Question

Find the area bounded by the parabolas y = 6x − x² and y = x² − 2x.

Answer 1

We have, $y=6x-x^2$ and $y=x^2-2x$

$$\begin{gathered} \mathrm{Solving}\mathrm{the}\mathrm{two}\mathrm{equations},\mathrm{we}\mathrm{get}, \\ \therefore x^2-2x=6x-x^2 \\ \Rightarrow 2x^2-8x=0 \\ \Rightarrow x=0\mathrm{or}x=4 \\ \Rightarrow y=0\mathrm{or}y=8 \\ \mathrm{Consider}\mathrm{a}\mathrm{vertical}\mathrm{strip}\mathrm{of}\mathrm{length}=\left|y_2-y_1\right|\mathrm{and}\mathrm{width}=dx\mathrm{in}\mathrm{the}\mathrm{shaded}\mathrm{region} \\ \mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{shaded}\mathrm{region}={\int }_0^4\left|y_2-y_1\right|dx \\ ={\int }_0^4\left(y_2-y_1\right)dx\left\{\because \left|y_2-y_1\right|\Rightarrow y_2-y_1\mathrm{as}{\mathrm{y}}_2>y_1\right\} \\ ={\int }_0^4\left[\left(6x-x^2\right)-\left(x^2-2x\right)\right]dx \\ ={\int }_0^4\left(8x-2x^2\right)dx \\ ={\left[\frac{8x^2}{2}-\frac{2x^3}{3}\right]}_0^4 \\ ={\left[4x^2-\frac{2x^3}{3}\right]}_0^4 \\ =\left(4\times 4\times 4\right)-\frac{2}{3}\times \left(4\times 4\times 4\right) \\ =\frac{1}{3}\times \left(4\times 4\times 4\right) \\ =\frac{64}{3}\mathrm{sq}\mathrm{units} \\ \mathrm{Area}\mathrm{enclosed}\mathrm{by}\mathrm{the}\mathrm{two}\mathrm{parabolas}=\frac{64}{3}\mathrm{sq}\mathrm{units} \end{gathered}$$