Question Number 151860 by maged last updated on 23/Aug/21 $$\underset{\mathrm{1}} {\overset{\mathrm{5}} {\int}}\mid\mathrm{2}−\mid\mathrm{3}−\boldsymbol{\mathrm{x}}\mid\mid\mathrm{dx} \\ $$ Commented by lyubita last updated on 23/Aug/21 $${It}\:{means}\:{area}\:{bounded}\:{by}\:{the}\:{graph} \\ $$$${and}\:{x}\:{axis}\:{between}\:{x}\:=\:\mathrm{1}\:{and}\:{x}\:=\:\mathrm{5} \\…
Question Number 86326 by ~blr237~ last updated on 28/Mar/20 $${Let}\:{f}\:{a}\:{continue}\:{function}\:{acknowleding} \\ $$$$\alpha\:{as}\:{a}\:{fix}\:{point}\:{on}\:\left[\mathrm{0},\mathrm{1}\right].{F}\:\:{a}\:{function}\:{such}\:{as}\:\frac{{dF}}{{dx}}={f}\left({x}\right) \\ $$$$\forall\:{n}\:,\:\:{u}_{{n}+\mathrm{1}} =\frac{{F}\left({u}_{{n}} \right)−{F}\left(\alpha\right)}{{u}_{{n}} −\alpha}\: \\ $$$${Prove}\:{that}\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{u}_{{n}} \:=\alpha \\ $$ Terms of…
Question Number 86324 by TawaTawa1 last updated on 28/Mar/20 $$\int_{\:\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} \:\frac{\mathrm{1}\:}{\:\sqrt{\boldsymbol{\mathrm{sin}}\:\boldsymbol{\mathrm{x}}}\:\:\:+\:\:\sqrt{\boldsymbol{\mathrm{cos}}\:\boldsymbol{\mathrm{x}}}}\:\boldsymbol{\mathrm{dx}} \\ $$ Commented by MJS last updated on 28/Mar/20 $$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{think}\:\mathrm{we}\:\mathrm{can}\:\mathrm{exactly}\:\mathrm{solve}\:\mathrm{this}… \\ $$ Commented…
Question Number 151851 by Tawa11 last updated on 23/Aug/21 Answered by OlafThorendsen last updated on 23/Aug/21 $$\mathrm{I}\:=\:\int_{−\frac{\pi}{\mathrm{2}}} ^{+\frac{\pi}{\mathrm{2}}} \left({x}^{\mathrm{2}} +\mathrm{ln}\left(\frac{\pi+{x}}{\pi−{x}}\right)\right)\mathrm{cos}{x}\:{dx}\:\:\:\left(\mathrm{1}\right) \\ $$$$\mathrm{Let}\:{u}\:=\:−{x}\:: \\ $$$$\mathrm{I}\:=\:\int_{−\frac{\pi}{\mathrm{2}}} ^{+\frac{\pi}{\mathrm{2}}}…
Question Number 86313 by john santu last updated on 28/Mar/20 $$\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\:\frac{{dx}}{\:\sqrt{\mathrm{1}+\mathrm{tan}\:{x}}} \\ $$ Commented by john santu last updated on 28/Mar/20 $$\frac{\sqrt{\mathrm{sin}\:{x}}}{\:\sqrt{\mathrm{sin}\:{x}+\mathrm{cos}\:{x}}}\:+\:\frac{\sqrt{\mathrm{cos}\:{x}}}{\:\sqrt{\mathrm{sin}\:{x}+\mathrm{cos}\:{x}}}\:=\:\mathrm{1} \\…
Question Number 151838 by talminator2856791 last updated on 23/Aug/21 $$\: \\ $$$$\int_{\mathrm{0}} ^{\:\infty} \:\frac{\left({x}^{\mathrm{log}\left(\lfloor\left(\lfloor{x}\rfloor!\right)^{\left(\mathrm{log}\left(\lfloor{x}−\mathrm{1}\rfloor!\right)\right)^{−\mathrm{1}} } \rfloor\right)+\mathrm{1}} +\mathrm{1}\right)^{{x}} }{\lfloor{x}^{\mathrm{log}\left({x}^{{x}} \right)+\mathrm{1}} \rfloor!+\mathrm{1}}\:{dx} \\ $$$$\: \\ $$ Terms…
Question Number 86302 by john santu last updated on 28/Mar/20 $$\int\:\frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}}\:=? \\ $$ Commented by abdomathmax last updated on 28/Mar/20 $${I}\:=\int\:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)\sqrt{{x}^{\mathrm{2}} \:+\mathrm{4}}}\:{changement}\:{x}\:=\mathrm{2}{sh}\left({t}\right)\:{give}…
Question Number 20733 by Joel577 last updated on 02/Sep/17 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{volume}\:\mathrm{of}\:\mathrm{the}\:\mathrm{solid}\:\mathrm{of}\:\mathrm{revolution} \\ $$$$\mathrm{obtained}\:\mathrm{by}\:\mathrm{revolving}\:\mathrm{area}\:\mathrm{bounded}\:\mathrm{by} \\ $$$${x}\:=\:\mathrm{4}\:+\:\mathrm{6}{y}\:−\:\mathrm{2}{y}^{\mathrm{2}} ,\:{x}\:=\:−\mathrm{4},\:{x}\:=\:\mathrm{0}\:\mathrm{about} \\ $$$$\mathrm{the}\:{y}−\mathrm{axis} \\ $$ Answered by mrW1 last updated on…
Question Number 86269 by M±th+et£s last updated on 27/Mar/20 $$\int\frac{{sec}^{\mathrm{2}} \left({x}\right)}{\left({tan}\left({x}\right)−\mathrm{1}\right)^{\mathrm{4}} \left({tan}\left({x}\right)−\mathrm{2}\right)}\:{dx} \\ $$ Commented by john santu last updated on 28/Mar/20 $${u}\:=\:\mathrm{tan}\:{x}−\mathrm{2}\: \\ $$$$\Rightarrow\:\int\:\frac{{du}}{\left({u}+\mathrm{1}\right)^{\mathrm{4}}…
Question Number 86258 by lémùst last updated on 27/Mar/20 $${calculate}\:{I}=\int_{\mathrm{1}} ^{+\infty} \frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}^{\mathrm{4}} −{x}^{\mathrm{2}} +\mathrm{1}}{dx} \\ $$ Commented by mathmax by abdo last updated on…