The Arithmetic mean of three mutually exclusive regions formed by $y = \sin x, y = \cos x, x = 0, x = \frac{π}{2}$ and x – axis is

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$\frac{\sqrt{2}}{3}$

$\frac{2}{\sqrt{3}}$

$\frac{2}{3}$

$\sqrt{\frac{2}{3}}$

answer is A.

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Detailed Solution

The required areas are the areas of shaded regions as below

The given curves are $y = \sin x, y = \cos x$

The area of the shaded region $A_{1}$ is the area of the region bouned by the curves $y = \sin x, y = \cos x$ and the lines $x = 0, x = \frac{π}{4}$ and $x -$ axis.

Hence,
$\begin{array}{rcl} A_{1} & = & {[\sin x + \cos x]}_{0}^{\frac{π}{4}} \\ = & \sqrt{2} - 1 \end{array}$

The area of the shaded region $A_{2}$ is the area of the region bouned by the curves $y = \sin x, y = \cos x$ and the lines $x = 0, x = \frac{π}{2}$ and $x -$ axis.

Hence,
$\begin{array}{rcl} A_{2} & = & \int_{0}^{\frac{π}{4}} \sin x d x + \int_{\frac{π}{4}}^{\frac{π}{2}} \cos x d x \\ = & {[- \cos x]}_{0}^{\frac{π}{4}} + {[\sin x]}_{\frac{π}{4}}^{\frac{π}{2}} \\ = & - \frac{1}{\sqrt{2}} + 1 + 1 - \frac{1}{\sqrt{2}} \\ = & 2 - \sqrt{2} \end{array}$

The area of the shaded region $A_{3}$ is the area of the region bouned by the curves $y = \sin x, y = \cos x$ and the lines $x = \frac{π}{4}, x = \frac{π}{2}$ and $x -$ axis.

Hence,
$\begin{array}{rcl} A_{3} & = & \int_{\frac{π}{4}}^{\frac{π}{2}} \sin x - \cos x d x \\ = & {[- \cos x - \sin x]}_{\frac{π}{4}}^{\frac{π}{2}} \\ = & - 1 + \sqrt{2} \end{array}$

The Arithmetic means of the above areas is $\frac{A_{1} + A_{2} + A_{3}}{3}$

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