Mathematics - Area Under Curves Question with Solution | TestHub

Here, area between 0 to $b$ is $R_1$ and $b$ to $$\begin{array}{rl}& 1\text{ is }{R}_2\\ & \therefore {\int }_0^b(1-x{)}^{2}dx-{\int }_b^1(1-x{)}^{2}dx=\frac{1}{4}\\ & \Rightarrow {\left(\frac{(1-x{)}^3}{-3}\right)}_0^b-{\left(\frac{(1-x{)}^3}{-3}\right)}_b^1=\frac{1}{4}\\ & \Rightarrow -\frac{1}{3}\left\{(1-b{)}^3-1\right\}+\frac{1}{3}\left\{0-(1-b{)}^3\right\}\\ & =\frac{1}{4}\end{array}$$ $$\begin{array}{rl}& \Rightarrow -\frac{2}{3}(1-b{)}^3=-\frac{1}{3}+\frac{1}{4}=-\frac{1}{12}\\ & \Rightarrow (1-b{)}^3=\frac{1}{8}\\ & \Rightarrow (1-b)=\frac{1}{2}\Rightarrow b=\frac{1}{2}\\ & \end{array}$$