Question

Find the area of the region bounded by the parabola y² = 2x + 1 and the line x − y − 1 = 0.

Answer 1

We have, $y^2=2x+1$ and $x-y-1=0$

To find the intersecting points of the curves ,we solve both the equations.

$$\begin{gathered} y^2=2\left(1+y\right)+1 \\ \Rightarrow y^2-2-2y-1=0 \\ \Rightarrow y^2-2y-3=0 \\ \Rightarrow \left(y-3\right)\left(y+1\right)=0 \\ \Rightarrow y=3\mathrm{or}y=-1 \\ \therefore x=4\mathrm{or}0 \\ \mathrm{Consider}\mathrm{a}\mathrm{horizantal}\mathrm{strip}\mathrm{of}\mathrm{length}\left|x_2-x_1\right|\mathrm{and}\mathrm{width}dy\mathrm{where}P\left(x_2\mathit{,}y\right)\mathrm{lies}\mathrm{on}\mathrm{straight}\mathrm{line}\mathrm{and}Q\left(x_1\mathit{,}y\right)\mathrm{lies}\mathrm{on}\mathrm{the}\mathrm{parabola}. \\ \mathrm{Area}\mathrm{of}\mathrm{approximating}\mathrm{rectangle}=\left|x_2-x_1\right|dy,\mathrm{and}\mathrm{it}\mathrm{moves}\mathrm{from}y=-1\mathrm{to}y=3 \\ \mathrm{Required}\mathrm{area}=\mathrm{area}\left(OADO\right)={\int }_{-1}^3\left|x_2-x_1\right|dy \\ ={\int }_{-1}^3\left|x_2-x_1\right|dy\left\{\because \left|x_2-x_1\right|=x_2-x_1\mathrm{as}{x}_2>x_1\right\} \\ ={\int }_{-1}^3\left\{\left(1+y\right)-\frac{1}{2}\left(y^2-1\right)\right\}dy \\ ={\int }_{-1}^3\left\{1+y-\frac{1}{2}{y}^2+\frac{1}{2}\right\}dy \\ ={\int }_{-1}^3\left\{\frac{3}{2}+y-\frac{1}{2}{y}^2\right\}dy \\ ={\left[\frac{3}{2}y+\frac{y^2}{2}-\frac{1}{6}{y}^3\right]}_{-1}^3 \\ =\left[\frac{9}{2}+\frac{9}{2}-\frac{27}{6}\right]-\left[\frac{-3}{2}+\frac{1}{2}+\frac{1}{6}\right] \\ =\left[\frac{9}{2}\right]+\left[\frac{5}{6}\right] \\ =\frac{16}{3}\mathrm{sq}\mathrm{units} \\ \mathrm{Area}\mathrm{enclosed}\mathrm{by}\mathrm{the}\mathrm{line}\mathrm{and}\mathrm{given}\mathrm{parabola}=\frac{16}{3}\mathrm{sq}\mathrm{units} \end{gathered}$$