Integral of $$$2 \sqrt{y}$$$

Find $$$\int 2 \sqrt{y}\, dy$$$.

Apply the constant multiple rule $$$\int c f{\left(y \right)}\, dy = c \int f{\left(y \right)}\, dy$$$ with $$$c=2$$$ and $$$f{\left(y \right)} = \sqrt{y}$$$:

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

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

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

Therefore,

$$\int{2 \sqrt{y} d y} = \frac{4 y^{\frac{3}{2}}}{3}$$

Add the constant of integration:

$$\int{2 \sqrt{y} d y} = \frac{4 y^{\frac{3}{2}}}{3}+C$$

$$$\int 2 \sqrt{y}\, dy = \frac{4 y^{\frac{3}{2}}}{3} + C$$$A