The ratio of the areas between the curves y = cos x and y = cos 2x and x-axis from x = 0 to x = π/3 is
(a) 1 : 2
(b) 2 : 1
(c) $\sqrt{3}$ : 1
(d) none of these

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Solution

(d) none of these

$The line x = \frac{π}{3} meets the curve y = \cos x at B (\frac{π}{3}, \frac{1}{2})$
Area between the curve y = cos x and x-axis from x =0 and x = $\frac{π}{3}$ is,
$A_{1} = \int_{0}^{\frac{π}{3}} y_{1} d x [Where, y_{1} = \cos (x)] = \int_{0}^{\frac{π}{3}} \cos (x) d x = {[\sin (x)]}_{0}^{\frac{π}{3}} = \sin (\frac{π}{3}) - \sin (0) = \frac{\sqrt{3}}{2}$

$The line x = \frac{π}{3} meets the curve y = \cos 2 x at B' (\frac{π}{3}, - \frac{1}{2})$

Area between the curve y = cos 2x and x-axis from x =0 and x = $\frac{π}{3}$ is,
$A_{2} = \int_{0}^{\frac{π}{4}} y_{2} d x - \int_{\frac{π}{4}}^{\frac{π}{3}} y_{2} d x [Where, y_{2} = \cos (2 x)] = \int_{0}^{\frac{π}{4}} \cos (2 x) d x - \int_{\frac{π}{4}}^{\frac{π}{3}} \cos (2 x) d x = {[\frac{1}{2} \sin (2 x)]}_{0}^{\frac{π}{4}} - {[\frac{1}{2} \sin (2 x)]}_{\frac{π}{4}}^{\frac{π}{3}} = \frac{1}{2} [\sin (\frac{π}{2}) - \sin (0)] - \frac{1}{2} [\sin (\frac{2 π}{3}) - \sin (\frac{π}{2})] = \frac{1}{2} - \frac{1}{2} [\frac{\sqrt{3}}{2} - 1] = \frac{1}{2} - \frac{\sqrt{3}}{4} + \frac{1}{2} = 1 - \frac{\sqrt{3}}{4} = \frac{4 - \sqrt{3}}{4}$

Therefore the ratios will be

$A_{1} : A_{2} = \frac{A_{1}}{A_{2}} = \frac{\frac{\sqrt{3}}{2}}{\frac{4 - \sqrt{3}}{4}} = \frac{2 \sqrt{3}}{4 - \sqrt{3}}$

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Similar questions

0 \leq x \leq 2 π

\frac{π}{3}

(\frac{1}{y})

y = cos x, y = cos 2 x

x = 0, x = \frac{π}{3}

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