{2}\, dx$$$.

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{1}{2}$$$ and $$$f{\left(x \right)} = \sqrt{x}$$$:

$${\color{red}{\int{\frac{\sqrt{x}}{2} d x}}} = {\color{red}{\left(\frac{\int{\sqrt{x} d x}}{2}\right)}}$$

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

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

Therefore,

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

Add the constant of integration:

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

$$$\int \frac{\sqrt{x}}{2}\, dx = \frac{x^{\frac{3}{2}}}{3} + C$$$A