The ratio of the areas between the curves y = cos x and y = cos 2x and x-axis from x = 0 to x = $\pi$/3 is ________ .

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none of these \text{ The line }x = $\pi$/3\text{ meets the curve }y = \cos x \text{ at } B\left( $\pi$/3, $\frac{1}{2}$ \right) Area between the curve y = cos x and x-axis from x =0 and x =$\pi$/3 is, A_1 = \int_0^$\pi$/3 y_1 d x .............\left[\text{Where, }y_1 = \cos\left( x \right) \right] = \int_0^$\pi$/3 \cos\left( x \right) d x = \left[ \sin\left( x \right) \right]_0^$\pi$/3 = \sin\left( $\pi$/3 \right) - \sin\left( 0 \right) = $\sqrt$(3)/2 \text{ The line } x = $\pi$/3\text{ meets the curve }y = \cos 2x\text{ at }B'\left( $\pi$/3, - $\frac{1}{2}$ \right) Area between the curve y = cos 2x and x-axis from x =0 and x = $\pi$/3 is, A_2 = \int_0^$\pi$/4 y_2 d x - \int_$\pi$/4^$\pi$/3 y_2 d x ............\left[\text{Where, }y_2 = \cos \left( 2x \right) \right] = \int_0^$\pi$/4 \cos \left( 2x \right) d x - \int_$\pi$/4^$\pi$/3 \cos \left( 2x \right) d x = \left[ $\frac{1}{2}$\sin \left( 2x \right) \right]_0^$\pi$/4 - \left[ $\frac{1}{2}$\sin \left( 2x \right) \right]_$\pi$/4^$\pi$/3 = $\frac{1}{2}$\left[ \sin \left( $\pi$/2 \right) - \sin \left( 0 \right) \right] - $\frac{1}{2}$\left[ \sin \left( 2$\pi$/3 \right) - \sin \left( $\pi$/2 \right) \right] = $\frac{1}{2}$ - $\frac{1}{2}$\left[ $\sqrt$(3)/2 - 1 \right] = $\frac{1}{2}$ - $\sqrt$(3)/4 + $\frac{1}{2}$ = 1 - $\sqrt$(3)/4 = 4 - $\sqrt$(3)/4 Therefore the ratios will be A_1 : A_2 = A_1/A_2 = \frac{$\sqrt$(3)/2}{4 - $\sqrt$(3)/4} = 2$\sqrt$(3)/4 - $\sqrt$(3) [Solution diagram: 44036_solution_1.png] [Solution diagram: 44036_solution_2.png]

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