Question Number 69027 by rajesh4661kumar@gmail.com last updated on 18/Sep/19
Commented by mathmax by abdo last updated on 18/Sep/19
![I =∫_0 ^(2π) sinxdx −∫_0 ^(2π) ∣sinx∣dx but ∫_0 ^(2π) sinx dx =_(x=π+t) ∫_(−π) ^π sin(π+t)dt =−∫_(−π) ^π sint dt =0(odd function) ∫_0 ^(2π) ∣sinx∣dx =_(x=π+t) ∫_(−π) ^π ∣sint∣dt =2 ∫_0 ^π sint dt =2[−cost]_0 ^π =2{1−(−1)} =4](Q69042.png)
$${I}\:=\int_{\mathrm{0}} ^{\mathrm{2}\pi} {sinxdx}\:−\int_{\mathrm{0}} ^{\mathrm{2}\pi} \mid{sinx}\mid{dx}\:\:{but} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:{sinx}\:{dx}\:=_{{x}=\pi+{t}} \:\:\int_{−\pi} ^{\pi} {sin}\left(\pi+{t}\right){dt}\:=−\int_{−\pi} ^{\pi} \:{sint}\:{dt}\:=\mathrm{0}\left({odd}\:{function}\right) \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}\pi} \mid{sinx}\mid{dx}\:=_{{x}=\pi+{t}} \:\:\int_{−\pi} ^{\pi} \mid{sint}\mid{dt}\:=\mathrm{2}\:\int_{\mathrm{0}} ^{\pi} {sint}\:{dt}\:=\mathrm{2}\left[−{cost}\right]_{\mathrm{0}} ^{\pi} \\ $$$$=\mathrm{2}\left\{\mathrm{1}−\left(−\mathrm{1}\right)\right\}\:=\mathrm{4} \\ $$
Commented by mathmax by abdo last updated on 18/Sep/19
$$\Rightarrow\:{I}\:=\mathrm{0}−\mathrm{4}\:=−\mathrm{4}\:. \\ $$
Answered by MJS last updated on 18/Sep/19
![sin x −∣sin x∣= { ((0; 0≤x<π)),((2sin x; π≤x≤2π)) :} ⇒ ∫_0 ^(2π) sin x −∣sin x∣ dx=2∫_π ^(2π) sin x dx=−2∫_0 ^π sin x dx= =2[cos x]_0 ^π =−4](Q69029.png)
$$\mathrm{sin}\:{x}\:−\mid\mathrm{sin}\:{x}\mid=\begin{cases}{\mathrm{0};\:\mathrm{0}\leqslant{x}<\pi}\\{\mathrm{2sin}\:{x};\:\pi\leqslant{x}\leqslant\mathrm{2}\pi}\end{cases} \\ $$$$\Rightarrow\:\underset{\mathrm{0}} {\overset{\mathrm{2}\pi} {\int}}\mathrm{sin}\:{x}\:−\mid\mathrm{sin}\:{x}\mid\:{dx}=\mathrm{2}\underset{\pi} {\overset{\mathrm{2}\pi} {\int}}\mathrm{sin}\:{x}\:{dx}=−\mathrm{2}\underset{\mathrm{0}} {\overset{\pi} {\int}}\mathrm{sin}\:{x}\:{dx}= \\ $$$$=\mathrm{2}\left[\mathrm{cos}\:{x}\right]_{\mathrm{0}} ^{\pi} =−\mathrm{4} \\ $$