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Question Number 55775 by gunawan last updated on 04/Mar/19

∫_( 0) ^1 ∣sin 2π x∣ dx =

$$\:\underset{\:\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\:\mid\mathrm{sin}\:\mathrm{2}\pi\:{x}\mid\:{dx}\:= \\ $$

Commented by maxmathsup by imad last updated on 04/Mar/19

let I =∫_0 ^1 ∣sin(2πx)∣dx ⇒ I =_(2πx =t) ∫_0 ^(2π) ∣sin(t)∣(dt/(2π)) =(1/(2π)){ ∫_0 ^π sint dt ) +∫_π ^(2π) ∣sint∣dt} but ∫_π ^(2π) ∣sint∣dt =_(t =π+u) ∫_0 ^π sinu du ⇒ I =(1/π) ∫_0 ^π sint dt =(1/π)[−cost]_0 ^π =(2/π) ⇒ I =(2/π) .

$${let}\:{I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \mid{sin}\left(\mathrm{2}\pi{x}\right)\mid{dx}\:\:\Rightarrow\:{I}\:=_{\mathrm{2}\pi{x}\:={t}} \:\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\mid{sin}\left({t}\right)\mid\frac{{dt}}{\mathrm{2}\pi} \\ $$$$\left.=\frac{\mathrm{1}}{\mathrm{2}\pi}\left\{\:\int_{\mathrm{0}} ^{\pi} \:{sint}\:{dt}\:\right)\:+\int_{\pi} ^{\mathrm{2}\pi} \:\mid{sint}\mid{dt}\right\}\:\:{but}\:\int_{\pi} ^{\mathrm{2}\pi} \mid{sint}\mid{dt}\:=_{{t}\:=\pi+{u}} \int_{\mathrm{0}} ^{\pi} {sinu}\:{du}\:\Rightarrow \\ $$$${I}\:=\frac{\mathrm{1}}{\pi}\:\int_{\mathrm{0}} ^{\pi} \:{sint}\:{dt}\:=\frac{\mathrm{1}}{\pi}\left[−{cost}\right]_{\mathrm{0}} ^{\pi} \:=\frac{\mathrm{2}}{\pi}\:\Rightarrow\:{I}\:=\frac{\mathrm{2}}{\pi}\:. \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 04/Mar/19

∫_0 ^(1/2) ∣sin2πx∣dx+∫_(1/2) ^1 ∣sin2πx∣dx ∫_0 ^(1/2) sin2πx dx+∫_(1/2) ^1 −sin2πx dx −∣((cos2πx)/(2π))∣_0 ^(1/2) +∣((cos2πx)/(2π))∣_(1/2) ^1 =((−cosπ+cos0)/(2π))+((cos2π−cosπ)/(2π)) =((1+1+1+1)/(2π))=(2/π) pls check...

$$\underset{\mathrm{0}} {\overset{\frac{\mathrm{1}}{\mathrm{2}}} {\int}}\mid{sin}\mathrm{2}\pi{x}\mid{dx}+\overset{\mathrm{1}} {\int}_{\frac{\mathrm{1}}{\mathrm{2}}} \mid{sin}\mathrm{2}\pi{x}\mid{dx} \\ $$$$\underset{\mathrm{0}} {\overset{\frac{\mathrm{1}}{\mathrm{2}}} {\int}}{sin}\mathrm{2}\pi{x}\:{dx}+\underset{\frac{\mathrm{1}}{\mathrm{2}}} {\overset{\mathrm{1}} {\int}}−{sin}\mathrm{2}\pi{x}\:{dx} \\ $$$$−\mid\frac{{cos}\mathrm{2}\pi{x}}{\mathrm{2}\pi}\mid_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} +\mid\frac{{cos}\mathrm{2}\pi{x}}{\mathrm{2}\pi}\mid_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \: \\ $$$$=\frac{−{cos}\pi+{cos}\mathrm{0}}{\mathrm{2}\pi}+\frac{{cos}\mathrm{2}\pi−{cos}\pi}{\mathrm{2}\pi} \\ $$$$=\frac{\mathrm{1}+\mathrm{1}+\mathrm{1}+\mathrm{1}}{\mathrm{2}\pi}=\frac{\mathrm{2}}{\pi} \\ $$$${pls}\:{check}... \\ $$

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