Find the area of the region enclosed by the parabola $x^{2}=y$, the line $y=x+2$ and $x$-axis
The area of the region enclosed by the parabola, $x^{2}=y$, the line, $y=x+2$, and $x$-axis is represented by the shaded region $\mathrm{OACO}$ as
The point of intersection of the parabola, x2 = y, and the line, y = x + 2, is A (–1, 1) and C(2, 4).
Area of OACO $=\int_{-1}^{2}(x+2) d x-\int_{-1}^{2} x^{2} d x$
$\Rightarrow$ Area of $\mathrm{OACO}=\left[\frac{x^{2}}{2}+2 x\right]_{-1}^{2}-\frac{1}{3}\left[x^{3}\right]_{-1}^{2}$
$\Rightarrow$ Area of $\mathrm{OACO}=\left[\left\{\frac{(2)^{2}}{2}+2(2)\right\}-\left\{\frac{(-1)^{2}}{2}+2(-1)\right\}\right]-\frac{1}{3}\left[(2)^{3}-(-1)^{3}\right]$
$\Rightarrow$ Area of $\mathrm{OACO}=\left[2+4-\left(\frac{1}{2}-2\right)\right]-\frac{1}{3}(8+1)$
$\Rightarrow$ Area of $\mathrm{OACO}=6+\frac{3}{2}-3$
$\Rightarrow$ Area of $\mathrm{OACO}=3+\frac{3}{2}=\frac{9}{2}$ square units