Find the area enclosed between the curves is  $y=x^2+1,y=2x-2$ and the ordinates $x=–1$ and $x=2.$

 $y=x^2+1$ is a top open parabola with vertex $(0, 1)$

$y=2x-2\Rightarrow \frac{x}{1}+\frac{y}{-2}=1$

cuts$x-axis$ at $(1, 0)$and $(0, –2)$

Area  $A={\int }_{-1}^2 \left(x^2+1\right)-(2x-2)dx$

$A={\int }_{-1}^2 \left(x^2-2x+3\right)dx$

$A={\left[\frac{x^3}{3}-2\frac{x^2}{2}+3x\right]}_{-1}^2$

$A=\left[\frac{8}{3}-4+6\right]-\left[\frac{-1}{3}-1-3\right]=\left[\frac{8}{3}+2+\frac{1}{3}+4\right]$

$A=\frac{9}{3}+6=3+6\Rightarrow A=9\text{ sq.units }$

$\therefore$  The area enclosed between the curves is $y^2=x^2+1\text{ and }y=2x-2\text{ is }9\text{ sq. units }$