Evaluating $\int_{1/4}^{1/\pi}\left\lfloor\frac{2}{x}\right\rfloor \text{dx}$

Question

I'd like to get some help in solving this floor function as the rest of the internet doesn't seem to be very helpful.

$$\int_{1/4}^{1/\pi}\left\lfloor\frac{2}{x}\right\rfloor \text{dx}$$

What really stumped me, is the floor function. I can't wrap my head around that and need help explaining why it works. I have seen a couple of videos dealing with functions that aren't in the denominator, but this confuses me.

Answer 1

For $x\in\left(\frac 14,\frac 27\right)$, $\left\lfloor\frac 2x\right\rfloor=7$. For $x\in\left(\frac 27,\frac 2{2\pi}\right)$, $\left\lfloor\frac 2x\right\rfloor=6$. If you can visualize this, then I think you can find the answer :)

Answer 2

$${I}=\underset{\mathrm{1}/\mathrm{4}} {\overset{\mathrm{1}/\pi} {\int}}\lfloor\frac{\mathrm{2}}{{x}}\rfloor{dx} \\ $$ $$\left[\lfloor\frac{\mathrm{2}}{\mathrm{1}/\pi}\rfloor;\lfloor\frac{\mathrm{2}}{\mathrm{1}/\mathrm{4}}\rfloor\right]=\left[\mathrm{6};\mathrm{8}\right] \\ $$ $$\lfloor\frac{\mathrm{2}}{{x}}\rfloor=\mathrm{6}\Rightarrow{x}=\left[\frac{\mathrm{1}}{\pi};\frac{\mathrm{2}}{\mathrm{7}}\right) \\ $$ $$\lfloor\frac{\mathrm{2}}{{x}}\rfloor=\mathrm{7}\Rightarrow{x}=\left[\frac{\mathrm{2}}{\mathrm{7}};\frac{\mathrm{1}}{\mathrm{4}}\right) \\ $$ $$\lfloor\frac{\mathrm{2}}{{x}}\rfloor=\mathrm{8}\Rightarrow{x}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$ $${I}=\mathrm{6}\left(\frac{\mathrm{2}}{\mathrm{7}}−\frac{\mathrm{1}}{\pi}\right)+\mathrm{7}\left(\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{2}}{\mathrm{7}}\right) \\ $$ $${I}=\frac{\mathrm{41}}{\mathrm{28}}−\frac{\mathrm{6}}{\pi} \\ $$