Question Number 206830 by Fabricista15 last updated on 27/Apr/24 $${c}\:=\:\sqrt{\left(\int_{{a}_{\mathrm{0}} } ^{{a}_{\mathrm{1}} } \sqrt{\mathrm{1}+\left[{f}'\left({x}\right)\right]^{\mathrm{2}} }{dx}\right)^{\mathrm{2}} +\left(\int_{{b}_{\mathrm{0}} } ^{{b}_{\mathrm{1}} } \sqrt{\mathrm{1}+\left[{f}'\left({x}\right)\right]^{\mathrm{2}} }{dx}\right)^{\mathrm{2}} } \\ $$$${c}\:=\:\sqrt{{L}_{\mathrm{1}} ^{\mathrm{2}}…
Question Number 206829 by 2kdw last updated on 27/Apr/24 $$\:\:\:\:\:\oint\frac{{x}}{{x}+\mathrm{2}}{dx}^{\mathrm{2}} \:\:\:\:{is}\:{wrong}? \\ $$ Commented by mr W last updated on 27/Apr/24 $${there}\:{are}\:{things}\:{to}\:{which}\:{you}\:{can} \\ $$$${say}\:{they}\:{are}\:{wrong},\:{for}\:{example} \\…
Question Number 206858 by Ghisom last updated on 27/Apr/24 $$\mathrm{prove}\:\mathrm{that} \\ $$$${H}_{{n}} =\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\frac{{t}^{{n}} −\mathrm{1}}{{t}−\mathrm{1}}{dt} \\ $$ Answered by mathzup last updated on 28/Apr/24…
Question Number 206789 by BaliramKumar last updated on 25/Apr/24 Answered by A5T last updated on 25/Apr/24 $$\sqrt[{{n}}]{{a}_{\mathrm{1}} {a}_{\mathrm{2}} …{a}_{{n}} }={P} \\ $$$$\sqrt[{\mathrm{2}{n}}]{{a}_{{n}+\mathrm{1}} {a}_{{n}+\mathrm{2}} …{a}_{\mathrm{3}{n}} }={Q}…
Question Number 206791 by akolade last updated on 25/Apr/24 $$\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}−\sqrt{\mathrm{x}}}.\mathrm{ln}^{\mathrm{2}} \mathrm{x}\:\mathrm{dx} \\ $$ Commented by Frix last updated on 25/Apr/24 $$\mathrm{Question}\:\mathrm{206754} \\ $$…
Question Number 206754 by mathzup last updated on 23/Apr/24 $${find}\:\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}−\sqrt{{x}}}{ln}^{\mathrm{2}} \left({x}\right){dx} \\ $$ Answered by Berbere last updated on 24/Apr/24 $${x}={u}^{\mathrm{2}} \Rightarrow{dx}=\mathrm{2}{udu} \\…
Question Number 206721 by ajfour last updated on 23/Apr/24 $$\int\frac{{xdx}}{{x}+\mathrm{4}}=?\:\:\:\:\:\:\:{please} \\ $$ Answered by A5T last updated on 23/Apr/24 $$\frac{{x}}{{x}+\mathrm{4}}=\frac{{x}+\mathrm{4}−\mathrm{4}}{{x}+\mathrm{4}}=\mathrm{1}−\frac{\mathrm{4}}{{x}+\mathrm{4}} \\ $$$$\Rightarrow\int\frac{{x}}{{x}+\mathrm{4}}{dx}=\int\mathrm{1}{dx}−\mathrm{4}\int\frac{\mathrm{1}}{{x}+\mathrm{4}}{dx} \\ $$$$={x}−\mathrm{4}{ln}\mid{x}+\mathrm{4}\mid+{c} \\…
Question Number 206645 by Red1ight last updated on 21/Apr/24 $$\mathrm{Let}\:{f}\left({x}\right)={x}\left({x}−\mathrm{10}\right) \\ $$$$\mathrm{and}\:\mathrm{let}\:\mathrm{A}\:\mathrm{be}\:\mathrm{the}\:\mathrm{region}\:\mathrm{enclosed}\:\mathrm{within} \\ $$$$\mathrm{the}\:\mathrm{following}\:\mathrm{points} \\ $$$$\left(\mathrm{2},\mathrm{7}\right),\left(\mathrm{8},\mathrm{7}\right),\left(\mathrm{2},\mathrm{4}\right),\left(\mathrm{8},\mathrm{4}\right) \\ $$$$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{average}\:\mathrm{arc}\:\mathrm{length}\:\mathrm{of}\:\mathrm{a}\centerdot{f}\left({x}\right) \\ $$$$\mathrm{inside}\:\mathrm{A},{a}\in\mathbb{R}^{−} \\ $$ Commented by mr…
Question Number 206642 by universe last updated on 21/Apr/24 Answered by MathematicalUser2357 last updated on 26/Apr/24 $$\mathrm{The}\:\mathrm{calculation}\:\mathrm{was}\:\mathrm{aborted}\:\mathrm{because}\:\mathrm{it}\:\mathrm{took}\:\mathrm{too} \\ $$$$\mathrm{long}.\:\mathrm{Please}\:\mathrm{make}\:\mathrm{sure}\:\mathrm{that}\:\mathrm{your}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{incorrect},\:\mathrm{or} \\ $$$$\mathrm{try}\:\mathrm{tosimplify}\:\mathrm{your}\:\mathrm{query}.\:{Have}\:{you}\:{tried}\:{using}\:{the} \\ $$$${option}\:{to}\:{simplify}\:{expressions}??? \\ $$…
Question Number 206639 by mathlove last updated on 21/Apr/24 Commented by Frix last updated on 21/Apr/24 $$=\underset{\frac{\pi}{\mathrm{12}}} {\overset{\frac{\pi}{\mathrm{11}}} {\int}}\left(\frac{\sqrt{{s}}}{\mathrm{2}\left(\sqrt{{c}}+\sqrt{{s}}\right)}−\frac{\sqrt{{cs}}}{{c}+\sqrt{{s}}\left(\sqrt{{c}}+\mathrm{1}\right)}+\frac{\left(\sqrt{{c}}−\mathrm{1}\right){s}+\sqrt{{cs}}}{\mathrm{2}\left(\left(\sqrt{{s}}+\mathrm{1}\right){c}+\left(\sqrt{{c}}+\mathrm{1}\right){s}\right)}\right){dx} \\ $$$$\mathrm{With}\:{c}=\mathrm{cos}\:{x}\:\wedge{s}=\mathrm{sin}\:{x} \\ $$$$\mathrm{I}\:\mathrm{doubt}\:\mathrm{we}\:\mathrm{can}\:\mathrm{solve}\:\mathrm{this}. \\ $$…