# The diagram shows two curves with equations y=sin x and y=sin2x for x values between 0 and pi. the curves meet at the origin and at the points P and Q. a) find P and Q? b)find the areas of the shaded regions A1 and A2?

Jun 11, 2018

We need to set the two curves equal.

$\sin x = \sin \left(2 x\right)$

$0 = \sin \left(2 x\right) - \sin x$

$0 = 2 \sin x \cos x - \sin x$

$0 = \sin x \left(2 \cos x - 1\right)$

$\sin x = 0 \mathmr{and} \cos x = \frac{1}{2}$

$x = 0 , \pi , \frac{\pi}{3}$

Now we set up our integral expressions for area.

${A}_{1} = {\int}_{0}^{\frac{\pi}{3}} \sin \left(2 x\right) - \sin x \mathrm{dx}$

${A}_{1} = {\left[- \frac{1}{2} \cos \left(2 x\right) + \cos x\right]}_{0}^{\frac{\pi}{3}}$

${A}_{1} = - \frac{1}{2} \cos \left(\frac{2 \pi}{3}\right) + \cos \left(\frac{\pi}{3}\right) + \frac{1}{2} \cos \left(0\right) - \cos \left(0\right)$

${A}_{1} = \frac{1}{4} + \frac{1}{2} + \frac{1}{2} - 1$

${A}_{1} = \frac{1}{4}$ square units.

Now onto ${A}_{2}$.

${A}_{2} = {\int}_{\frac{\pi}{3}}^{\pi} \sin x - \sin \left(2 x\right) \mathrm{dx}$

${A}_{2} = {\left[- \cos x + \frac{1}{2} \cos \left(2 x\right)\right]}_{\frac{\pi}{3}}^{\pi}$

A_2 = -cos(pi) + 1/2cos(2pi) - (-cos(pi/3) + 1/2cos(2(pi/3))

${A}_{2} = 1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{4}$

${A}_{2} = \frac{9}{4}$

The total area is $\frac{5}{2}$ square units.

Hopefully this helps!