Question Number 209924 by OmoloyeMichael last updated on 26/Jul/24 $$\boldsymbol{{If}}\:\:\boldsymbol{{f}}\left(\boldsymbol{{x}}\right)=\left(\boldsymbol{{x}}!\right)\centerdot\left(\boldsymbol{{x}}!!\right)\centerdot\left(\boldsymbol{{x}}!!!\right)\:\: \\ $$$$\boldsymbol{{find}}\:\:\frac{\boldsymbol{{d}}}{\boldsymbol{{dx}}}\left(\boldsymbol{{f}}\left(\boldsymbol{{x}}\right)\right)=? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 209624 by mnjuly1970 last updated on 16/Jul/24 $$ \\ $$$$ \\ $$$$\mathrm{I}=\int_{\mathrm{0}} ^{\:\infty} \int_{\mathrm{0}} ^{\:\infty} \:\frac{\:\mathrm{1}}{\mathrm{1}+\:{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \:+{x}^{\mathrm{2}} {y}^{\mathrm{2}} }\:{dxdy}=? \\ $$$$\:{using}\:\:\:\:{polar}\:\:{system}… \\…
Question Number 209308 by Erico last updated on 06/Jul/24 $$\mathrm{Donner}\:\mathrm{l}'\acute {\mathrm{e}quivalence}\:\mathrm{simple} \\ $$$$\mathrm{de}\:\mathrm{I}_{\mathrm{n}} =\underset{\:\mathrm{0}} {\int}^{\:\mathrm{1}} \frac{{t}^{{n}} }{{t}^{{n}} −{t}+\mathrm{1}}{dt} \\ $$ Answered by mathzup last updated…
Question Number 209228 by universe last updated on 04/Jul/24 Answered by Frix last updated on 04/Jul/24 $$\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\left(\mathrm{cos}\:{x}\:+\mathrm{sin}\:{x}\right)^{{n}} {dx}=\sqrt{\mathrm{2}}\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\mathrm{sin}^{{n}} \:\left({x}+\frac{\pi}{\mathrm{4}}\right)\:{dx}= \\ $$$$=\sqrt{\mathrm{2}}\underset{\frac{\pi}{\mathrm{4}}}…
Question Number 209229 by Tawa11 last updated on 04/Jul/24 Commented by klipto last updated on 06/Jul/24 $$\boldsymbol{\mathrm{take}}\:\boldsymbol{\mathrm{in}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{both}}\:\boldsymbol{\mathrm{side}} \\ $$$$\mathrm{1}.\:\boldsymbol{\mathrm{iny}}=\boldsymbol{\mathrm{e}}^{\boldsymbol{\mathrm{x}}} \boldsymbol{\mathrm{inx}} \\ $$$$\frac{\boldsymbol{\mathrm{d}}\left(\boldsymbol{\mathrm{iny}}\right)}{\boldsymbol{\mathrm{dx}}}=\boldsymbol{\mathrm{v}}\frac{\boldsymbol{\mathrm{du}}}{\boldsymbol{\mathrm{dx}}}+\boldsymbol{\mathrm{u}}\frac{\boldsymbol{\mathrm{dv}}}{\boldsymbol{\mathrm{dx}}} \\ $$$$\frac{\mathrm{1}}{\boldsymbol{\mathrm{y}}}\:\frac{\boldsymbol{\mathrm{dy}}}{\boldsymbol{\mathrm{dx}}}=\boldsymbol{\mathrm{e}}^{\boldsymbol{\mathrm{x}}} \boldsymbol{\mathrm{inx}}+\frac{\mathrm{e}^{\mathrm{x}}…
Question Number 209232 by alcohol last updated on 04/Jul/24 $${u}_{\mathrm{0}} \:=\:{a},\:{u}_{{n}+\mathrm{1}} \:=\:\sqrt{{u}_{{n}} {v}_{{n}} } \\ $$$$\left.{v}_{\mathrm{0}} \:=\:{b}\:\in\:\right]\mathrm{0},\mathrm{1}\left[\:,\:{v}_{{n}+\mathrm{1}} \:=\:\frac{\mathrm{1}}{\mathrm{2}\left({u}_{{n}} +{v}_{{n}} \right)}\right. \\ $$$$\bullet\:{show}\:{that}\:{a}\leqslant{u}_{{n}} \leqslant{u}_{{n}+\mathrm{1}} \leqslant{v}_{{n}} \leqslant{v}_{{n}+\mathrm{1}}…
Question Number 209041 by alcohol last updated on 30/Jun/24 Commented by alcohol last updated on 30/Jun/24 $${find}\:{r} \\ $$ Commented by mr W last updated…
Question Number 208976 by efronzo1 last updated on 30/Jun/24 Answered by mr W last updated on 30/Jun/24 $${say}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={u},\:{xy}={v} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} \geqslant\mathrm{2}\sqrt{{x}^{\mathrm{2}} {y}^{\mathrm{2}}…
Question Number 208908 by alcohol last updated on 26/Jun/24 $${I}\left({a},{b}\right)\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} {t}^{{a}} \left(\mathrm{1}−{t}\right)^{{b}} {dt} \\ $$$${Note}\::\:{I}\left({a},{b}\right)\:=\:\frac{{a}}{{b}+\mathrm{1}}{I}\left({a}−\mathrm{1},{b}+\mathrm{1}\right) \\ $$$${show}\:{that} \\ $$$$\bullet\:{I}\left({a}+\mathrm{1},\:{b}\right)\:+\:{I}\left({a},{b}+\mathrm{1}\right)\:=\:{I}\left({a},\:{b}\right) \\ $$$$\bullet\:{find}\:{B}\left({a}+\mathrm{1},\:{b}+\mathrm{1}\right)\:{interms}\:{of}\:{B}\left({a},{b}\right) \\ $$$$\bullet\:{use}\:{I}\left({a},{b}\right)\:=\:{I}\left({b},{a}\right)\:{and}\:{deduce}\:{I}\left(\frac{\mathrm{5}}{\mathrm{2}},\frac{\mathrm{3}}{\mathrm{2}}\right) \\…
Question Number 208855 by efronzo1 last updated on 25/Jun/24 Answered by mr W last updated on 25/Jun/24 $${let}\:{t}=\mathrm{1}+\mid{x}\mid\geqslant\mathrm{1} \\ $$$$\frac{\mathrm{2024}^{−{t}} −\lambda}{\mathrm{2024}^{−{t}} −\lambda^{−\mathrm{1}} }=\lambda\:\mathrm{2024}^{{t}} \\ $$$$\mathrm{2024}^{−{t}}…