Question

Make a sketch of the region {(x, y) : 0 ≤ y ≤ x² + 3; 0 ≤ y ≤ 2x + 3; 0 ≤ x ≤ 3} and find its area using integration.

Answer 1

$$\begin{gathered} R=\left\{\left(x,y\right):0\le y\le x^2+3,0\le y\le 2x+3,0\le x\le 3\right\} \\ {R}_1=\left\{\left(x,y\right):0\le y\le x^2+3\right\} \\ {R}_2=\left\{\left(x,y\right):,0\le y\le 2x+3\right\} \\ {R}_3=\left\{\left(x,y\right):0\le x\le 3\right\} \\ \Rightarrow \mathrm{R}={\mathrm{R}}_1\cap {\mathrm{R}}_2\cap {\mathrm{R}}_3 \end{gathered}$$

y = x² + 3 is a upward opening parabola with vertex A(0, 3).
Thus R₁ is the region above x-axis and below the parabola
y = 2x + 3 is a straight line passing through A(0, 3) and cuts y-axis on (−3/2, 0).
Hence R₂ is the region above x-axis and below the line
x = 3 is a straight line parallel to y-axis, cutting x-axis at E(3, 0).
Hence R₃ is the region above x-axis and to the left of the line x = 3.
Point of intersection of the parabola and y = 2x + 3 is given by solving the two equations
$$\begin{gathered} y=x^2+3 \\ y=2x+3 \\ \Rightarrow x^2+3=2x+3 \\ \Rightarrow x^2-2x=0 \\ \Rightarrow x\left(x-2\right)=0 \\ \Rightarrow x=0\mathrm{or}x=2 \\ \Rightarrow y=3\mathrm{or}y=7 \\ \Rightarrow A\left(0,3\right)\mathrm{and}B\left(2,7\right)a\mathrm{re}\mathrm{points}\mathrm{of}\mathrm{intersection} \\ \mathrm{Also},x=3\mathrm{cuts}\mathrm{the}\mathrm{parabola}\mathrm{at}C\left(3,12\right) \\ \mathrm{and}x=3\mathrm{cuts}y=2x+3\mathrm{at}D\left(3,9\right) \\ \mathrm{We}\mathrm{require}\mathrm{thearea}\mathrm{of}\mathrm{shaded}\mathrm{region}. \\ \mathrm{Total}\mathrm{shaded}\mathrm{area}={\int }_0^2\left(x^2+3\right)dx+{\int }_2^3\left(2x+3\right)dx \\ ={\left[\frac{x^3}{3}+3x\right]}_0^2+{\left[\frac{2x^2}{2}+3x\right]}_2^3 \\ ={\left[\frac{x^3}{3}+3x\right]}_0^2+{\left[x^2+3x\right]}_2^3 \\ =\left[\frac{8}{3}+6\right]+\left[9+9-4-6\right] \\ =\frac{8}{3}+6+8 \\ =\frac{8+42}{3} \\ =\frac{50}{3}\mathrm{sq}.\mathrm{units} \end{gathered}$$