Find the area of the region bounded by the curves $y = x$, $y = 1$ and $y = \dfrac{x^2}{4}$, lying in the first quadrant.
The curves $y = \dfrac{x^2}{4}$ and $y = x$ intersect at the points $O \left(0,0\right)$ and $B \left(4,4\right)$.
The curves $y = \dfrac{x^2}{4}$ and $y=1$ intersect at the point $C \left(2,1\right)$ in the first quadrant.
The lines $y = x$ and $y = 1$ intersect at the point $A \left(1,1\right)$.
The regions bounded by the given curves in the first quadrant are OAC (shaded blue) and ABC (shaded red).
Let the area of region OAC be $A_1$.
Area of region OCD $= A_2 = \left|I_1\right|$ where
$\begin{aligned}
I_1 & = \int \limits_{0}^{2} y \; dx \hspace{2em} \left[\text{where } y = \dfrac{x^2}{4}\right] \\\\
& = \int \limits_{0}^{2} \dfrac{x^2}{4} \; dx \\\\
& = \dfrac{1}{4} \times \left[\dfrac{x^3}{3}\right]_{0}^{2} = \dfrac{1}{4} \times \dfrac{8}{3} = \dfrac{2}{3}
\end{aligned}$
$\therefore$ $\;$ $A_2 = \dfrac{2}{3}$ $\;$ sq units $\;\;\; \cdots \; (1)$
$\begin{aligned}
\text{Area of region OACD (trapezium)} = A_3 & = \int \limits_{0}^{1} x \; dx + \int \limits_{1}^{2} 1 \; dx \\\\
& = \left[\dfrac{x^2}{2}\right]_{0}^{1} + \left[x\right]_{1}^{2} \\\\
& = \dfrac{1}{2} + 1 = \dfrac{3}{2} \; \text{sq units} \;\;\; \cdots \; (2)
\end{aligned}$
$\begin{aligned}
\text{Now, } A_1 & = A_3 - A_2 \\\\
& = \dfrac{3}{2} - \dfrac{2}{3} \hspace{2em} \left[\text{from equations }(1) \text{ and }(2)\right] \\\\
& = \dfrac{5}{6} \; \text{sq units} \;\;\; \cdots \; (3)
\end{aligned}$
Let the area of region ABC be $A_4$.
Let area of region OBF (triangle) $= A_5 = \left|I_2\right|$ $\;$ where
$\begin{aligned}
I_2 & = \int \limits_{0}^{4} y \; dx \hspace{2em} \left[\text{where } y = x\right] \\\\
& = \int \limits_{0}^{4} x \; dx = \left[\dfrac{x^2}{2}\right]_{0}^{4} = \dfrac{16}{2} = 8
\end{aligned}$
$\therefore$ $\;$ $A_5 = 8$ $\;$ sq units $\;\;\; \cdots \; (4)$
Let area of region bounded by $y = \dfrac{x^2}{4}$ and the X axis be $= A_6 = \left|I_3\right|$ $\;$ where
$\begin{aligned}
I_3 & = \int \limits_{0}^{4} y \; dx \\\\
& = \int \limits_{0}^{4} \dfrac{x^2}{4} \; dx = \dfrac{1}{4} \times \left[\dfrac{x^3}{3}\right]_{0}^{4} = \dfrac{1}{4} \times \dfrac{64}{3} = \dfrac{16}{3}
\end{aligned}$
$\therefore$ $\;$ $A_6 = \dfrac{16}{3}$ $\;$ sq units $\;\;\; \cdots \; (5)$
$\begin{aligned}
\text{Now, }A_4 & = A_5 - A_6 - A_1 \\\\
& = 8 - \dfrac{16}{3} - \dfrac{5}{6} \hspace{2em} \left[\text{from equations } (3), (4) \text{ and } (5)\right] \\\\
& = \dfrac{11}{6} \; \text{sq units}
\end{aligned}$