Question

Find the area of the region bounded by y = $\sqrt{x}$ and y = x.

Answer 1

$$\begin{gathered} y=\sqrt{x}...\left(1\right)\mathrm{is}\mathrm{a}\mathrm{parabola}\mathrm{opening}\mathrm{side}\mathrm{ways},\mathrm{with}\mathrm{vertex}\mathrm{at}\mathrm{O}(0,0)\mathrm{and}+\mathrm{ve}x-\mathrm{axis}\mathrm{as}\mathrm{axis}\mathrm{of}\mathrm{symmetry} \\ x=y...\left(2\right)\mathrm{is}\mathrm{a}\mathrm{straight}\mathrm{line}\mathrm{passsing}\mathrm{through}\mathrm{O}(0,0)\mathrm{and}\mathrm{at}\mathrm{angle}{45}^{\mathrm{o}}\mathrm{with}\mathrm{the}x-\mathrm{axis} \\ \mathrm{Solving}\left(1\right)\mathrm{and}\left(2\right) \\ {y}^2=x=y \\ \Rightarrow y^2=y \\ \Rightarrow y(y-1)=0 \\ \Rightarrow y=0\mathrm{or}y=1\mathrm{and}x=0\mathrm{or}x=1 \\ \mathrm{Thus},\mathrm{the}\mathrm{line}\mathrm{intersects}\mathrm{the}\mathrm{parabola}\mathrm{at}\mathrm{O}(0,0)\mathrm{and}\mathrm{A}(1,1) \\ \mathrm{Consider}\mathrm{a}\mathrm{approximating}\mathrm{rectangle}\mathrm{of}\mathrm{length}=\left|y_2-y_1\right|\mathrm{and}\mathrm{width}=dx \\ \mathrm{Area}\mathrm{of}\mathrm{approximating}\mathrm{rectangle}=\left|y_2-y_1\right|dx \\ \mathrm{Approximating}\mathrm{rectangle}\mathrm{moves}\mathrm{from}x=0\mathrm{to}x=1 \\ \therefore \mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{shaded}\mathrm{region}={\int }_0^1\left|y_2-y_1\right|dx={\int }_0^1\left(y_2-y_1\right)dx\left[\mathrm{As},y_2>y_1,\left|y_2-y_1\right|=y_2-y_1\right] \\ \Rightarrow A={\int }_0^1\left(\sqrt{x}-x\right)dx \\ \Rightarrow A={\left[\frac{x^{{\displaystyle \frac{3}{2}}}}{{\displaystyle \frac{3}{2}}}-\frac{x^2}{2}\right]}_0^1 \\ \Rightarrow A=\left[\frac{1^{{\displaystyle \frac{3}{2}}}}{{\displaystyle \frac{3}{2}}}-\frac{1^2}{2}-0\right] \\ \Rightarrow A=\frac{2}{3}-\frac{1}{2}=\frac{1}{6}\mathrm{sq}.\mathrm{units} \\ \therefore \mathrm{Area}\mathrm{bound}\mathrm{by}\mathrm{the}\mathrm{parabola}\mathrm{and}\mathrm{straight}\mathrm{line}=\frac{1}{6}\mathrm{sq}.\mathrm{units} \end{gathered}$$