Integration Questions

Question Number 152088 by rexford last updated on 25/Aug/21

$$\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}} \mid{sinx}−{cosx}\mid \\$$$${please}\:{help}\:{me}\:{out} \\$$

Answered by Olaf_Thorendsen last updated on 25/Aug/21

$$\mathrm{I}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mid\mathrm{sin}{x}−\mathrm{cos}{x}\mid\:{dx} \\$$$$\mathrm{I}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mid\sqrt{\mathrm{2}}\mathrm{sin}\left(\frac{\pi}{\mathrm{4}}−{x}\right)\mid\:{dx} \\$$$$\mathrm{I}\:=\:\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \sqrt{\mathrm{2}}\mathrm{sin}\left(\frac{\pi}{\mathrm{4}}−{x}\right)\:{dx} \\$$$$\mathrm{I}\:=\:\mathrm{2}\sqrt{\mathrm{2}}\left[\mathrm{cos}\left(\frac{\pi}{\mathrm{4}}−{x}\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \\$$$$\mathrm{I}\:=\:\mathrm{2}\sqrt{\mathrm{2}}\left(\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)\:=\:\mathrm{2}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right) \\$$

Answered by puissant last updated on 25/Aug/21

$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \left({cosx}−{sinx}\right){dx}+\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{2}}} \left({sinx}−{cosx}\right){dx} \\$$$$=\left[{sinx}+{cosx}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} +\left[−{cosx}−{sinx}\right]_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{2}}} \\$$$$=\:\sqrt{\mathrm{2}}−\mathrm{1}+\left(−\mathrm{1}+\sqrt{\mathrm{2}}\right) \\$$$$\\$$$$=\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{2} \\$$$$\\$$$$=\mathrm{2}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right) \\$$

Commented by peter frank last updated on 25/Aug/21

$$\mathrm{good} \\$$

Commented by rexford last updated on 26/Aug/21

$${thank}\:{you} \\$$