Question Number 111965 by mnjuly1970 last updated on 05/Sep/20 $$\:\:\:\:\:\:….{advanced}\:\:{calculus}… \\ $$$${evaluate}: \\ $$$$ \\ $$$$\:\:\:\:{i}::\:\int_{\mathrm{0}} ^{\:\mathrm{1}} {xH}_{{x}} {dx}\:=???\:\: \\ $$$$\:\:\:\:\:{ii}::\underset{{n}=\mathrm{1}\:} {\overset{\infty} {\sum}}\frac{{H}_{{n}} }{{n}^{\mathrm{2}} \mathrm{2}^{{n}\:}…
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Question Number 177456 by cortano1 last updated on 05/Oct/22 Answered by a.lgnaoui last updated on 05/Oct/22 $${posons}\:\:{x}=\mathrm{sin}\:{t}\:\:\:\:{alors}\:{y}=\mathrm{cos}\:{t} \\ $$$$\frac{\mathrm{1}}{\mathrm{sin}\:{t}}+\frac{\mathrm{4}}{\mathrm{cos}\:{t}}=\frac{\mathrm{cos}\:{t}+\mathrm{4sin}\:{t}}{\mathrm{sin}\:{t}\mathrm{cos}\:{t}} \\ $$$$\left.{le}\:{rapoort}\:\:{est}\:{minimale}\:\:{si}\:\:\:\mathrm{scos}\:{t}+\mathrm{4sin}\:{t}\:{mnimale}\right) \\ $$$$\begin{cases}{\mathrm{cos}\:{t}+\mathrm{4sin}\:{t}=\mathrm{0}}\\{\mathrm{sin}\:{t}\mathrm{cos}\:{t}\neq\mathrm{0}}\end{cases} \\ $$$$\mathrm{4sin}\:{t}=−\mathrm{cos}\:{t}\:\:\:\mathrm{tan}\:{t}=−\frac{\mathrm{1}}{\mathrm{4}}=\frac{\mathrm{sin}\:{t}}{\mathrm{cos}\:{t}}…
Question Number 177320 by mnjuly1970 last updated on 03/Oct/22 Answered by a.lgnaoui last updated on 04/Oct/22 $$\sqrt{\mathrm{2}}\:+\left(\mathrm{1}+\sqrt{\mathrm{2}}\:\right){x}+{x}^{\mathrm{2}} =\left({x}+\frac{\mathrm{1}+\sqrt{\mathrm{2}}}{\mathrm{2}}\right)^{\mathrm{2}} −\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} =\left({x}+\frac{\mathrm{1}+\sqrt{\mathrm{2}}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)\left({x}+\frac{\mathrm{1}+\sqrt{\mathrm{2}}}{\mathrm{2}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right) \\ $$$$\frac{\mathrm{1}}{\left({x}+\frac{\mathrm{1}+\sqrt{\mathrm{2}}}{\mathrm{2}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)\left({x}+\frac{\mathrm{1}+\sqrt{\mathrm{2}}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)}=−\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}}\left[\frac{\mathrm{1}}{\left({x}+\frac{\mathrm{1}+\sqrt{\mathrm{2}}\:−\sqrt{\mathrm{3}}}{\mathrm{2}}\right)}+\frac{\mathrm{1}}{\left({x}+\frac{\mathrm{1}+\sqrt{\mathrm{2}}\:+\sqrt{\mathrm{3}}}{\mathrm{2}}\right)}\right] \\ $$$$\int_{\mathrm{0}} ^{\infty}…
Question Number 111584 by ajfour last updated on 04/Sep/20 Answered by ajfour last updated on 04/Sep/20 $${C}\left(\mathrm{0},{y}\right)\:\:\:;\:{A}\left({x},\mathrm{0}\right)\:\:;\:\:{B}\left({z},{z}^{\mathrm{2}} \right) \\ $$$$\sqrt{\left({x}−{z}\right)^{\mathrm{2}} +{z}^{\mathrm{4}} }+\sqrt{{z}^{\mathrm{2}} +\left({y}−{z}^{\mathrm{2}} \right)^{\mathrm{2}} }\:=\:{a}…
Question Number 177071 by cortano1 last updated on 30/Sep/22 Commented by Frix last updated on 30/Sep/22 $$\mathrm{I}\:\mathrm{think}\:\mathrm{the}\:\mathrm{limit}\:\mathrm{doesn}'\mathrm{t}\:\mathrm{exist} \\ $$ Commented by Frix last updated on…
Question Number 45994 by peter frank last updated on 19/Oct/18 Answered by MrW3 last updated on 20/Oct/18 $${we}\:{know}\:\frac{{x}+{y}}{\mathrm{2}}\geqslant\sqrt{{xy}}\:{when}\:{x},{y}\geqslant\mathrm{0}. \\ $$$$ \\ $$$$\mathrm{2}\left({a}+{b}\right)={L}\Rightarrow{a}+{b}=\frac{{L}}{\mathrm{2}} \\ $$$${A}_{{rectangular}} ={ab}\leqslant\left(\frac{{a}+{b}}{\mathrm{2}}\right)^{\mathrm{2}}…
Question Number 45993 by peter frank last updated on 19/Oct/18 Commented by math khazana by abdo last updated on 20/Oct/18 $$\frac{{d}}{{dx}}\left({e}^{{x}^{\mathrm{2}} } \right)={lim}_{{h}\rightarrow\mathrm{0}} \:\frac{{e}^{\left({x}+{h}\right)^{\mathrm{2}} }…
Question Number 176809 by cortano1 last updated on 27/Sep/22 $$\:\:\begin{cases}{\mathrm{x}=\mathrm{cos}\:^{\mathrm{3}} \emptyset}\\{\mathrm{y}=\mathrm{sin}\:^{\mathrm{3}} \emptyset}\end{cases}\:\Rightarrow\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{dx}^{\mathrm{2}} }\:=? \\ $$ Answered by mr W last updated on 27/Sep/22 $$\frac{{dy}}{{d}\phi}=−\mathrm{3}\:\mathrm{cos}^{\mathrm{2}}…
Question Number 111213 by mnjuly1970 last updated on 02/Sep/20 Answered by mathmax by abdo last updated on 02/Sep/20 $$\mathrm{let}\:\mathrm{S}_{\mathrm{n}} =\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{2n}} \:\mathrm{cos}^{\mathrm{2}} \left(\frac{\mathrm{k}\pi}{\mathrm{2n}+\mathrm{1}}\right)\:\Rightarrow\mathrm{S}_{\mathrm{n}} =\frac{\mathrm{1}}{\mathrm{2}}\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{2n}}…