Question Number 213844 by universe last updated on 18/Nov/24 $$\:\int_{−\mathrm{1}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} \int_{\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }} ^{\sqrt{\mathrm{2}−{x}^{\mathrm{2}} −{y}^{\mathrm{2}} }} \sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} }\:{dzdydx} \\…
Question Number 213776 by mnjuly1970 last updated on 16/Nov/24 $$ \\ $$$$\:\:\:\:\:\:\:\:\mathrm{F}{ind}\:\:{the}\:\:{value}\:{of}\:\:{the}\:{following} \\ $$$$\:\:\:\:\:\:\:\:\:\:{expression}. \\ $$$$\:\:\:\:\: \\ $$$$\:\:\:\:\Omega=\:\:\:\frac{\:\mathrm{I}{m}\left(\:\mathrm{Li}_{\mathrm{2}} \:\left(\mathrm{2}\right)\right)}{\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \:\mathrm{ln}\left(\mathrm{sin}\left({x}\:\right)\right)\:{dx}}\:\:=\:? \\ $$ Answered by…
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Question Number 213724 by Spillover last updated on 14/Nov/24 Commented by TonyCWX08 last updated on 16/Nov/24 $${I}\:{see}\:{who}\:{you}\:{are}… \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 213660 by mnjuly1970 last updated on 13/Nov/24 $$ \\ $$$$\:\:\:\:\:{prove}\:{that}\:… \\ $$$$\mathrm{lim}_{{n}\rightarrow\infty} \int_{\mathrm{0}} ^{\:\mathrm{3}} \frac{\:{x}^{\mathrm{2}} \:\left(\mathrm{1}−{x}\:\right){x}^{{n}} \:}{\mathrm{1}+\:{x}^{\mathrm{2}{n}} }\:{dx}\overset{?} {=}\mathrm{0} \\ $$$$\:\:\:\:\:−−−−−−−−−−− \\ $$$$…
Question Number 213662 by Spillover last updated on 13/Nov/24 Answered by mathmax last updated on 14/Nov/24 $${I}=\int_{\mathrm{0}} ^{\infty} \frac{{x}^{\mathrm{2}} }{\mathrm{1}+{e}^{{x}} }{dx}\:\Rightarrow{I}=\int_{\mathrm{0}} ^{\infty} \frac{{e}^{−{x}} {x}^{\mathrm{2}} }{\mathrm{1}+{e}^{−{x}}…
Question Number 213659 by efronzo1 last updated on 13/Nov/24 $$\:\:\:\:\underset{−\infty} {\overset{\infty} {\int}}\:\frac{\mid\mathrm{24x}−\mathrm{24}\mid−\mathrm{20}}{\mathrm{22}^{\mathrm{x}} +\mathrm{22}}\:\mathrm{dx}\:=? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 213656 by efronzo1 last updated on 12/Nov/24 Answered by golsendro last updated on 13/Nov/24 $$\:\:\left(\mathrm{i}\right)\:\mathrm{g}\left(\mathrm{4}−\mathrm{x}\right)=\:−\mathrm{g}\left(\mathrm{x}\right)\: \\ $$$$\:\:\:\:\:\:\:\:\underset{\mathrm{0}} {\overset{\mathrm{4}} {\int}}\:\mathrm{g}\left(\mathrm{x}\right)\mathrm{dx}\:=\:\underset{\mathrm{0}} {\overset{\mathrm{4}} {\int}}\:\mathrm{g}\left(\mathrm{4}−\mathrm{x}\right)\mathrm{dx}\: \\ $$$$\:\:\:\:\:\:\:\:\underset{\mathrm{0}}…