Question

Find the area of the region bounded by the curves y = x − 1 and (y − 1)² = 4 (x + 1).

Answer 1

We have, y = x − 1 and (y − 1)² = 4 (x + 1)

$$\begin{gathered} \therefore {\left(x-1-1\right)}^2=4\left(x+1\right) \\ \Rightarrow {\left(x-2\right)}^2=4\left(x+1\right) \\ \Rightarrow x^2+4-4x=4x+4 \\ \Rightarrow x^2+4-4x-4x-4=0 \\ \Rightarrow x^2-8x=0 \\ \Rightarrow x=0\mathrm{or}x=8 \\ \therefore y=-1\mathrm{or}7 \\ \mathrm{Consider}\mathrm{a}\mathrm{horizantal}\mathrm{strip}\mathrm{of}\mathrm{length}\left|x_2-x_1\right|\mathrm{and}\mathrm{width}dy\mathrm{where}P\left(x_2\mathit{,}y\right)\mathrm{lies}\mathrm{on}\mathrm{straight}\mathrm{line}\mathrm{and}Q\left(x_1\mathit{,}y\right)\mathrm{lies}\mathrm{on}\mathrm{the}\mathrm{parabola}. \\ \mathrm{Area}\mathrm{of}\mathrm{approximating}\mathrm{rectangle}=\left|x_2-x_1\right|dy,\mathrm{and}\mathrm{it}\mathrm{moves}\mathrm{from}y=-1\mathrm{to}y=7 \\ \mathrm{Required}\mathrm{area}=\mathrm{area}\left(OADO\right)={\int }_{-1}^7\left|x_2-x_1\right|dy \\ ={\int }_{-1}^7\left|x_2-x_1\right|dy\left\{\because \left|x_2-x_1\right|=x_2-x_1\mathrm{as}{x}_2>x_1\right\} \\ ={\int }_{-1}^7\left[\left(1+y\right)-\frac{1}{4}\left\{{\left(y-1\right)}^2-4\right\}\right]dy \\ ={\int }_{-1}^7\left\{1+y-\frac{1}{4}{\left(y-1\right)}^2+1\right\}dy \\ ={\int }_{-1}^7\left\{2+y-\frac{1}{4}{\left(y-1\right)}^2\right\}dy \\ ={\left[2y+\frac{y^2}{2}-\frac{1}{12}{\left(y-1\right)}^3\right]}_{-1}^7 \\ =\left[14+\frac{49}{2}-\frac{1}{12}\times 6\times 6\times 6\right]-\left[-2+\frac{1}{2}+\frac{1}{12}\times 2\times 2\times 2\right] \\ =\left[14+\frac{49}{2}-18\right]-\left[-2+\frac{1}{2}+\frac{2}{3}\right] \\ =\left[\frac{41}{2}\right]+\left[\frac{5}{6}\right] \\ =\frac{64}{3}\mathrm{sq}\mathrm{units} \\ \mathrm{Area}\mathrm{enclosed}\mathrm{by}\mathrm{the}\mathrm{line}\mathrm{and}\mathrm{given}\mathrm{parabola}=\frac{64}{3}\mathrm{sq}\mathrm{units} \end{gathered}$$