Find the area of the region enclosed by the parabola $y^2 = x$ and the line $x + y = 2$.
The parabola $y^2= x$ and the line $x + y = 2$ intersect at the points $A \left(1,1\right)$ and $B \left(4,-2\right)$
Area enclosed between the parabola and the line is the region OABO.
$\text{Area} \left(OABO\right) = \text{Area} \left(RPB\right) - \left[\text{Area} \left(OAQO\right) + \text{Area} \left(QAR\right) + \text{Area} \left(OPBO\right)\right]$ $\;\;\; \cdots \; (1)$
Let $\text{Area} \left(RPB\right) = \left|I_1\right|$
$\begin{aligned}
I_1 & = \int \limits_{y=-2}^{y=2} x \; dy \hspace{2em} \text{ where } \; x = 2 - y \\\\
& = \int \limits_{-2}^{2} \left(2 - y\right) \; dy \\\\
& = \left[2 y - \dfrac{y^2}{2}\right]_{-2}^{2} \\\\
& = 4 - \dfrac{4}{2} - \left(- 4 - \dfrac{4}{2}\right) = 8
\end{aligned}$
$\therefore$ $\;$ $\text{Area} \left(RPB\right) = 8$ sq units $\;\;\; \cdots \; (2)$
Let $\text{Area} \left(OAQO\right) = \left|I_2\right|$
$\begin{aligned}
I_2 & = \int \limits_{y=0}^{y=1} x \; dy \hspace{2em} \text{ where } \; x = y^2 \\\\
& = \int \limits_{0}^{1} y^2 \; dy \\\\
& = \left[\dfrac{y^3}{3}\right]_{0}^{1} \\\\
& = \dfrac{1}{3}
\end{aligned}$
$\therefore$ $\;$ $\text{Area} \left(OAQO\right) = \dfrac{1}{3}$ sq units $\;\;\; \cdots \; (3)$
Let $\text{Area} \left(QAR\right) = \left|I_3\right|$
$\begin{aligned}
I_3 & = \int \limits_{y=1}^{y=2} x \; dy \hspace{2em} \text{ where } \; x = 2 - y \\\\
& = \int \limits_{1}^{2} \left(2 - y\right) \; dy \\\\
& = \left[2 y - \dfrac{y^2}{2}\right]_{1}^{2} \\\\
& = 4 - \dfrac{4}{2} - \left(2 - \dfrac{1}{2}\right) = \dfrac{1}{2}
\end{aligned}$
$\therefore$ $\;$ $\text{Area} \left(QAR\right) = \dfrac{1}{2}$ sq units $\;\;\; \cdots \; (4)$
Let $\text{Area} \left(OPBO\right) = \left|I_4\right|$
$\begin{aligned}
I_4 & = \int \limits_{y=-2}^{y=0} x \; dy \hspace{2em} \text{ where } \; x = y^2 \\\\
& = \int \limits_{-2}^{0} y^2 \; dy \\\\
& = \left[\dfrac{y^3}{3}\right]_{-2}^{0} \\\\
& = 0 - \left(\dfrac{-8}{3}\right) = \dfrac{8}{3}
\end{aligned}$
$\therefore$ $\;$ $\text{Area} \left(OPBO\right) = \dfrac{8}{3}$ sq units $\;\;\; \cdots \; (5)$
$\therefore$ $\;$ In view of equations $(2)$, $(3)$, $(4)$ and $(5)$, equation $(1)$ becomes
$\text{Area} \left(OABO\right) = 8 - \left(\dfrac{1}{3} + \dfrac{1}{2} + \dfrac{8}{3}\right) = \dfrac{9}{2}$ sq units