Question Number 121649 by mnjuly1970 last updated on 10/Nov/20 $$\:\:\:\:\:\:\:\:{please}\:\:{evaluate}\::: \\ $$$$\:\:\:\:\:\:\:{i}:\:\:\:\Omega_{\mathrm{1}} \:\overset{?} {=}\int_{\mathrm{0}} ^{\:\infty} {e}^{−{x}} {ln}\left(\Gamma\left(\mathrm{1}−{e}^{−{x}} \right)\right){dx} \\ $$$$\:\:\:\:\:\:\:{ii}:\:\:\:\Omega_{\mathrm{2}} \overset{?} {=}\int_{\mathrm{0}} ^{\:\infty} \left(\frac{{sin}\left({x}\right)}{{x}}\right)^{\mathrm{3}} {dx}…
Question Number 56069 by necx1 last updated on 09/Mar/19 $${Find}\:{the}\:{expansion}\:{and}\:{the}\:{convergemce} \\ $$$${of}\:{the}\:{following}\:{in}\:{the}\:{power}\:{of}\:{x}: \\ $$$$\left({i}\right){sinx} \\ $$$$\left({ii}\right)\mathrm{ln}\:\left(\mathrm{1}+{x}\right) \\ $$$$\left({iii}\right)\mathrm{tan}^{−\mathrm{1}} {x} \\ $$ Commented by tanmay.chaudhury50@gmail.com last…
Question Number 187053 by horsebrand11 last updated on 13/Feb/23 $$\:{How}\:{do}\:{you}\:{make}\:{a}\:{curve}\: \\ $$$$\:{y}={ax}^{\mathrm{3}} +{bx}^{\mathrm{2}} +{cx}+{d}\:{with}\:{a}\:{critical} \\ $$$$\:{point}\:{of}\:\left(\mathrm{1},\mathrm{0}\right)\:{and}\:\left(−\mathrm{2},\mathrm{27}\right)\:? \\ $$ Answered by cortano12 last updated on 13/Feb/23…
Question Number 121400 by rs4089 last updated on 07/Nov/20 $${ind}\:{maximum}\:{value}\:{of}\:\sqrt{{x}^{\mathrm{4}} +{y}^{\mathrm{4}} } \\ $$ Answered by MJS_new last updated on 07/Nov/20 $$\mathrm{for}\:{x},\:{y}\:\in\mathbb{R}:\:\mathrm{0}\leqslant\sqrt{{x}^{\mathrm{4}} +{y}^{\mathrm{4}} }<+\infty \\…
Question Number 121368 by abdullahquwatan last updated on 07/Nov/20 $$\mathrm{find}\:\mathrm{minimun}\:\mathrm{and}\:\mathrm{maksimum} \\ $$$${y}=\frac{{x}^{\mathrm{3}} −\mathrm{1}}{{x}^{\mathrm{2}} −\mathrm{3}} \\ $$ Answered by MJS_new last updated on 07/Nov/20 $${x}^{\mathrm{3}} −\mathrm{1}=\mathrm{0}\:\Rightarrow\:\mathrm{zero}\:\mathrm{at}\:{x}=\mathrm{1}…
Question Number 121361 by rs4089 last updated on 07/Nov/20 Answered by TANMAY PANACEA last updated on 07/Nov/20 $${u}={x}^{\mathrm{3}} {sin}^{−\mathrm{1}} \left(\frac{{y}}{{x}}\right)−{y}^{\mathrm{3}} {sin}^{−\mathrm{1}} \left(\frac{{x}}{{y}}\right) \\ $$$$={x}^{\mathrm{3}} \left\{{sin}^{−\mathrm{1}}…
Question Number 121362 by mnjuly1970 last updated on 07/Nov/20 $$\:\:\:\:\:\:\:…\:{advanced}\:\:{calculus}… \\ $$$$\:\:\:\:{evaluate}\:: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{K}=\underset{{n}\in\mathbb{N}} {\sum}\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}{n}} \left(\mathrm{1}+{cos}\left(\frac{\pi}{\mathrm{2}^{{n}} }\right)\right)}\:=? \\ $$ Commented by mindispower last updated on…
Question Number 121353 by john santu last updated on 06/Nov/20 Answered by liberty last updated on 07/Nov/20 $$\mathrm{Let}\:\mathrm{x}\:\mathrm{be}\:\mathrm{the}\:\mathrm{x}−\mathrm{coordinate}\:\mathrm{of}\:\mathrm{end}\:\mathrm{point}\: \\ $$$$\mathrm{that}\:\mathrm{lies}\:\mathrm{on}\:\mathrm{the}\:\mathrm{x}−\mathrm{axis}\:\mathrm{and}\:\mathrm{let}\:\mathrm{the}\:\mathrm{other} \\ $$$$\mathrm{endpoint}\:\mathrm{be}\:\left(\mathrm{0},\mathrm{y}\right).\:\mathrm{Thus}\:\mathrm{there}\:\mathrm{is}\:\mathrm{a}\:\mathrm{function} \\ $$$$\mathrm{f}\:\mathrm{which}\:\mathrm{gives}\:\mathrm{y}\:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{x}.\:\mathrm{Since} \\…
Question Number 121282 by mnjuly1970 last updated on 06/Nov/20 $$\:\:\:\:\:\:\:\:…\:\mathrm{advanced}\:\:\mathrm{mathematics}… \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\mathrm{prove}\:\:\mathrm{that}\:\::: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Psi=\int_{\mathrm{0}} ^{\:\mathrm{1}} \Gamma\left(\mathrm{2}−{x}\right)\Gamma\left(\mathrm{1}+{x}\right){dx}=\frac{\mathrm{7}}{\pi^{\mathrm{2}} }\:\zeta\left(\mathrm{3}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…\mathrm{m}.\mathrm{n}.\mathrm{july}.\mathrm{1970}… \\ $$$$ \\ $$…
Question Number 121264 by benjo_mathlover last updated on 06/Nov/20 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{point}\:\mathrm{on}\:\mathrm{the}\:\mathrm{curve} \\ $$$$\mathrm{x}=\mathrm{2y}^{\mathrm{2}} \:\mathrm{closest}\:\mathrm{to}\:\mathrm{the}\:\mathrm{point} \\ $$$$\left(\mathrm{10},\mathrm{0}\right) \\ $$ Answered by liberty last updated on 06/Nov/20 $$\mathrm{Let}\:\mathrm{L}\:\mathrm{be}\:\mathrm{the}\:\mathrm{square}\:\mathrm{of}\:\mathrm{distance}\:\mathrm{between}…