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Question Number 109453 by mathdave last updated on 23/Aug/20

Answered by mathmax by abdo last updated on 23/Aug/20

$$\mathrm{I}={\int }_0^{4\pi }\mid \mathrm{cosx}\mid \mathrm{dx}\Rightarrow \mathrm{I}{=}_{\mathrm{x}=\mathrm{t}+2\pi }{\int }_{-2\pi }^{2\pi }\mid \mathrm{cost}\mid \mathrm{dt}$$$$={\int }_{-2\pi }^0\mid \mathrm{cost}\mid \mathrm{dt}(\to \mathrm{t}=-\mathrm{u})+{\int }_0^{2\pi }\mid \mathrm{cost}\mid \mathrm{dt}$$$$={\int }_0^{2\pi }\mid \mathrm{cosu}\mid \mathrm{du}+{\int }_0^{2\pi }\mid \mathrm{cost}\mid \mathrm{dt}=2{\int }_0^{2\pi }\mid \mathrm{cost}\mid \mathrm{dt}$$$$=2{\int }_0^{\pi }\mid \mathrm{cost}\mid \mathrm{dt}+2{\int }_{\pi }^{2\pi }\mid \mathrm{cost}\mid \mathrm{dt}(\to \mathrm{t}=\mathrm{u}+\pi )$$$$=2{\int }_0^{\pi }\mid \mathrm{cost}\mid \mathrm{dt}+2{\int }_0^{\pi }\mid \mathrm{cost}\mid \mathrm{dt}=4{\int }_0^{\pi }\mid \mathrm{cost}\mid \mathrm{dt}$$$$=4\{{\int }_0^{\frac{\pi }{2}}\mathrm{cost}\mathrm{dt}+{\int }_{\frac{\pi }{2}}^{\pi }-\mathrm{cost}\mathrm{dt}\}$$$$=4\{{\left[\mathrm{sint}\right]}_0^{\frac{\pi }{2}}-{\left[\mathrm{sint}\right]}_{\frac{\pi }{2}}^{\pi }\}=4\{1-(-1)\}=8\mathrm{all}\mathrm{answer}\mathrm{given}\mathrm{is}\mathrm{false}$$

Commented by Her_Majesty last updated on 24/Aug/20

$$whysocomplicated?$$$$\underset{0}{\stackrel{4\pi }{\int }}\mid cosx\mid dx=8\underset{0}{\stackrel{\pi /2}{\int }}cosxdx=8$$

Commented by malwan last updated on 24/Aug/20

$$fantasticyourmjs$$

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