Question

Find the area enclosed by the parabolas y = 4x − x² and y = x² − x.

Answer 1

We have, $y=4x-x^2$ and $y=x^2-x$
The points of intersection of two curves is obtained by solving the simultaneous equations

$$\begin{gathered} \therefore x^2-x=4x-x^2 \\ \Rightarrow 2x^2-5x=0 \\ \Rightarrow x=0\mathrm{or}x=\frac{5}{2} \\ \Rightarrow y=0\mathrm{or}y=\frac{15}{4} \\ \Rightarrow O\left(0,0\right)\mathrm{and}D\left(\frac{5}{2},\frac{15}{4}\right)\mathrm{are}\mathrm{points}\mathrm{of}\mathrm{intersection}\mathrm{of}\mathrm{two}\mathrm{parabolas}. \\ \mathrm{In}\mathrm{the}\mathrm{shaded}\mathrm{area}\mathit{}CBDC,\mathrm{consider}P(x,y_{\mathit{2}})\mathrm{on}y=4x-x^2\mathrm{and}Q(x,y_1)\mathrm{on}y=x^2-x \\ \mathrm{Area}\left(OBDCO\right)\hspace{0.17em}=\mathrm{area}\left(OBCO\right)+\mathrm{area}\left(CBDC\right) \\ ={\int }_0^1\left|y\right|dx+{\int }_1^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left|y_2-y_1\right|dx \\ ={\int }_0^{1}ydx+{\int }_1^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left({\mathrm{y}}_2-{\mathrm{y}}_1\right)dx\left\{\because y>0\Rightarrow \left|y\right|=y\mathrm{and}\left|y_2-y_1\right|\Rightarrow y_2-y_1\mathrm{as}{y}_2>y_1\right\} \\ ={\int }_0^1\left(4x-x^2\right)dx+{\int }_1^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left\{\left(4x-x^2\right)-\left(x^2-x\right)\right\}dx \\ ={\left[\frac{4x^2}{2}-\frac{x^3}{3}\right]}_0^1+{\int }_1^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left(5x-2x^2\right)dx \\ ={\left[2x^2-\frac{x^3}{3}\right]}_0^1+{\left[\frac{5x^2}{2}-\frac{2x^3}{3}\right]}_1^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \\ =\left(2-\frac{1}{3}\right)+\left[\frac{5}{2}{\left(\frac{5}{2}\right)}^2-\frac{2}{3}{\left(\frac{5}{2}\right)}^3-\frac{5}{2}+\frac{2}{3}\right] \\ =\left(\frac{5}{3}\right)+\left[{\left(\frac{5}{2}\right)}^3\left(1-\frac{2}{3}\right)-\frac{11}{6}\right] \\ =\frac{5}{3}+{\left(\frac{5}{2}\right)}^3\frac{1}{3}-\frac{11}{6} \\ =\frac{10-11}{6}+\frac{125}{24} \\ =\frac{121}{24}\mathrm{sq}\mathrm{units}...\left(1\right) \\ \mathrm{Area}\left(OCV\mathit{\text{'}}O\right)={\int }_0^1\left|y\right|dx={\int }_0^1-ydx\left\{\because y<0\Rightarrow \left|y\right|=-y\right\} \\ ={\int }_0^1-\left(x^2-x\right)dx \\ ={\int }_0^1\left(x-x^2\right)dx \\ ={\left[\frac{x^2}{2}-\frac{x^3}{3}\right]}_0^1 \\ =\frac{1}{2}-\frac{1}{3}=\frac{1}{6}\mathrm{sq}\mathrm{units}...\left(2\right) \\ \mathrm{From}\left(1\right)\mathrm{and}\left(2\right) \\ \mathrm{Shaded}\mathrm{area}=\mathrm{area}\left(OBDCO\right)\mathrm{and}\mathrm{area}\left(OCV\mathit{\text{'}}O\right) \\ =\frac{121}{24}+\frac{1}{6} \\ =\frac{125}{24}\mathrm{sq}\mathrm{units} \end{gathered}$$