find-the-area-between-the-function-y-2sin2x-1-and-the-x-axis-on-pi-pi-2-

Question Number 84135 by M±th+et£s last updated on 09/Mar/20

find the area between the function y=2sin2x −1 and the x−axis on [−π,(π/2)]

$${find}\:{the}\:{area}\:{between}\:{the}\:{function}\: \\ $$$${y}=\mathrm{2}{sin}\mathrm{2}{x}\:−\mathrm{1}\:{and}\:\:{the}\:{x}−{axis}\:\:{on}\:\left[−\pi,\frac{\pi}{\mathrm{2}}\right] \\ $$

Answered by Rio Michael last updated on 09/Mar/20

$$\int_{−\pi} ^{\frac{\pi}{\mathrm{2}}} \left(\mathrm{2sin}\:\mathrm{2}{x}−\mathrm{1}\right){dx} \\ $$

Commented by MJS last updated on 09/Mar/20

$$\mathrm{this}\:\mathrm{is}\:\mathrm{not}\:\mathrm{the}\:\mathrm{area} \\ $$

Answered by MJS last updated on 09/Mar/20

for the area we need the zeros 2sin 2x −1=0 sin 2x =(1/2) ∧ −π≤x≤(π/2) ⇒ x∈{−((11π)/(12)); −((7π)/(12)); (π/(12)); ((5π)/(12))} y=2sin (−2π) −1=−1<0 ⇒ area = −∫_(−π) ^(−((11π)/(12))) ydx+∫_(−((11π)/(12))) ^(−((7π)/(12))) ydx−∫_(−((7π)/(12))) ^(π/(12)) ydx+∫_(π/(12)) ^((5π)/(12)) ydx−∫_((5π)/(12)) ^(π/2) ydx ∫ydx=∫(2sin 2x −1)dx=−x−cos 2x +C ⇒ area = (π/6)−2+4(√3) ≈5.45180

$$\mathrm{for}\:\mathrm{the}\:\mathrm{area}\:\mathrm{we}\:\mathrm{need}\:\mathrm{the}\:\mathrm{zeros} \\ $$$$\mathrm{2sin}\:\mathrm{2}{x}\:−\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{sin}\:\mathrm{2}{x}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\wedge\:−\pi\leqslant{x}\leqslant\frac{\pi}{\mathrm{2}} \\ $$$$\Rightarrow\:{x}\in\left\{−\frac{\mathrm{11}\pi}{\mathrm{12}};\:−\frac{\mathrm{7}\pi}{\mathrm{12}};\:\frac{\pi}{\mathrm{12}};\:\frac{\mathrm{5}\pi}{\mathrm{12}}\right\} \\ $$$${y}=\mathrm{2sin}\:\left(−\mathrm{2}\pi\right)\:−\mathrm{1}=−\mathrm{1}<\mathrm{0} \\ $$$$\Rightarrow \\ $$$$\mathrm{area}\:=\:−\underset{−\pi} {\overset{−\frac{\mathrm{11}\pi}{\mathrm{12}}} {\int}}{ydx}+\underset{−\frac{\mathrm{11}\pi}{\mathrm{12}}} {\overset{−\frac{\mathrm{7}\pi}{\mathrm{12}}} {\int}}{ydx}−\underset{−\frac{\mathrm{7}\pi}{\mathrm{12}}} {\overset{\frac{\pi}{\mathrm{12}}} {\int}}{ydx}+\underset{\frac{\pi}{\mathrm{12}}} {\overset{\frac{\mathrm{5}\pi}{\mathrm{12}}} {\int}}{ydx}−\underset{\frac{\mathrm{5}\pi}{\mathrm{12}}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}{ydx} \\ $$$$\int{ydx}=\int\left(\mathrm{2sin}\:\mathrm{2}{x}\:−\mathrm{1}\right){dx}=−{x}−\mathrm{cos}\:\mathrm{2}{x}\:+{C} \\ $$$$\Rightarrow \\ $$$$\mathrm{area}\:=\:\frac{\pi}{\mathrm{6}}−\mathrm{2}+\mathrm{4}\sqrt{\mathrm{3}}\:\approx\mathrm{5}.\mathrm{45180} \\ $$

Commented by M±th+et£s last updated on 10/Mar/20

$${thank}\:{you}\:{sir} \\ $$

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