Question Number 109453 by mathdave last updated on 23/Aug/20
Answered by mathmax by abdo last updated on 23/Aug/20
$$\mathrm{I}\:=\int_{\mathrm{0}} ^{\mathrm{4}\pi} \mid\mathrm{cosx}\mid\:\mathrm{dx}\:\Rightarrow\:\mathrm{I}\:=_{\mathrm{x}\:=\mathrm{t}+\mathrm{2}\pi} \:\:\:\:\int_{−\mathrm{2}\pi} ^{\mathrm{2}\pi} \mid\mathrm{cost}\mid\:\mathrm{dt}\: \\ $$$$=\int_{−\mathrm{2}\pi} ^{\mathrm{0}} \:\mid\mathrm{cost}\mid\:\mathrm{dt}\left(\rightarrow\mathrm{t}=−\mathrm{u}\right)\:+\int_{\mathrm{0}} ^{\mathrm{2}\pi} \mid\mathrm{cost}\mid\:\mathrm{dt} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \mid\mathrm{cosu}\mid\mathrm{du}\:+\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\mid\mathrm{cost}\mid\:\mathrm{dt}\:=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{2}\pi} \mid\mathrm{cost}\mid\mathrm{dt} \\ $$$$=\mathrm{2}\:\int_{\mathrm{0}} ^{\pi} \mid\mathrm{cost}\mid\:\mathrm{dt}\:+\mathrm{2}\int_{\pi} ^{\mathrm{2}\pi} \mid\mathrm{cost}\mid\:\mathrm{dt}\left(\rightarrow\mathrm{t}=\mathrm{u}+\pi\right) \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\pi} \:\mid\mathrm{cost}\mid\mathrm{dt}\:+\mathrm{2}\int_{\mathrm{0}} ^{\pi} \:\mid\mathrm{cost}\mid\:\mathrm{dt}\:=\mathrm{4}\int_{\mathrm{0}} ^{\pi} \:\mid\mathrm{cost}\mid\:\mathrm{dt} \\ $$$$=\mathrm{4}\left\{\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{cost}\:\mathrm{dt}\:+\int_{\frac{\pi}{\mathrm{2}}} ^{\pi} −\mathrm{cost}\:\mathrm{dt}\right\} \\ $$$$=\mathrm{4}\left\{\:\left[\mathrm{sint}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} −\left[\mathrm{sint}\right]_{\frac{\pi}{\mathrm{2}}} ^{\pi} \right\}\:=\mathrm{4}\left\{\mathrm{1}−\left(−\mathrm{1}\right)\right\}\:=\mathrm{8}\:\:\:\mathrm{all}\:\mathrm{answer}\:\mathrm{given}\:\mathrm{is}\:\mathrm{false} \\ $$
Commented by Her_Majesty last updated on 24/Aug/20
$${why}\:{so}\:{complicated}? \\ $$$$\underset{\mathrm{0}} {\overset{\mathrm{4}\pi} {\int}}\mid{cosx}\mid{dx}=\mathrm{8}\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}{cosxdx}=\mathrm{8} \\ $$
Commented by malwan last updated on 24/Aug/20
$${fantastic}\:{your}\:{mjs} \\ $$