Question Number 160252 by mnjuly1970 last updated on 26/Nov/21 $$ \\ $$$$\:\:\:\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:{ln}^{\:\mathrm{2}} \left(\mathrm{1}−{x}\:\right){ln}\left({x}\right)}{{x}}{dx}=? \\ $$$$ \\ $$ Answered by TheSupreme last updated on…
Question Number 160223 by mnjuly1970 last updated on 26/Nov/21 $$ \\ $$$$\:\:\:\:\:\Omega:=\int_{\mathrm{0}} ^{\:\infty} \frac{\:{x}}{\left(\:{e}^{\:{x}} \:+{e}^{\:−{x}} \right)^{\:\mathrm{3}} }\:{dx}\:=? \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 160159 by mnjuly1970 last updated on 25/Nov/21 $$ \\ $$$$\:\:\:\:\:\:\:{prove}\:\:{that}\:.. \\ $$$$ \\ $$$$\:\:\int_{\mathrm{0}} ^{\:\infty} \frac{\:{x}}{{cosh}^{\:\mathrm{3}} \:\left({x}\:\right)}\:{dx}\:=\:\mathrm{G}\:−\:\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\blacksquare \\ $$$$\:\:\:\:\:\mathrm{G}:\:\:{catalan}\:{constant} \\ $$$$ \\ $$…
Question Number 94579 by pete last updated on 20/May/20 $$\mathrm{A}\:\mathrm{study}\:\mathrm{indicates}\:\mathrm{that}\:{x}\:\mathrm{months}\:\mathrm{from}\:\mathrm{now} \\ $$$$\mathrm{the}\:\mathrm{population}\:\mathrm{of}\:\mathrm{a}\:\mathrm{certain}\:\mathrm{town}\:\mathrm{will}\:\mathrm{be}\: \\ $$$$\mathrm{decreasing}\:\mathrm{at}\:\mathrm{the}\:\mathrm{rate}\:\mathrm{of}\:\mathrm{5}+\mathrm{3}{x}^{\frac{\mathrm{2}}{\mathrm{3}}} \:\mathrm{people} \\ $$$$\mathrm{per}\:\mathrm{month}.\:\mathrm{By}\:\mathrm{how}\:\mathrm{much}\:\mathrm{will}\:\mathrm{the}\:\mathrm{population} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{town}\:\mathrm{increase}\:\mathrm{per}\:\mathrm{the}\:\mathrm{next}\:\mathrm{8}\:\mathrm{months}. \\ $$$$\boldsymbol{{I}}\:\boldsymbol{{need}}\:\boldsymbol{{help}}\:\boldsymbol{{with}}\:\boldsymbol{{the}}\:\boldsymbol{{above}}\:\boldsymbol{{question}},\:\boldsymbol{{please}}. \\ $$ Answered by…
Question Number 29037 by abdo imad last updated on 03/Feb/18 $${let}\:{give}\:{the}\:{sequence}\:\:\left({y}_{{n}} \right)\:/{y}_{\mathrm{0}} \left({x}\right)=\mathrm{1}\:\:{and} \\ $$$${y}_{{n}} \left({x}\right)=\:\mathrm{1}+\:\int_{\mathrm{0}} ^{{x}} \left({y}_{{n}−\mathrm{1}} \left({t}\right)\right)^{\mathrm{2}} {dt}\:,\:{let}\:{suppose}\:{x}\in\left[\mathrm{0},\mathrm{1}\right]\:{prove} \\ $$$${that}\:\left({y}_{{n}} \right)\:{is}\:{increasing}\:{majored}\:{by}\:\frac{\mathrm{1}}{\mathrm{1}−{x}}\:{if}\:{y}={lim}_{{n}\rightarrow+\infty} {y}_{{n}} \\…
Question Number 29031 by abdo imad last updated on 03/Feb/18 $${find}\:{all}\:{function}\:{f}\:\in{C}^{\mathrm{1}} \left({R}^{\mathrm{2}} ,{R}\right)\:{wich}\:{verify} \\ $$$$\frac{\partial{f}}{\partial{x}}\:−\frac{\partial{f}}{\partial{y}}=\mathrm{0}\:\:\:\forall\left({x},{y}\right)\in{R}^{\mathrm{2}} . \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 28949 by amit96 last updated on 02/Feb/18 $${xy}\frac{{x}^{\mathrm{4}} −{y}^{\mathrm{4}} }{{x}^{\mathrm{4}} +{y}^{\mathrm{4}} }\:\:{and}\:\mathrm{0}\:{for}\:{origin} \\ $$$${then}\:{funtion}\:{is} \\ $$$$\mathrm{1}.{continuous} \\ $$$$\mathrm{2}.{mixpartial}\:{are}\:{not}\:{equal}\:{at}\:{origin} \\ $$$$\mathrm{3}.{limit}\:{at}\:{origin}\:{is}\:\mathrm{1} \\ $$ Terms…
Question Number 159982 by ArielVyny last updated on 23/Nov/21 $$\left(\mathrm{1}+{bf}\left({x}\right)\right){f}''\left({x}\right)=\frac{{p}}{\lambda{a}} \\ $$$${solve}\:{this}\:{equation}:\:{find}\:\:{f}\left({x}\right) \\ $$ Answered by mr W last updated on 23/Nov/21 $${let}\:{y}'={u} \\ $$$${y}''=\frac{{du}}{{dx}}=\frac{{du}}{{dy}}×\frac{{dy}}{{dx}}={u}\frac{{du}}{{dy}}…
Question Number 28903 by amit96 last updated on 01/Feb/18 Commented by abdo imad last updated on 01/Feb/18 $${let}\:{put}\:{I}=\:\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \:\left[\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right]{dx}\:\:\:{and}\:{use}\:{the}\:{ch}.\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} }={t}\:\Leftrightarrow{x}^{\mathrm{2}} =\frac{\mathrm{1}}{{t}} \\ $$$$\Leftrightarrow{x}=\frac{\mathrm{1}}{\:\sqrt{{t}}}\:\Rightarrow\:{dx}=\frac{−\mathrm{1}}{\mathrm{2}{t}\sqrt{{t}}}{dt}\:\:\:{and}\:\:{I}=\:−\int_{\mathrm{1}}…
Question Number 28894 by amit96 last updated on 01/Feb/18 Commented by abdo imad last updated on 01/Feb/18 $${let}\:{put}\:{f}\left({x}\right)={F}\left({x},{y}\right)=\:\int_{\mathrm{0}} ^{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} } {cos}^{\mathrm{2}} \left({t}+{x}\right){dt}\:\:{we}\:{have}\:{by}\:{ch}. \\ $$$${t}+{x}={u}\:\:\:\:\Rightarrow{f}\left({x}\right)=\:\int_{{x}}…