Integration Questions

Question Number 90784 by jagoll last updated on 26/Apr/20

$$\underset{\mathrm{0}} {\overset{\mathrm{2}} {\int}}\:\sqrt{\frac{\mathrm{4}−{x}}{{x}}}−\sqrt{\frac{{x}}{\mathrm{4}−{x}}}\:{dx}\:? \\$$

Commented by jagoll last updated on 26/Apr/20

$$\int_{\mathrm{0}} ^{\mathrm{2}} \:\frac{\left(\sqrt{\mathrm{4}−{x}}\right)^{\mathrm{2}} −\left(\sqrt{{x}}\right)^{\mathrm{2}} }{\sqrt{{x}}\:\left(\sqrt{\mathrm{4}−{x}}\:\right)\:}\:=\: \\$$$$\int_{\mathrm{0}} ^{\mathrm{2}} \:\frac{\mathrm{4}−\mathrm{2}{x}}{\sqrt{\mathrm{4}{x}−{x}^{\mathrm{2}} }}\:{dx}\:=\:\int_{\mathrm{0}} ^{\mathrm{2}} \:\frac{{d}\left(\mathrm{4}{x}−{x}^{\mathrm{2}} \right)}{\sqrt{\mathrm{4}{x}−{x}^{\mathrm{2}} }} \\$$$$\left.=\:\mathrm{2}\:\sqrt{\mathrm{4}{x}−{x}^{\mathrm{2}} }\:\right]_{\mathrm{0}} ^{\mathrm{2}} \:=\:\mathrm{2}\:\sqrt{\mathrm{8}−\mathrm{4}} \\$$$$=\:\mathrm{2}×\mathrm{2}\:=\:\mathrm{4} \\$$

Commented by abdomathmax last updated on 27/Apr/20

$${A}=\int_{\mathrm{0}} ^{\mathrm{2}} \sqrt{\frac{\mathrm{4}−{x}}{{x}}}{dx}−\int_{\mathrm{0}} ^{\mathrm{2}} \sqrt{\frac{{x}}{\mathrm{4}−{x}}}{dx}\:={H}−{K} \\$$$${changement}\:\sqrt{\frac{\mathrm{4}−{x}}{{x}}}={t}\:{give}\:\frac{\mathrm{4}−{x}}{{x}}={t}^{\mathrm{2}} \:\Rightarrow \\$$$$\mathrm{4}−{x}\:={t}^{\mathrm{2}} {x}\:\Rightarrow\left(\mathrm{1}+{t}^{\mathrm{2}} \right){x}\:=\mathrm{4}\:\Rightarrow{x}=\frac{\mathrm{4}}{{t}^{\mathrm{2}} \:+\mathrm{1}}\:\Rightarrow \\$$$${H}\:=−\int_{\mathrm{1}} ^{+\infty} {t}×\left(−\frac{\mathrm{4}\left(\mathrm{2}{t}\right)}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }\right){dt} \\$$$$=\mathrm{8}\:\int_{\mathrm{1}} ^{+\infty} \:\frac{{t}^{\mathrm{2}} +\mathrm{1}−\mathrm{1}}{\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }{dt}\:=\mathrm{8}\:\int_{\mathrm{1}} ^{+\infty} \frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }−\mathrm{8}\int_{\mathrm{1}} ^{+\infty} \:\frac{{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} } \\$$$$\int_{\mathrm{1}} ^{+\infty} \:\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }\:=\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{4}}\:=\frac{\pi}{\mathrm{4}} \\$$$$\int_{\mathrm{1}} ^{+\infty} \:\frac{{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }\:=_{{t}={tan}\theta} \:\:\:\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{1}+{tan}^{\mathrm{2}} \theta}{\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right)^{\mathrm{2}} }{d}\theta \\$$$$=\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{2}}} {cos}^{\mathrm{2}} \theta\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{2}}} \left(\mathrm{1}+{cos}\left(\mathrm{2}\theta\right)\right){d}\theta \\$$$$=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\pi}{\mathrm{4}}\:+\frac{\mathrm{1}}{\mathrm{4}}\left[{sin}\left(\mathrm{2}\theta\right)\right]_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{2}}} \:=\frac{\pi}{\mathrm{8}}+\frac{\mathrm{1}}{\mathrm{4}}\left(−\mathrm{1}\right)\:\Rightarrow \\$$$${H}\:=\mathrm{2}\pi−\mathrm{8}\left\{\frac{\pi}{\mathrm{8}}−\frac{\mathrm{1}}{\mathrm{4}}\right\}=\pi+\mathrm{2}\:\:\:{and}\:{K}=\int_{\mathrm{0}} ^{\mathrm{2}} \sqrt{\frac{{x}}{\mathrm{4}−{x}}}{dx} \\$$$${we}\:{do}\:{the}\:{changement}\:\sqrt{\frac{{x}}{\mathrm{4}−{x}}}={t}\:\Rightarrow \\$$$$\frac{{x}}{\mathrm{4}−{x}}\:={t}^{\mathrm{2}} \:\Rightarrow{x}\:=\mathrm{4}{t}^{\mathrm{2}} −{t}^{\mathrm{2}} {x}\:\Rightarrow\left(\mathrm{1}+{t}^{\mathrm{2}} \right){x}\:=\mathrm{4}{t}^{\mathrm{2}} \:\Rightarrow \\$$$${x}\:=\frac{\mathrm{4}{t}^{\mathrm{2}} }{{t}^{\mathrm{2}} \:+\mathrm{1}}\:\Rightarrow\frac{{dx}}{{dt}}\:=\frac{\mathrm{8}{t}\left({t}^{\mathrm{2}} +\mathrm{1}\right)−\mathrm{4}{t}^{\mathrm{2}} \left(\mathrm{2}{t}\right)}{\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }=\frac{\mathrm{8}{t}}{\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} } \\$$$${K}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:{t}×\frac{\mathrm{8}{t}}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }{dt}\:=\mathrm{8}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{t}^{\mathrm{2}} \:+\mathrm{1}−\mathrm{1}}{\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }{dt} \\$$$$=\mathrm{8}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dt}}{{t}^{\mathrm{2}} \:+\mathrm{1}}−\mathrm{8}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dt}}{\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} } \\$$$$=\mathrm{8}×\frac{\pi}{\mathrm{4}}−\mathrm{8}\left\{\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\frac{\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right){d}\theta}{\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right)^{\mathrm{2}} }\right\}\:\:\left({t}={tan}\theta\right) \\$$$$=\mathrm{2}\pi\:−\mathrm{8}\left\{\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {cos}^{\mathrm{2}} \theta\:{d}\theta\right\} \\$$$$=\mathrm{2}\pi−\mathrm{4}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \left(\mathrm{1}+{cos}\left(\mathrm{2}\theta\right)\right){d}\theta \\$$$$=\pi−\mathrm{2}\:\left[{sin}\left(\mathrm{2}\theta\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:=\pi−\mathrm{2}\left\{\mathrm{1}\right\}\:=\pi−\mathrm{2} \\$$$${A}\:={H}−{K}\:=\pi\:+\mathrm{2}−\pi\:+\mathrm{2}\:=\mathrm{4} \\$$$$\\$$$$\\$$$$\\$$