Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 115026 by jm2bok last updated on 23/Sep/20

$$\mathrm{Solve}:{\int }_{1/\pi }^{1/2}\ln \lfloor \frac{1}{x}\rfloor dx$$

Answered by PRITHWISH SEN 2 last updated on 23/Sep/20

$$\mathrm{when}$$$$\frac{1}{\pi }⩽\mathrm{x}⩽\frac{1}{3}\Rightarrow 3⩽\frac{1}{\mathrm{x}}⩽\pi \Rightarrow \ln \lfloor \frac{1}{\mathrm{x}}\rfloor =\ln 3$$$$\frac{1}{3}<\mathrm{x}⩽\frac{1}{2}\Rightarrow 2⩽\mathrm{x}<3\Rightarrow \ln \lfloor \frac{1}{\mathrm{x}}\rfloor =\ln 2$$$$\therefore \mathrm{the}\mathrm{integration}$$$$={\int }_{\frac{1}{\pi }}^{\frac{1}{3}}\ln \left(3\right)\mathrm{dx}+{\int }_{\frac{1}{3}}^{\frac{1}{2}}\ln \left(2\right)\mathrm{dx}$$$$=(\frac{1}{3}-\frac{1}{\pi })\ln 3+(\frac{1}{2}-\frac{1}{3})\ln 2\backsim 0.13202947375$$

Answered by mathmax by abdo last updated on 23/Sep/20

$$\mathrm{I}={\int }_{\frac{1}{\pi }}^{\frac{1}{2}}\ln \left\{\left[\frac{1}{\mathrm{x}}\right]\right\}\mathrm{dx}\mathrm{we}\mathrm{do}\mathrm{the}\mathrm{changement}\frac{1}{\mathrm{x}}=\mathrm{t}\Rightarrow$$$$\mathrm{I}={\int }_{\pi }^2\ln \left\{\left[\mathrm{t}\right]\right\}(-\frac{\mathrm{dt}}{{\mathrm{t}}^2})={\int }_2^{\pi }\frac{\ln \left[\mathrm{t}\right]}{{\mathrm{t}}^2}\mathrm{dt}(\pi \sim 3,14)\Rightarrow$$$$\mathrm{I}={\int }_2^3\frac{\ln \left[\mathrm{t}\right]}{{\mathrm{t}}^2}\mathrm{dt}+{\int }_3^{\pi }\frac{\ln \left[\mathrm{t}\right]}{{\mathrm{t}}^2}\mathrm{dt}=2{\int }_2^3\frac{\mathrm{dt}}{{\mathrm{t}}^2}+3{\int }_3^{\pi }\frac{\mathrm{dt}}{{\mathrm{t}}^2}$$$$=2{[-\frac{1}{\mathrm{t}}]}_2^3+3{[-\frac{1}{\mathrm{t}}]}_3^{\pi }=2(\frac{1}{2}-\frac{1}{3})+3(\frac{1}{3}-\frac{1}{\pi })$$$$=1-\frac{2}{3}+1-\frac{3}{\pi }=2-\frac{2}{3}-\frac{3}{\pi }=\frac{4}{3}-\frac{3}{\pi }$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com