Question

Sketch the region bounded by the curves y = x² + 2, y = x, x = 0 and x = 1. Also, find the area of this region.

Answer 1

We have, $y=x^2+2$ and $y=x$
We see that parabola and the line $y=x$ do not intersect
$x=1$ is a line parallel to y axis

Point of intersection between parabola and $x=1$ is
$$\begin{gathered} \mathrm{Putting}x=1\mathrm{in}y=x^2+2,\mathrm{we}\mathrm{get}, \\ y=1+2=3 \\ \mathrm{Point}\mathrm{of}\mathrm{intersection}\mathrm{of}\mathrm{two}\mathrm{lines}\mathrm{is}\mathrm{given}\mathrm{by} \\ \mathrm{Putting}x=1\mathrm{in}y=x,\mathrm{we}\mathrm{get}, \\ y=1 \\ \mathrm{Consider}\mathrm{a}\mathrm{vetical}\mathrm{strip}\mathrm{of}\mathrm{length}\left|y_2-y_1\right|a\mathrm{nd}\mathrm{width}=dx\mathrm{such}\mathrm{that}P\left(x,y_2\right)\mathrm{lies}\mathrm{on}\mathrm{parabola}\mathrm{and}Q\left(x,y_1\right)\mathrm{lies}\mathrm{on}y=x \\ \mathrm{Shaded}\mathrm{area}={\int }_0^1\left|y_2-y_1\right|dx \\ ={\int }_0^1\left(y_2-y_1\right)dx\left\{\because \left|y_2-y_1\right|\Rightarrow y_2-y_1\mathrm{as}{y}_2>y_1,\right\} \\ ={\int }_0^1\left\{\left(x^2+2\right)-\left(x\right)\right\}dx \\ ={\int }_0^1\left(x^2+2-x\right)dx \\ ={\left[\frac{x^3}{3}-\frac{x^2}{2}+2x\right]}_0^1 \\ =\frac{1}{3}-\frac{1}{2}+2 \\ =\frac{2-3+12}{6} \\ =\frac{11}{6}\mathrm{sq}\mathrm{units} \\ \mathrm{Thus},\mathrm{area}\mathrm{enclosed}\mathrm{by}\mathrm{parabola}\mathrm{and}\mathrm{given}\mathrm{two}\mathrm{lines}=\frac{11}{6}\mathrm{s}\mathrm{q}\mathrm{u}\mathrm{n}\mathrm{i}\mathrm{t}\mathrm{s} \end{gathered}$$