Question

Draw the rough sketch of y² + 1 = x, x ≤ 2. Find the area enclosed by the curve and the line x = 2.

Answer 1

$$\begin{gathered} y^2+1=x,x\le 2\mathrm{is}\mathrm{a}\mathrm{parabola}\mathrm{with}\mathrm{vertex}(1,0)\mathrm{and}\mathrm{symmetrical}\mathrm{about}+\mathrm{ve}\mathrm{side}\mathrm{of}x-\mathrm{axis} \\ x=2\mathrm{is}\mathrm{a}\mathrm{line}\mathrm{parallel}\mathrm{to}\mathit{}y-\mathrm{axis}\mathrm{and}\mathrm{cutting}x-\mathrm{axis}\mathrm{at}(2,0) \\ \mathrm{Consider},\mathrm{a}\mathrm{vertical}\mathrm{section}\mathrm{of}\mathrm{height}=\left|y\right|\mathrm{and}\mathrm{width}=dx\mathrm{in}\mathrm{the}\mathrm{first}\mathrm{quadrant} \\ \Rightarrow \mathrm{area}\mathrm{of}\mathrm{corresponding}\mathrm{rectangle}=\left|\mathrm{y}\right|dx \\ \mathrm{Since},\mathrm{the}\mathrm{corresponding}\mathrm{rectangle}\mathrm{is}\mathrm{moving}\mathrm{from}x=1\mathrm{to}x=2\mathrm{and}\mathrm{the}\mathrm{curve}\mathrm{being}\mathrm{symmetrical} \\ \therefore A=\mathrm{area}ABCA=2\times \mathrm{area}ABDA \\ \Rightarrow A=2{\int }_1^2\left|y\right|dx \\ \Rightarrow A=2{\int }_1^{2}ydx\left[\left|y\right|=y\mathrm{as}y>0\right] \\ \Rightarrow A=2{\int }_1^2\sqrt{x-1}dx\left[y^2+1=x\Rightarrow y=\sqrt{x-1}\right] \\ \Rightarrow A=2{\left[\frac{{\left(x-1\right)}^{{\displaystyle \frac{3}{2}}}}{{\displaystyle \frac{3}{2}}}\right]}_1^2 \\ \Rightarrow A=\frac{4}{3}\left(1^{\frac{3}{2}}-0\right) \\ \Rightarrow A=\frac{4}{3}\mathrm{sq}.\mathrm{units} \\ \therefore \mathrm{Enclosed}\mathrm{area}=\frac{4}{3}\mathrm{s}\mathrm{q}.\mathrm{u}\mathrm{n}\mathrm{i}\mathrm{t}\mathrm{s} \end{gathered}$$