Question Number 113110 by gopikrishnan last updated on 11/Sep/20
$$\overset{\mathrm{1}} {\int}_{\mathrm{0}} \sqrt{{x}\left({x}−\mathrm{1}\right){dx}} \\ $$
Commented by 1549442205PVT last updated on 11/Sep/20
$$\mathrm{The}\:\mathrm{function}\:\sqrt{\mathrm{x}\left(\mathrm{x}−\mathrm{1}\right)}\:\mathrm{is}\:\mathrm{defined}\:\mathrm{on} \\ $$$$\mathrm{set}\:\mathrm{X}=\left(−\infty,\mathrm{0}\right]\cup\left[\mathrm{1},+\infty\right)\:\mathrm{and}\:\mathrm{isn}'\mathrm{t} \\ $$$$\mathrm{defined}\:\mathrm{on}\:\left(\mathrm{0},\mathrm{1}\right),\mathrm{so}\:\overset{\mathrm{1}} {\int}_{\mathrm{0}} \sqrt{{x}\left({x}−\mathrm{1}\right){dx}} \\ $$$$\mathrm{don}'\mathrm{t}\:\mathrm{exist} \\ $$$$\boldsymbol{\mathrm{You}}\:\boldsymbol{\mathrm{should}}\:\boldsymbol{\mathrm{see}}\:\boldsymbol{\mathrm{your}}\:\boldsymbol{\mathrm{question}}\:\boldsymbol{\mathrm{again}} \\ $$
Commented by gopikrishnan last updated on 11/Sep/20
$${ok}\:{sir} \\ $$
Answered by MJS_new last updated on 11/Sep/20
$$\mathrm{I}\:\mathrm{think}\:\mathrm{you}\:\mathrm{mean}\:\int\sqrt{{x}\left({x}−\mathrm{1}\right)}\:{dx} \\ $$$$\int\sqrt{{x}\left({x}−\mathrm{1}\right)}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\sqrt{\frac{{x}}{{x}−\mathrm{1}}}\:\rightarrow\:{dx}=−\mathrm{2}\sqrt{{x}\left({x}−\mathrm{1}\right)^{\mathrm{3}} }{dt}\right] \\ $$$$=−\mathrm{2}\int\frac{{t}^{\mathrm{2}} }{\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{3}} }{dx}= \\ $$$$\:\:\:\:\:\left[\mathrm{Ostrogradski}'\mathrm{s}\:\mathrm{method}\right] \\ $$$$=\frac{{t}\left({t}^{\mathrm{2}} +\mathrm{1}\right)}{\mathrm{4}\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{4}}\int\frac{{dt}}{{t}^{\mathrm{2}} −\mathrm{1}}= \\ $$$$=\frac{{t}\left({t}^{\mathrm{2}} +\mathrm{1}\right)}{\mathrm{4}\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{8}}\mathrm{ln}\:\frac{{t}−\mathrm{1}}{{t}+\mathrm{1}}\:= \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{2}{x}−\mathrm{1}\right)\sqrt{{x}\left({x}−\mathrm{1}\right)}+\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\:\left(\sqrt{{x}}−\sqrt{{x}−\mathrm{1}}\right)\:+{C} \\ $$$$\Rightarrow\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\sqrt{{x}\left({x}−\mathrm{1}\right)}{dx}=\frac{\pi}{\mathrm{8}}\mathrm{i} \\ $$
Commented by gopikrishnan last updated on 11/Sep/20
$${Thank}\:{u}\:{sir} \\ $$
Answered by abdomsup last updated on 11/Sep/20
$${i}\:{think}\:\:{is}\:\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{{x}\left(\mathrm{1}−{x}\right)}{dx} \\ $$$${we}\:{do}\:{the}\:{changement}\:\sqrt{{x}}={t}\:\Rightarrow \\ $$$${I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} {t}\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\left(\mathrm{2}{t}\right){dt} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \:{t}^{\mathrm{2}} \sqrt{\mathrm{1}−{t}^{\mathrm{2}} }{dt} \\ $$$$=_{{t}={sin}\theta} \:\:\:\:\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{sin}^{\mathrm{2}} \theta\:{cos}\theta\:{cos}\theta\:{d}\theta \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\frac{\mathrm{1}}{\mathrm{2}}{sin}\left(\mathrm{2}\theta\right)\right)^{\mathrm{2}} \:{d}\theta \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{1}−{cos}\left(\mathrm{4}\theta\right)}{\mathrm{2}}\:{d}\theta \\ $$$$=\frac{\pi}{\mathrm{8}}\:−\frac{\mathrm{1}}{\mathrm{4}}\left[\frac{\mathrm{1}}{\mathrm{4}}{sin}\left(\mathrm{4}\theta\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:=\frac{\pi}{\mathrm{8}}−\mathrm{0} \\ $$$$\Rightarrow{I}=\frac{\pi}{\mathrm{8}} \\ $$
Commented by gopikrishnan last updated on 11/Sep/20
$${Thank}\:{u}\:{sir} \\ $$