Question Number 137251 by mnjuly1970 last updated on 31/Mar/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:……\mathscr{A}{dvanced}\:\:…\:\:{calculus}…… \\ $$$$\:\:\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}^{\mathrm{2}} {ln}\left({x}\right){ln}\left(\mathrm{1}−{x}\right){dx}=??? \\ $$$$ \\ $$ Answered by Ar Brandon last updated…
Question Number 6152 by sanusihammed last updated on 16/Jun/16 $${Differentiate}\:\:\:{s}\:=\:{ut}\:+\:\frac{\mathrm{1}}{\mathrm{2}}{at}^{\mathrm{2}} \:\:\:\:\:\:{from}\:{the}\:{first}\:{principle} \\ $$$$ \\ $$ Answered by Rasheed Soomro last updated on 16/Jun/16 $${s}\:=\:{ut}\:+\:\frac{\mathrm{1}}{\mathrm{2}}{at}^{\mathrm{2}} \:\:\:…
Question Number 137226 by SLVR last updated on 31/Mar/21 Answered by MJS_new last updated on 31/Mar/21 $${f}\left({x}\right)=\mathrm{2ln}\:{x} \\ $$$${x}^{\mathrm{3}} −\mathrm{6}{x}^{\mathrm{2}} +\mathrm{11}{x}−\mathrm{6}=\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)\left({x}−\mathrm{3}\right) \\ $$$$\mathrm{the}\:\mathrm{curves}\:\mathrm{intersect}\:\mathrm{at}\:{x}=\mathrm{1} \\ $$$$\underset{\mathrm{0}}…
Question Number 137190 by mnjuly1970 last updated on 30/Mar/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:….{advanced}\:…….\:\:{calculus}….. \\ $$$$\:\:\:\:\:\:\:\:{prove}\:{that}::: \\ $$$$\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\mathrm{1}} {ln}\left({ln}\left(\frac{\mathrm{1}}{{x}}\right)\right).\frac{{dx}}{\:\sqrt{{ln}\left(\frac{\mathrm{1}}{{x}}\right)}}\:=−\sqrt{\pi}\:\left(\gamma+{ln}\left(\mathrm{4}\right)\right) \\ $$$$ \\ $$ Answered by Dwaipayan Shikari last…
Question Number 137129 by Chhing last updated on 30/Mar/21 $$ \\ $$$$\:\left(\mathrm{x}^{\mathrm{2}} −\mathrm{1}\right)\mathrm{y}''−\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)\mathrm{y}=\mathrm{0}\:\:\:,\:\:\mathrm{n}\in\mathbb{N} \\ $$$$\mathrm{Find}\:\mathrm{solution}\:\mathrm{that}\:\mathrm{can}\:\mathrm{be}\:\mathrm{expanded}\:\mathrm{in}\:\mathrm{series} \\ $$$$\mathrm{help}\:\mathrm{me} \\ $$ Answered by mathmax by abdo last…
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Question Number 71550 by mhmd last updated on 17/Oct/19 Answered by MJS last updated on 17/Oct/19 $$\mathrm{it}'\:\mathrm{wrong},\:\mathrm{it}\:\mathrm{must}\:\mathrm{be} \\ $$$$\begin{pmatrix}{{n}+\mathrm{1}}\\{{k}}\end{pmatrix}\:=\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}\:+\begin{pmatrix}{{n}}\\{{k}−\mathrm{1}}\end{pmatrix} \\ $$$$\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}\:+\begin{pmatrix}{{n}}\\{{k}−\mathrm{1}}\end{pmatrix}\:=\frac{{n}!}{{k}!\left({n}−{k}\right)!}+\frac{{n}!}{\left({k}−\mathrm{1}\right)!\left({n}−{k}+\mathrm{1}\right)!}= \\ $$$$=\frac{{n}!\left({n}−{k}+\mathrm{1}\right)}{\left({k}−\mathrm{1}\right)!\left({n}−{k}+\mathrm{1}\right)!{k}}+\frac{{n}!{k}}{\left({k}−\mathrm{1}\right)!\left({n}−{k}+\mathrm{1}\right)!{k}}= \\ $$$$=\frac{{n}!\left({n}−{k}+\mathrm{1}\right)+{n}!{k}}{\left({k}−\mathrm{1}\right)!\left({n}−{k}+\mathrm{1}\right)!{k}}=\frac{{n}!\left({n}+\mathrm{1}\right)}{{k}!\left({n}+\mathrm{1}−{k}\right)!}=…
Question Number 71548 by Cmr 237 last updated on 17/Oct/19 $$\mathrm{let}\:\mathrm{f}:\boldsymbol{\mathrm{U}}\subset\mathbb{R}^{\mathrm{n}} \rightarrow\mathbb{R}^{\mathrm{p}} \:\mathrm{be}\:\mathrm{an}\:\mathrm{application} \\ $$$$\mathrm{where}\:\boldsymbol{\mathrm{U}}\:\mathrm{is}\:\mathrm{an}\:\mathrm{open}\:\mathrm{set} \\ $$$$\mathrm{prove}\:\mathrm{that}\:\forall\mathrm{x}\in\boldsymbol{{U}},\exists\mathrm{h}\in\mathbb{R}^{\mathrm{n}} \mathrm{such}\:\mathrm{as} \\ $$$$\mathrm{x}+\mathrm{h}\in\boldsymbol{{U}} \\ $$ Answered by mind…
Question Number 71551 by mhmd last updated on 17/Oct/19 $${whats}\:{the}\:{mean}\:{R}^{++} \:?{and}\:{R}^{−−} ? \\ $$ Answered by MJS last updated on 17/Oct/19 $$\mathrm{I}\:\mathrm{only}\:\mathrm{know}\:\mathbb{R}^{+} =\left\{{x}\in\mathbb{R}\mid{x}>\mathrm{0}\right\}\:\mathrm{and} \\ $$$$\mathbb{R}^{−}…
Question Number 71508 by Cmr 237 last updated on 16/Oct/19 $$\mathrm{montrer}\:\mathrm{que}:\forall\mathrm{a},\mathrm{b}\in\mathbb{R}\:\mathrm{ona} \\ $$$$\mathrm{2}\mid\mathrm{ab}\mid\leqslant\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} \\ $$$$\mathrm{Endeduire}\:\mathrm{que}\:\forall\mathrm{x}_{\mathrm{1}} ,…,\mathrm{x}_{\mathrm{n}} \in\mathbb{R}\:\mathrm{on}\:\mathrm{a}: \\ $$$$\left(\underset{\mathrm{i}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\mid\mathrm{x}_{\mathrm{i}} \mid\right)^{\mathrm{2}} \leqslant\mathrm{n}\underset{\mathrm{i}=\mathrm{1}} {\overset{\mathrm{n}}…