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Question Number 169048 Answers: 0 Comments: 0
$${find}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{lnx}}{{x}^{\mathrm{2}} −{x}+\mathrm{2}}{dx} \\ $$
Question Number 169044 Answers: 1 Comments: 0
$${does}\:{the}\:{series}\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{e}^{−\mathrm{2}\left({n}−\mathrm{1}\right)} \:{sin}\left(\frac{{n}\pi}{\mathrm{2}}\right)\:{is}\:{converge}\:{or}\:{diverge}\:? \\ $$
Question Number 169038 Answers: 2 Comments: 1
Question Number 169042 Answers: 0 Comments: 0
Question Number 169013 Answers: 2 Comments: 1
Question Number 169009 Answers: 1 Comments: 10
Question Number 169004 Answers: 1 Comments: 3
$$\underset{{x}\rightarrow\frac{\pi}{\mathrm{2}}} {\mathrm{lim}}\left(\frac{{sinx}−\mathrm{1}}{{x}−\frac{\pi}{\mathrm{2}}}\right)=? \\ $$
Question Number 169000 Answers: 1 Comments: 0
Question Number 168982 Answers: 1 Comments: 1
Question Number 168980 Answers: 0 Comments: 7
Question Number 168979 Answers: 0 Comments: 0
$${f}\left({x},{y},{z}\right)\:=\:\left(\mathrm{3}{x}^{\mathrm{2}} {y},{x}^{\mathrm{3}} +{y}^{\mathrm{3}} ,\:\mathrm{2}{z}\right) \\ $$$$\mathrm{prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{function}\:\mathrm{has}\:\mathrm{a}\:\mathrm{potential} \\ $$$$\mathrm{to}\:\mathrm{be}\:\mathrm{determined}. \\ $$
Question Number 168978 Answers: 1 Comments: 0
$$\mathrm{Evaluate}\: \\ $$$$\left(\mathrm{a}\right)\:\int\frac{{t}−\mathrm{2}}{{t}−\mathrm{3}\sqrt{\mathrm{2}{t}−\mathrm{4}}\:+\mathrm{2}}\:{dt}\: \\ $$$$\left(\mathrm{b}\right)\:\int\frac{\mathrm{3}{z}}{\left(\mathrm{1}−\mathrm{4}{z}−\mathrm{2}{z}^{\mathrm{2}} \right)^{\mathrm{2}} }\:{dz} \\ $$
Question Number 168977 Answers: 0 Comments: 0
$$\mathrm{check}\:\mathrm{that}\:\mathrm{the}\:\mathrm{function} \\ $$$${u}\left({x},{t}\right)\:=\:\mathrm{exp}\left\{−\frac{{n}^{\mathrm{2}} \alpha^{\mathrm{2}} \pi^{\mathrm{2}} }{{L}^{\mathrm{2}} }{t}\right\}\:\mathrm{sin}\frac{{n}\pi{x}}{{L}} \\ $$$${n}\:=\:\mathrm{1},\mathrm{2},...\:\mathrm{satisfy}\:\mathrm{the}\:\mathrm{heat}\:\mathrm{equation} \\ $$$$\boldsymbol{\mathrm{heat}}\:\boldsymbol{\mathrm{equation}} \\ $$$$\alpha^{\mathrm{2}} \frac{\partial^{\mathrm{2}} {u}}{\partial{x}^{\mathrm{2}} }\:=\:\frac{\partial{u}}{\partial{t}},\:\mathrm{0}\:<\:{x}\:<\:{L} \\ $$
Question Number 168952 Answers: 1 Comments: 0
Question Number 168948 Answers: 1 Comments: 0
Question Number 168946 Answers: 2 Comments: 1
Question Number 168942 Answers: 0 Comments: 2
Question Number 168956 Answers: 4 Comments: 0
$$ \\ $$$$\:\:\:\:{solve} \\ $$$$\:\:\lfloor\:{x}\:\rfloor\:+\:\lfloor\:\frac{{x}}{\mathrm{6}}\:\rfloor\:=\frac{\mathrm{2}}{\mathrm{3}}\:{x} \\ $$$$ \\ $$
Question Number 168937 Answers: 1 Comments: 0
$$ \\ $$$$\underline{\mathcal{D}{etermine}\:{m}\:\&\:{n}\:{such}\:{that}:} \\ $$$$\:\:\underset{\underset{{digit}-{sum}\left({n}^{\mathrm{2}} \right)={m}} {\&}} {\:{digit}-{sum}\left({m}^{\mathrm{2}} \right)={n}} \\ $$$$\:^{\blacksquare} {digit}-{sum}\left(\overline {{abc}..}.\right)={a}+{b}+{c}+...\:\:\:\:\:\:\:\: \\ $$
Question Number 168968 Answers: 0 Comments: 0
Question Number 168966 Answers: 2 Comments: 0
$$\:\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\int_{\mathrm{0}} ^{\:{x}} \:\left(\int_{\mathrm{0}} ^{\:{u}^{\mathrm{2}} } \mathrm{tan}^{−\mathrm{1}} \left(\mathrm{1}+{t}\right){dt}\right){dt}\:}{{x}−{x}\:\mathrm{cos}\:{x}}\:=? \\ $$
Question Number 168964 Answers: 2 Comments: 0
$$\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\underset{\mathrm{0}} {\overset{{x}} {\int}}\:\frac{\mathrm{sin}\:\mathrm{2}{t}}{\:\sqrt{\mathrm{4}+{t}^{\mathrm{2}} }\:\underset{\mathrm{0}} {\overset{{x}} {\int}}\:\left(\sqrt{\mathrm{1}+{t}}−\mathrm{1}\right){dt}}\:{dt}\:=?\: \\ $$
Question Number 168926 Answers: 2 Comments: 2
Question Number 168921 Answers: 1 Comments: 0
Question Number 168911 Answers: 1 Comments: 1
Question Number 168910 Answers: 0 Comments: 3
$${Resolve} \\ $$$$\left.\mathrm{1}\right)\:\left({x}−{y}\right){ydx}−{x}^{\mathrm{2}} {dy}=\mathrm{0} \\ $$$$\left.\mathrm{2}\right)\left(\mathrm{2}{x}−{y}\right){dx}+\left(\mathrm{4}{x}−\mathrm{2}{y}+\mathrm{3}\right){dy}=\mathrm{0} \\ $$
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