The ratio of the areas bounded by the curves | vedclass

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(D) Let $A_1$ be the area bounded by $y = \cos x$,the $x$-axis,$x = 0$,and $x = \pi/3$. $A_1 = \int_{0}^{\pi/3} \cos x \, dx = [\sin x]_{0}^{\pi/3} =

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$2 : 1$

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(D) Let $A_1$ be the area bounded by $y = \cos x$,the $x$-axis,$x = 0$,and $x = \pi/3$. $A_1 = \int_{0}^{\pi/3} \cos x \, dx = [\sin x]_{0}^{\pi/3} = \sin(\pi/3) - \sin(0) = \frac{\sqrt{3}}{2}$. Let $A_2$ be the area bounded by $y = \cos 2x$,the $x$-axis,$x = 0$,and $x = \pi/3$. $A_2 = \int_{0}^{\pi/3} \cos 2x \, dx = [\frac{1}{2} \sin 2x]_{0}^{\pi/3} = \frac{1}{2} \sin(2\pi/3) - \frac{1}{2} \sin(0) = \frac{1}{2} 	imes \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{4}$. Therefore,the ratio $A_1 : A_2 = \frac{\sqrt{3}}{2} : \frac{\sqrt{3}}{4} = 2 : 1$.

The ratio of the areas bounded by the curves $y = \cos x$ and $y = \cos 2x$ between $x = 0,$ $x = \pi /3$ and the $x$-axis is:

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