Question

Find the area of the region included between
the parabola $y=x^2+3$ and the line $y=x+3$

Answer 1

Solving $y=x^2+3$

$y=x+3$

$\therefore x^2+3=x+3$

$x^2-x=0$, $x(x-1)=0$

$x=0,1$

$\therefore y=0,4$

Area included $={\int }_0^{1}ydx-{\int }_0^{1}ydx$

$={\int }_0^1(x-3)dx-{\int }_0^1(x^2+3)dx$

$={(\frac{x^2}{2}+3x)}_0^1-{(\frac{x^3}{3}+3x)}_0^1$

$=(\frac{1}{2}+3)-(\frac{1}{3}+3)$

$=\frac{1}{2}-\frac{1}{3}$

$=\frac{1}{6}$ sq, units.