Find the area of the region enclosed by the parabola ${x^2} = y$ and the line $y = x + 2$.

${x^2} = y$ is an upward parabola line $y = x + 2$

$$\left. \begin{array}{l}x\\y\end{array} \right|\left. \begin{array}{l}0\\2\end{array} \right|\left. \begin{array}{l}1\\3\end{array} \right|\left. \begin{array}{l} - 1\\1\end{array} \right|\left. \begin{array}{l} - 2\\0\end{array} \right|$$

They meet at pts (2,4) and (-1,1)
By solving the equation
${x^2} = y$, $y = x + 2$

$\Rightarrow$ ${x^2} - x - 2 = 0$

$\Rightarrow$ $x = 20$or $- 1$

$y = 4$ or 1
Required area
$= \int_{ - 1}^2$ Area of line $- \int_{ - 1}^2$ Area of parabola

$= \int_{ - 1}^2 {(x + 2)} dx - \int_{ - 1}^2 {{x^2}} dx$
$$\left. {\left. { = \frac{{{x^2}}}{2}} \right]_{ - 1}^2 + \left. {2x} \right]_{ - 1}^2 - \frac{{{x^3}}}{3}} \right]_{ - 1}^2$$

$= \frac{4}{2} - \frac{{{{( - 1)}^2}}}{2} + 2 \times 2 - (2 \times - 1) - \left( {\frac{{{2^2}}}{3} - \frac{{{{( - 1)}^3}}}{3}} \right)$
$= 2 - \frac{1}{2} + 4 + 2 - \left( {\frac{8}{3} + \frac{1}{3}} \right)$

$= 8 - \frac{1}{2} - 3 = 5 - \frac{1}{2} = \frac{9}{2}$ sq units.