Integral of $$$\frac{4 k}{9}$$$

Find $$$\int \frac{4 k}{9}\, dk$$$.

Apply the constant multiple rule $$$\int c f{\left(k \right)}\, dk = c \int f{\left(k \right)}\, dk$$$ with $$$c=\frac{4}{9}$$$ and $$$f{\left(k \right)} = k$$$:

$${\color{red}{\int{\frac{4 k}{9} d k}}} = {\color{red}{\left(\frac{4 \int{k d k}}{9}\right)}}$$

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

$$\frac{4 {\color{red}{\int{k d k}}}}{9}=\frac{4 {\color{red}{\frac{k^{1 + 1}}{1 + 1}}}}{9}=\frac{4 {\color{red}{\left(\frac{k^{2}}{2}\right)}}}{9}$$

Therefore,

$$\int{\frac{4 k}{9} d k} = \frac{2 k^{2}}{9}$$

Add the constant of integration:

$$\int{\frac{4 k}{9} d k} = \frac{2 k^{2}}{9}+C$$

$$$\int \frac{4 k}{9}\, dk = \frac{2 k^{2}}{9} + C$$$A