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Question Number 32763 by San Sophanethsan069 last updated on 01/Apr/18

Commented by abdo imad last updated on 01/Apr/18

$$letputI={\int }_0^{2\pi }(\mid sinx\mid +\mid cosx\mid )dx.ch.x=\pi +tgive$$$$I={\int }_{-\pi }^{\pi }(\mid sint\mid +\mid cost\mid )dt=2{\int }_0^{\pi }(sint+\mid cost\mid )dt$$$$=2{\int }_0^{\pi }sintdt+2{\int }_0^{\pi }\mid cost\mid dtbut$$$${\int }_0^{\pi }sintdt={[-cost]}_0^{\pi }=2$$$${\int }_0^{\pi }\mid cost\mid dt={\int }_0^{\frac{\pi }{2}}cost-{\int }_{\frac{\pi }{2}}^{\pi }costdt$$$$={\left[sint\right]}_0^{\frac{\pi }{2}}-{\left[sint\right]}_{\frac{\pi }{2}}^{\pi }=1-(-1)=2\Rightarrow$$$$I=4+4=8.$$

Answered by MJS last updated on 02/Apr/18

$$\sin x,\cos x\mathrm{both}\mathrm{run}\mathrm{a}\mathrm{full}$$$$\mathrm{period}\mathrm{in}[0;2\pi ]\Rightarrow$$$$\Rightarrow \underset{0}{\stackrel{2\pi }{\int }}\mid \sin x\mid dx=\underset{0}{\stackrel{2\pi }{\int }}\mid \cos x\mid dx=$$$$=2\underset{0}{\stackrel{\pi }{\int }}\sin xdx$$$$\mathrm{so}{\mathrm{we}}^{\prime }\mathrm{re}\mathrm{looking}\mathrm{for}$$$$4\underset{0}{\stackrel{\pi }{\int }}\sin xdx=-4\cos x\underset{0}{\stackrel{\pi }{\mid }}=$$$$=-4\cos \pi +4\cos 0=8$$

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