Question Number 211008 by mnjuly1970 last updated on 26/Aug/24 Commented by Frix last updated on 26/Aug/24 $${y}=\alpha{x}+{p}? \\ $$ Commented by mnjuly1970 last updated on…
Question Number 210969 by universe last updated on 24/Aug/24 $$\:\:\:\:\mathrm{if}\:\mathrm{a}_{\mathrm{n}} \:=\:\mathrm{n}^{\mathrm{4}} \int_{\mathrm{n}} ^{\mathrm{n}+\mathrm{1}} \:\frac{\mathrm{x}\:\mathrm{dx}}{\mathrm{1}+\mathrm{x}^{\mathrm{5}} }\:\:\mathrm{then} \\ $$$$\:\:\:\:\left(\mathrm{1}\right)\:\Sigma\mathrm{a}_{\mathrm{n}} \:\mathrm{is}\:\mathrm{convergent}\:\mathrm{or}\:\mathrm{divergent}?? \\ $$$$\:\:\:\:\left(\mathrm{2}\right)\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{a}_{\mathrm{n}\:} \:=\:?? \\ $$ Terms…
Question Number 210933 by mnjuly1970 last updated on 22/Aug/24 $$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$\:\:\:\:\:\:\mathrm{I}=\:\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}\:} \frac{\:{sin}\left(\:\mathrm{25}{x}\:\right)}{{sinx}}\:{dx}=? \\ $$$$ \\ $$$$ \\…
Question Number 210917 by depressiveshrek last updated on 22/Aug/24 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{intersected}\:\mathrm{by}\:\mathrm{three} \\ $$$$\mathrm{circles}\:\mathrm{of}\:\mathrm{radius}\:\mathrm{1},\:\mathrm{centered}\:\mathrm{at}\:\mathrm{the} \\ $$$$\mathrm{origin},\:\mathrm{at}\:\left(\mathrm{1},\:\mathrm{0}\right)\:\mathrm{and}\:\left(\mathrm{1},\:\mathrm{1}\right)\:\mathrm{respectively}. \\ $$ Answered by mr W last updated on 22/Aug/24 Commented…
Question Number 210855 by zhou0429 last updated on 20/Aug/24 Answered by Frix last updated on 20/Aug/24 $${x},\:{y}\:>\mathrm{0}\:\wedge\:{x}\neq{y} \\ $$$${x}\mathrm{ln}\:{x}\:={y}\mathrm{ln}\:{y} \\ $$$$\mathrm{Let}\:{y}={px}\wedge{p}>\mathrm{0}\wedge{p}\neq\mathrm{1} \\ $$$${x}\mathrm{ln}\:{x}\:={px}\mathrm{ln}\:{px} \\ $$$$\mathrm{ln}\:{x}\:={p}\mathrm{ln}\:{p}\:+{p}\mathrm{ln}\:{x}…
Question Number 210820 by Ghisom last updated on 19/Aug/24 $$\mathrm{prove} \\ $$$$\underset{\mathrm{0}} {\overset{\infty} {\int}}\:\frac{\mathrm{arctan}\:\sqrt{{x}^{\mathrm{2}} +\mathrm{2}}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\sqrt{{x}^{\mathrm{2}} +\mathrm{2}}}\:{dx}=\frac{\pi^{\mathrm{2}} }{\mathrm{12}} \\ $$ Answered by BHOOPENDRA last updated…
Question Number 210723 by universe last updated on 17/Aug/24 $$\mathrm{the}\:\mathrm{volume}\:\mathrm{of}\:\mathrm{the}\:\mathrm{region}\:\mathrm{between}\:\mathrm{the}\: \\ $$$$\mathrm{planes}\:\mathrm{x}+\mathrm{y}+\mathrm{2z}=\mathrm{2}\:\mathrm{and}\:\mathrm{2x}+\mathrm{y}+\mathrm{z}\:=\:\mathrm{4}\:\mathrm{in}\: \\ $$$$\mathrm{the}\:\mathrm{first}\:\mathrm{octant}\:\mathrm{is} \\ $$ Answered by mr W last updated on 17/Aug/24 $${V}=\frac{\mathrm{1}}{\mathrm{6}}\left(\mathrm{2}×\mathrm{4}×\mathrm{4}−\mathrm{2}×\mathrm{2}×\mathrm{1}\right)=\frac{\mathrm{14}}{\mathrm{3}}…
Question Number 210601 by mnjuly1970 last updated on 13/Aug/24 $$ \\ $$$$\:\: \\ $$$$\:\:\:\:\:\:\:\:\:{Find}\:\:{the}\:\:{value}\:{of}\:: \\ $$$$ \\ $$$$\:\:\:\:\:\Omega\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \int_{\mathrm{0}} ^{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} \:\int_{\sqrt{\:{x}^{\:\mathrm{2}} \:+{y}^{\:\mathrm{2}} }}…
Question Number 210593 by efronzo1 last updated on 13/Aug/24 $$\:\:\int\:\frac{\mathrm{sin}\:\mathrm{x}\:\mathrm{cos}\:\mathrm{x}}{\:\sqrt[{\mathrm{3}}]{\mathrm{cos}\:\mathrm{2x}}\:+\:\sqrt{\mathrm{cos}\:\mathrm{2x}}}\:\mathrm{dx}\:=? \\ $$$$\:\:\int\:\frac{\mathrm{dx}}{\mathrm{sec}\:\mathrm{x}\:\sqrt[{\mathrm{2}}]{\mathrm{sin}\:\mathrm{x}}\:+\:\mathrm{cos}\:\mathrm{x}\:\sqrt[{\mathrm{3}}]{\mathrm{cosec}\:^{\mathrm{5}} \mathrm{x}}}\:=? \\ $$ Answered by Sutrisno last updated on 30/Aug/24 $${misal} \\ $$$${cos}\mathrm{2}{x}={y}^{\mathrm{6}}…
Question Number 210517 by depressiveshrek last updated on 11/Aug/24 $$\int\frac{\mathrm{1}}{\mathrm{sin}{x}−\mathrm{cos2}{x}}{dx} \\ $$ Commented by Frix last updated on 11/Aug/24 $$\mathrm{Simply}\:\mathrm{use}\:{t}=\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\:\mathrm{to}\:\mathrm{get} \\ $$$$\frac{\mathrm{2}}{\mathrm{3}\left(\mathrm{1}+\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\right)}+\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{9}}\mathrm{ln}\:\mid\frac{\mathrm{2}−\sqrt{\mathrm{3}}−\mathrm{tan}\:\frac{{x}}{\mathrm{2}}}{\mathrm{2}+\sqrt{\mathrm{3}}−\mathrm{tan}\:\frac{{x}}{\mathrm{3}}}\mid\:+{C} \\ $$ Commented…