If $g(x)=x\lfloor \frac{1}{x}\rfloor$, where $\lfloor \cdot \rfloor$ is the greatest integer function, find the value of ${\int }_{1/3}^{1}g(x)dx$.

We split the integral based on the steps of the floor function:For $1/2<x\le 1$, $1\le \frac{1}{x}<2$, so $\lfloor \frac{1}{x}\rfloor =1$.Then $g(x)=x\cdot 1=x$.For $1/3<x\le 1/2$, $2\le \frac{1}{x}<3$, so $\lfloor \frac{1}{x}\rfloor =2$.Then $g(x)=x\cdot 2=2x$.The integral is:
$$I={\int }_{1/3}^{1/2}2xdx+{\int }_{1/2}^{1}xdx$$
$$I={\left[x^2\right]}_{1/3}^{1/2}+{\left[\frac{x^2}{2}\right]}_{1/2}^1$$
$$I=\left(\frac{1}{4}-\frac{1}{9}\right)+\left(\frac{1}{2}-\frac{1}{8}\right)$$
$$I=\frac{5}{36}+\frac{3}{8}$$
$$I=\frac{10}{72}+\frac{27}{72}=\frac{37}{72}$$