Mathematics - Area Under Curves Question with Solution | TestHub

Let $A_1$ be the area bounded by $y=\cos x$, $x=0$, $x=\frac{\pi }{3}$ and the X-axis.

$A_1={\int }_0^{\pi /3}|\cos x|dx$

Since $\cos x>0$ in $\left[0,\frac{\pi }{3}\right]$,

$A_1={\int }_0^{\pi /3}\cos xdx=[\sin x{]}_0^{\pi /3}=\sin \left(\frac{\pi }{3}\right)-\sin (0)=\frac{\sqrt{3}}{2}$

Let $A_2$ be the area bounded by $y=\cos 2x$, $x=0$, $x=\frac{\pi }{3}$ and the X-axis.

$A_2={\int }_0^{\pi /3}|\cos 2x|dx$

Since $\cos 2x\ge 0$ in $\left[0,\frac{\pi }{4}\right]$ and $\cos 2x\le 0$ in $\left[\frac{\pi }{4},\frac{\pi }{3}\right]$,

$A_2={\int }_0^{\pi /4}\cos 2xdx-{\int }_{\pi /4}^{\pi /3}\cos 2xdx$

$A_2={\left[\frac{\sin 2x}{2}\right]}_0^{\pi /4}-{\left[\frac{\sin 2x}{2}\right]}_{\pi /4}^{\pi /3}$

$A_2=\frac{1}{2}\left(\sin \frac{\pi }{2}-\sin 0\right)-\frac{1}{2}\left(\sin \frac{2\pi }{3}-\sin \frac{\pi }{2}\right)$

$A_2=\frac{1}{2}(1-0)-\frac{1}{2}\left(\frac{\sqrt{3}}{2}-1\right)$

$A_2=\frac{1}{2}-\frac{\sqrt{3}}{4}+\frac{1}{2}=1-\frac{\sqrt{3}}{4}=\frac{4-\sqrt{3}}{4}$

The required ratio is:

$\frac{A_1}{A_2}=\frac{\frac{\sqrt{3}}{2}}{\frac{4-\sqrt{3}}{4}}=\frac{2\sqrt{3}}{4-\sqrt{3}}$

Thus, the ratio is $2\sqrt{3}:(4-\sqrt{3})$.

Answer: $2\sqrt{3}:(4-\sqrt{3})$