Question

Find the area of the region bounded by the curves $y=x^2+2,y=x,x=0$ and x = 3.

Answer 1

Given curve $y=x^2+2$ represents a parabola and it is symmetrical about Y-axis having vertex (0, 2).

The given region bounded by $y=x^2+2,y=x,x=0$ and x = 3, is represented by the shaded area.

The point of intersection of the curve $y=x^2+2$ and the line x = 3 is (3, 11). Required area (shown in shaded region)

= Area OABDO - Area OCDO = [Area under $y=x^2+2$ between x = 0, x = 3] - [Area under y = x between x = 0, x = 3]

$={\int }_0^3(x^2+2)dx-{\int }_0^{3}xdx=[\frac{x^3}{3}+2x{]}_0^3-[\frac{x^2}{2}{]}_0^3=[\frac{3^3}{3}+6-0]-[\frac{3^2}{2}-0]$

$=9+6-\frac{9}{2}=\frac{21}{2}squnit$

Area of triangle ABC = Area AMB + Area BMNC - Area ANC

${\int }_{-1}^1\frac{3}{2}(x+1)dx+{\int }_1^3(-\frac{1}{2}x+\frac{7}{2})dx-{\int }_{-1}^3(\frac{1}{2}x+\frac{1}{2})dx=\frac{3}{2}[\frac{x^2}{2}+x{]}_{-1}^1+[-\frac{x^2}{4}+\frac{7}{2}{]}_1^3-[\frac{x^2}{4}+\frac{1}{2}x]=\frac{3}{2}[\frac{1}{2}+1-\frac{1}{2}-(-1\left)\right]+[-\frac{9}{4}+\frac{21}{2}+\frac{1}{4}-\frac{7}{2}]-[\frac{9}{4}+\frac{3}{2}-\frac{1}{4}+\frac{1}{2}]=\frac{3}{2}\left[2\right]+[7-2]-[2+2]=3+5-4=4squnit$