91Ó°ÊÓ

Since \(R_1\) is in the first quadrant and we are given the curve \(y=\cos^{-1}x\), we need to find the x-values at which this curve intersects the x-axis (when \(y=0\)) and y-axis (when \(x=0\)). For the x-axis intersection, we have: \(y=\cos^{-1}x\) when \(y=0\), \(x=\cos(0)=1\). For the y-axis intersection, we have: \(y=\cos^{-1}x\) when \(x=0\), \(y=\cos^{-1}(0)=\frac{\pi}{2}\). So the limits of integration for \(R_1\) would be from \(x=0\) to \(x=1\) and for \(y=0\) to \(y=\frac{\pi}{2}\).

To find the area of \(R_1\), we will integrate the function \(y=\cos^{-1}x\) with respect to x, using the limits of integration found in Step 1: \(\displaystyle{\int_0^1 (\cos^{-1}x) dx}\). Integration by parts is required for this integral. Let \(u=\cos^{-1}x\) and \(dv=dx\) which implies \(du=-\frac{1}{\sqrt{1-x^2}}dx\) and \(v=x\). Now, we can apply integration by parts: \(\displaystyle{\int_0^1 (\cos^{-1}x) dx = x\cos^{-1}x\Big|_0^1 - \int_0^1 x(-\frac{1}{\sqrt{1-x^2}}dx)}\). Now simplify the expression: \(\displaystyle{x\cos^{-1}x\Big|_0^1 - (-\int_0^1 \frac{x}{\sqrt{1-x^2}}dx)}\). Calculate the definite integral: \(\displaystyle{[x\cos^{-1}x-\int \frac{x}{\sqrt{1-x^2}}dx]\Bigg|_0^1}=1\cos^{-1}1-0\cos^{-1}0-\int_0^1 \frac{x}{\sqrt{1-x^2}}dx}\) Area of \(R_1\) = \(\int_0^1 \frac{x}{\sqrt{1-x^2}}dx\)

Now that we have found the area of \(R_1\), we can use geometry to find the area of \(R_2\). As per the given figure, the area of \(R_2\) is the difference between the rectangle formed by \(x=0, x=1, y=0\) and \(y=\frac{\pi}{2}\) and the area of \(R_1\). Area of the rectangle = \((1-0)\times(\frac{\pi}{2}-0)=\frac{\pi}{2}\). Therefore, the area of \(R_2\) = Area of the rectangle - Area of \(R_1\) = \(\frac{\pi}{2} - \int_0^1 \frac{x}{\sqrt{1-x^2}}dx\).