The ratio of the areas bounded by the curves | vedclass

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Question

(C) Let $A_1$ be the area bounded by $y = \cos x$ from $x = 0$ to $x = \frac{\pi}{3}$.$A_1 = \int_{0}^{\pi/3} \cos x \, dx = [\sin x]_{0}^{\pi/3} = \s

Vedclass · Accepted Answer

$2: 1$

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(C) Let $A_1$ be the area bounded by $y = \cos x$ from $x = 0$ to $x = \frac{\pi}{3}$.$A_1 = \int_{0}^{\pi/3} \cos x \, dx = [\sin x]_{0}^{\pi/3} = \sin(\frac{\pi}{3}) - \sin(0) = \frac{\sqrt{3}}{2}$.Let $A_2$ be the area bounded by $y = \cos 2x$ from $x = 0$ to $x = \frac{\pi}{3}$.$A_2 = \int_{0}^{\pi/3} \cos 2x \, dx = [\frac{\sin 2x}{2}]_{0}^{\pi/3} = \frac{1}{2} (\sin(\frac{2\pi}{3}) - \sin(0)) = \frac{1}{2} (\frac{\sqrt{3}}{2}) = \frac{\sqrt{3}}{4}$.The ratio of the areas is $\frac{A_1}{A_2} = \frac{\sqrt{3}/2}{\sqrt{3}/4} = \frac{4}{2} = 2: 1$.

The ratio of the areas bounded by the curves $y = \cos x$ and $y = \cos 2x$ between $x = 0$ and $x = \frac{\pi}{3}$ with the $X$-axis is:

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