Applications of Definite Integration

Using integration, find the area of the region $\left\{\left(x,y\right) \bigg| \left|x-1\right| \leq y \leq \sqrt{5 - x^2} \right\}$

When $x > 1$, $\left|x - 1\right| = x - 1$

When $x < 1$, $\left|x - 1\right| = - \left(x - 1\right) =1 - x$

$\therefore$ $\;$ The given region is

$\left\{\left(x,y\right) \bigg| \left(1 - x\right) \leq y \leq \sqrt{5 - x^2}, \; \left(x - 1\right) \leq y \leq \sqrt{5 - x^2} \right\}$

The equations represented by the inequation $\left(x - 1\right) \leq y \leq \sqrt{5 - x^2}$ $\;$ are

$x - 1 = y$ $\;\;\; \cdots \; (1a)$

$y = \sqrt{5 - x^2}$ $\;\;\; \cdots \; (1b)$

The equation represented by the inequation $\left(1 - x\right) \leq y$ is

$1 - x = y$ $\;\;\; \cdots \; (1c)$

Required area is the shaded region CABC.

$\text{Area} \left(CABC\right) = \text{Area} \left(PABQP\right) - \left[\text{Area} \left(PAC\right) + \text{Area} \left(CBQ\right)\right]$ $\;\;\; \cdots \; (2)$

Let $\text{Area} \left(PABQP\right) = \left|I_1\right|$

$\begin{aligned} I_1 & = \int \limits_{-1}^{2} y \; dx \hspace{3em} \text{where } \; y = \sqrt{5 - x^2} \\\\ & = \int \limits_{-1}^{2} \sqrt{5 - x^2} \; dx \\\\ & = \int \limits_{-1}^{2} \sqrt{\left(\sqrt{5}\right)^2 - \left(x\right)^2} \; dx \\\\ & \left[\text{Note: } \int \limits_{p}^{q} \sqrt{a^2 - x^2} \; dx = \left[\dfrac{x}{2} \sqrt{a^2 - x^2} + \dfrac{a^2}{2} \sin^{-1} \left(\dfrac{x}{a}\right)\right]_{p}^{q}\right] \\\\ & = \left[\dfrac{x}{2} \times \sqrt{5 - x^2} + \dfrac{5}{2} \sin^{-1} \left(\dfrac{x}{\sqrt{5}}\right)\right]_{-1}^{2} \\\\ & = \dfrac{2}{2} \times \sqrt{5 - 4} + \dfrac{5}{2} \sin^{-1} \left(\dfrac{2}{\sqrt{5}}\right) - \left[\dfrac{-1}{2} \times \sqrt{5 - 1} + \dfrac{5}{2} \sin^{-1} \left(\dfrac{-1}{\sqrt{5}}\right)\right] \\\\ & = 1 + \dfrac{5}{2} \sin^{-1} \left(\dfrac{2}{\sqrt{5}}\right) + \dfrac{1}{2} \times 2 + \dfrac{5}{2} \sin^{-1} \left(\dfrac{1}{\sqrt{5}}\right) \\\\ & = 2 + \dfrac{5}{2} \left[\sin^{-1} \left(\dfrac{2}{\sqrt{5}}\right) + \sin^{-1} \left(\dfrac{1}{\sqrt{5}}\right)\right] \\\\ & \left[\text{Note: } \sin^{-1}\left(x\right) + \sin^{-1} \left(y\right) = \sin^{-1} \left(x \sqrt{1 - y^2} + y \sqrt{1 - x^2}\right)\right] \\\\ & = 2 + \dfrac{5}{2} \sin^{-1} \left(\dfrac{2}{\sqrt{5}} \times \sqrt{1 - \dfrac{1}{5}} + \dfrac{1}{\sqrt{5}} \times \sqrt{1 - \dfrac{4}{5}}\right) \\\\ & = 2 + \dfrac{5}{2} \sin^{-1} \left(\dfrac{2}{\sqrt{5}} \times \dfrac{2}{\sqrt{5}} + \dfrac{1}{\sqrt{5}} \times \dfrac{1}{\sqrt{5}}\right) \\\\ & = 2 + \dfrac{5}{2} \sin^{-1} \left(\dfrac{4}{5} + \dfrac{1}{5}\right) \\\\ & = 2 + \dfrac{5}{2} \sin^{-1} \left(1\right) \\\\ & = 2 + \dfrac{5}{2} \times \dfrac{\pi}{2} = 2 + \dfrac{5 \pi}{4} \end{aligned}$

$\therefore$ $\;$ $\text{Area} \left(PABQP\right) = 2 + \dfrac{5 \pi}{4}$ sq units $\;\;\; \cdots \; (3)$

Let $\text{Area} \left(PAC\right) = \left|I_2\right|$

$\begin{aligned} I_2 & = \int \limits_{-1}^{1} y \; dx \hspace{2em} \text{ where } \; y = 1-x \\\\ & = \int \limits_{-1}^{1} \left(1 - x\right) \; dx \\\\ & = \left[x - \dfrac{x^2}{2}\right]_{-1}^{1} \\\\ & = 1 - \dfrac{1}{2} - \left[-1 - \dfrac{1}{2}\right] = 2 \end{aligned}$

$\therefore$ $\;$ $\text{Area} \left(PAC\right) = 2$ sq units $\;\;\; \cdots \; (4)$

Let $\text{Area} \left(CBQ\right) = \left|I_3\right|$

$\begin{aligned} I_3 & = \int \limits_{1}^{2} y \; dx \hspace{2em} \text{ where } \; y = x-1 \\\\ & = \int \limits_{1}^{2} \left(x - 1\right) \; dx \\\\ & = \left[\dfrac{x^2}{2} - x\right]_{1}^{2} \\\\ & = \dfrac{4}{2} - 2 - \dfrac{1}{2} + 1 = \dfrac{1}{2} \end{aligned}$

$\therefore$ $\;$ $\text{Area} \left(CBQ\right) = \dfrac{1}{2}$ sq units $\;\;\; \cdots \; (5)$

$\therefore$ $\;$ In view of equations $(3)$, $(4)$ and $(5)$, equation $(2)$ becomes

$\text{Area} \left(CABC\right) = 2 + \dfrac{5 \pi}{4} - 2 - \dfrac{1}{2} = \dfrac{5 \pi}{4} - \dfrac{1}{2}$ sq units

Applications of Definite Integration

Find the area of the region $\left\{\left(x,y\right) \bigg| \; 0 \leq y \le x^2 + 1, \; 0 \leq y \le x + 1, \; 0 \leq x \leq 2 \right\}$

The equation representing the inequation $0 \leq y \leq x^2 + 1$ is the parabola $y = x^2 + 1$

The equation representing the inequation $0 \leq y \leq x + 1$ is the line $y = x + 1$

The equation representing the inequation $0 \leq x \leq 2$ is the line $x = 2$

The required area is the shaded region OABCDO.

$\text{Area} \left(OABCDO\right) = \text{Area} \left(OABEO\right) + \text{Area} \left(EBCDE\right)$ $\;\;\; \cdots \; (1)$

Let $\text{Area} \left(OABEO\right) = \left|I_1\right|$

$\begin{aligned} I_1 & = \displaystyle \int \limits_{0}^{1} y \; dx \text{ where } y = x^2 + 1 \\\\ & = \int \limits_{0}^{1} \left(x^2 + 1\right) \; dx \\\\ & = \left[\dfrac{x^3}{3} + x\right]_{0}^{1} = \dfrac{1}{3} + 1 = \dfrac{4}{3} \end{aligned}$

$\therefore$ $\;$ $\text{Area} \left(OABEO\right) = \dfrac{4}{3}$ sq units $\;\;\; \cdots \; (2)$

Let $\text{Area} \left(EBCDE\right) = \left|I_2\right|$

$\begin{aligned} I_2 & = \displaystyle \int \limits_{1}^{2} y \; dx \text{ where } y = x + 1 \\\\ & = \int \limits_{1}^{2} \left(x + 1\right) \; dx \\\\ & = \left[\dfrac{x^2}{2} + x\right]_{1}^{2} = \dfrac{4}{2} + 2 - \dfrac{1}{2} - 1 = \dfrac{5}{2} \end{aligned}$

$\therefore$ $\;$ $\text{Area} \left(EBCDE\right) = \dfrac{5}{2}$ sq units $\;\;\; \cdots \; (3)$

$\therefore$ $\;$ In view of equations $(2)$ and $(3)$, equation $(1)$ becomes

$\text{Area} \left(OABCDO\right) = \dfrac{4}{3} + \dfrac{5}{2} = \dfrac{23}{6}$ sq units

Applications of Definite Integration

Prove that the curves $y^2 = 4x$ and $x^2 = 4y$ divide the area of the square bounded by $x = 0$, $x = 4$, $y = 4$ and $y = 0$ into three equal areas.

The two parabolas intersect at the points $O \left(0,0\right)$ and $B \left(4,4\right)$.

The given parabolas and the lines intersect at the point $B \left(4,4\right)$

OABC is the square bounded by $x=0$, $x=4$, $y=4$ and $y=0$.

Let the area bounded by the curve $x^2 = 4y$, the X axis and the line $x = 4$ be $A_1$.

Then $A_1 = \left|I_1\right| = \displaystyle \int \limits_{0}^{4} y \; dx$ where $y = \dfrac{x^2}{4}$

$\begin{aligned} \therefore \; I_1 & = \int \limits_{0}^{4} \dfrac{x^2}{4} \; dx \\\\ & = \dfrac{1}{4} \times \left[\dfrac{x^3}{3}\right]_{0}^{4} = \dfrac{64}{12} = \dfrac{16}{3} \end{aligned}$

$\therefore$ $\;$ $A_1 = \dfrac{16}{3}$ sq units $\;\;\; \cdots \; (1)$

Let the area bounded by the curve $y^2 = 4x$, the X axis and the line $x = 4$ be $A_4$.

Then $A_4= \left|I_4\right| = \displaystyle \int \limits_{0}^{4} y \; dx$ where $y = 2 \sqrt{x}$

$\begin{aligned} \therefore \; I_4 & = \int \limits_{0}^{4} 2 \sqrt{x} \; dx \\\\ & = 2 \times \dfrac{2}{3} \times \left[\left(x\right)^{3/2}\right]_{0}^{4} = \dfrac{4}{3} \times 8 = \dfrac{32}{3} \end{aligned}$

$\therefore$ $\;$ $A_4 = \dfrac{32}{3}$ sq units $\;\;\; \cdots \; (2)$

$\therefore$ $\;$ Area $A_2 = A_4 - A_1 = \dfrac{32}{3} - \dfrac{16}{3} = \dfrac{16}{3}$ sq units $\;\;\; \cdots \; (3)$ $\hspace{2em}$ [From equations $(1)$ and $(2)$]

Let the area bounded by the curve $x^2 = 4y$, the Y axis and the line $y = 4$ be $A_5$.

Then $A_5 = \left|I_5\right| = \displaystyle \int \limits_{y=0}^{y=4} x \; dy$ where $x = 2 \sqrt{y}$

$\begin{aligned} \therefore \; I_5 & = \int \limits_{0}^{4} 2 \sqrt{y} \; dy \\\\ & = 2 \times \dfrac{2}{3} \times \left[\left(y\right)^{3/2}\right]_{0}^{4} = \dfrac{4}{3} \times 8 = \dfrac{32}{3} \end{aligned}$

$\therefore$ $\;$ $A_5 = \dfrac{32}{3}$ sq units $\;\;\; \cdots \; (4)$

$\therefore$ $\;$ Area $A_3 = A_5 - A_2 = \dfrac{32}{3} - \dfrac{16}{3} = \dfrac{16}{3}$ sq units $\;\;\; \cdots \; (5)$ $\hspace{2em}$ [From equations $(3)$ and $(4)$]

$\therefore$ $\;$ We have from equations $(1)$, $(3)$ and $(5)$, $A_1 = A_2 = A_3 = \dfrac{16}{3}$ sq units

$\implies$ The curves $y^2 = 4x$ and $x^2 = 4y$ divide the area of the square bounded by $x = 0$, $x = 4$, $y = 4$ and $y = 0$ into three equal areas.

Applications of Definite Integration

Find the area lying in the first quadrant enclosed by the X axis, the circle $x^2 + y^2 = 8x$ and the parabola $y^2 = 4x$.

The circle and the parabola intersect at the points $A \left(4,4\right)$ and $B \left(4,-4\right)$.

Equation of the circle is $x^2 + y^2 = 8x$

i.e. $y = \sqrt{8x - x^2}$

Let $y = f\left(x_1\right) = \sqrt{8x - x^2}$ $\;\;\; \cdots \; (1)$

Equation of the parabola is $y^2 = 4x$

i.e. $y = 2 \sqrt{x}$

Let $y = f\left(x_2\right) = 2 \sqrt{x}$ $\;\;\; \cdots \; (2)$

Let the area enclosed by the X axis, the circle and the parabola in the first quadrant $= A = \left|I\right|$ where

$I = \displaystyle \int \limits_{0}^{4} f \left(x_2\right) \; dx + \int \limits_{4}^{8} f \left(x_1\right) \; dx$ $\;\;\; \cdots \; (3)$

Since the circle is symmetric about its radius $AD$,

$\displaystyle \int \limits_{4}^{8} f \left(x_1\right) \; dx = \int \limits_{0}^{4} f \left(x_1\right) \; dx$ $\;\;\; \cdots \; (3a)$

$\therefore$ $\;$ In view of equation $(3a)$, equation $(3)$ can be written as

$I = \displaystyle \int \limits_{0}^{4} \left\{f \left(x_1\right) + f \left(x_2\right) \right\} \; dx$ $\;\;\; \cdots \; (3b)$

$\therefore$ $\;$ By equations $(1)$ and $(2)$, equation $(3b)$ becomes

$\begin{aligned} I & = \int \limits_{0}^{4} \left\{\sqrt{8x - x^2} + 2 \sqrt{x} \right\} \; dx \\\\ & = \int \limits_{0}^{4} \sqrt{8x - x^2} \; dx + 2 \int \limits_{0}^{4} \sqrt{x} \; dx \\\\ & = I_1 + I_2 \;\;\; \cdots \; (4) \end{aligned}$

$\begin{aligned} \text{Now, } I_1 & = \int \limits_{0}^{4} \sqrt{8x - x^2} \; dx \\\\ & = \int \limits_{0}^{4} \sqrt{-\left[\left(x^2 - 8x + 16\right) - 16\right]} \; dx \\\\ & = \int \limits_{0}^{4} \sqrt{\left(4\right)^2 - \left(x - 4\right)^2} \; dx \\\\ & \left[\text{Note: } \int \limits_{p}^{q} \sqrt{a^2 - x^2} \; dx = \left[\dfrac{x}{2} \sqrt{a^2 - x^2} + \dfrac{a^2}{2} \sin^{-1} \left(\dfrac{x}{a}\right)\right]_{p}^{q} \right] \\\\ & = \left[\left(\dfrac{x - 4}{2}\right)\sqrt{16 - \left(x-4\right)^2} + \dfrac{16}{2} \sin^{-1} \left(\dfrac{x - 4}{4}\right)\right]_{0}^{4} \\\\ & = 0 + 8 \sin^{-1} \left(0\right) - \left(-2\right) \sqrt{16 - 16} - 8 \sin^{-1} \left(-1\right) \\\\ & = 8 \times \dfrac{\pi}{2} = 4 \pi \;\;\; \cdots \; (5) \end{aligned}$

$\begin{aligned} I_2 & = \int \limits_{0}^{4} \sqrt{x} \; dx \\\\ & = \dfrac{2}{3} \left[\left(x\right)^{3/2}\right]_{0}^{4} = \dfrac{2}{3} \times \left(4\right)^{3/2} = \dfrac{16}{3} \;\;\; \cdots \; (6) \end{aligned}$

$\therefore$ $\;$ In view of equations $(5)$ and $(6)$, equation $(4)$ becomes

$I = 4 \pi + 2 \times \dfrac{16}{3} = 4 \pi + \dfrac{32}{3}$

$\therefore$ $\;$ Required area $= 4 \pi + \dfrac{32}{3}$ sq units

Applications of Definite Integration

Find the area of the region in the first quadrant enclosed by the X axis, the line $y = x$ and the circle $x^2 + y^2 = 32$

$x^2 + y^2 = 32$ is a circle with center at $O \left(0,0\right)$ and radius $\sqrt{32}$ units.

The line $y = x$ and the circle intersect at the the point $A \left(4,4\right)$.

Required area is the shaded region OAB.

From the figure, $\text{Area OAB} = \text{Area OAC} + \text{Area ABC}$ $\;\;\; \cdots \; (1)$

Let $\text{Area OAC} = \left|I_1\right|$. Here

$\begin{aligned} I_1 & = \int \limits_{0}^{4} y \; dx \hspace{2em} \left[\text{where } y = x\right] \\\\ & = \int \limits_{0}^{4} x \; dx = \left[\dfrac{x^2}{2}\right]_{0}^{4} = \dfrac{16}{2} = 8 \end{aligned}$

$\therefore$ $\text{Area OAC} = 8$ sq units $\;\;\; \cdots \; (2)$

Let $\text{Area ABC} = \left|I_2\right|$. Here

$\begin{aligned} I_2 & = \int \limits_{4}^{\sqrt{32}} y \; dx \hspace{2em} \left[\text{where } y = \sqrt{32 - x^2}\right] \\\\ & = \int \limits_{4}^{\sqrt{32}} \sqrt{32 - x^2} \; dx \\\\ & = \int \limits_{4}^{\sqrt{32}} \sqrt{\left(4 \sqrt{2}\right)^2 - \left(x\right)^2} \; dx \\\\ & \hspace{2em} \left[\text{Note: }\int \limits_{p}^{q} \sqrt{a^2 - x^2} \; dx = \left[\dfrac{x}{2} \sqrt{a^2 - x^2} + \dfrac{a^2}{2} \sin^{-1} \left(\dfrac{x}{a}\right)\right]_{a}^{b}\right] \\\\ & = \left[\dfrac{x}{2} \sqrt{32 - x^2} + \dfrac{32}{2} \sin^{-1} \left(\dfrac{x}{4 \sqrt{2}}\right)\right]_{4}^{\sqrt{32}} \\\\ & = \dfrac{\sqrt{32}}{2} \times \sqrt{32 - 32} + 16 \sin^{-1} \left(\dfrac{\sqrt{32}}{4 \sqrt{2}}\right) - \dfrac{4}{2} \sqrt{32 - 16} - 16 \sin^{-1} \left(\dfrac{4}{4 \sqrt{2}}\right) \\\\ & = 0 + 16 \sin^{-1} \left(1\right) - 2 \times 4 - 16 \sin^{-1} \left(\dfrac{1}{\sqrt{2}}\right) \\\\ & = 16 - \dfrac{\pi}{2} - 8 - 16 \times \dfrac{\pi}{4} = 4 \pi - 8 \end{aligned}$

$\therefore$ $\text{Area OAB} = 4 \pi - 8$ sq units $\;\;\; \cdots \; (3)$

$\therefore$ $\;$ In view of equations $(2)$ and $(3)$, equation $(1)$ becomes

$\text{Area OAB } = 8 + 4 \pi - 8 = 4 \pi$ sq units

Applications of Definite Integration

Determine the area of the region bounded by $y = 2x^2 + 10$ and $y = 4x + 16$

The parabola $y = 2x^2 + 10$ and the line $y = 4x + 16$ intersect at the points $A \left(-1,12\right)$ and $B \left(3,28\right)$.

The required area A is the shaded region ABC.

$\begin{aligned} \text{Shaded Area ABC} \left(A\right) & = \text{Area BDF } \left(A_1\right) - \left[\text{Area DAE} \left(A_2\right) \right.\\ & \hspace{2em} \left. + \text{Area EACO} \left(A_3\right) + \text{Area OCBF} \left(A_4\right) \right] \end{aligned}$ $\;\;\; \cdots \; (1)$

Now, $A_1 = \left|I_1\right|$ where

$\begin{aligned} I_1 & = \int \limits_{-4}^{3} y \; dx \hspace{2em} \text{where } y = 4x + 16 \\\\ & = \int \limits_{-4}^{3} \left(4x + 16\right) \; dx \\\\ & = \dfrac{4}{2} \left[x^2\right]_{-4}^{3} +16 \left[x\right]_{-4}^{3} = 2 \left(9-16\right) + 16 \left(3 + 4\right) = 98 \end{aligned}$

$\therefore$ $\;$ $A_1 = 98$ $\;$ sq units $\;\;\; \cdots \; (2)$

$A_2 = \left|I_2\right|$ where

$\begin{aligned} I_2 & = \int \limits_{-4}^{-1} y \; dx \hspace{2em} \text{where } y = 4x + 16 \\\\ & = \int \limits_{-4}^{-1} \left(4x + 16\right) \; dx \\\\ & = \dfrac{4}{2} \left[x^2\right]_{-4}^{-1} +16 \left[x\right]_{-4}^{-1} = 2 \left(1-16\right) + 16 \left(-1 + 4\right) = 18 \end{aligned}$

$\therefore$ $\;$ $A_2 = 18$ $\;$ sq units $\;\;\; \cdots \; (3)$

$A_3 = \left|I_3\right|$ where

$\begin{aligned} I_3 & = \int \limits_{-1}^{0} y \; dx \hspace{2em} \text{where } y = 2x^2 + 10 \\\\ & = \int \limits_{-1}^{0} \left(2x^2 + 10\right) \; dx \\\\ & = \dfrac{2}{3} \left[x^3\right]_{-1}^{0} + 10 \left[x\right]_{-1}^{0} = \dfrac{2}{3} \left[0 - \left(-1\right)^3\right] + 10 \left[0 - \left(-1\right)\right] = \dfrac{32}{3} \end{aligned}$

$\therefore$ $\;$ $A_3 = \dfrac{32}{3}$ $\;$ sq units $\;\;\; \cdots \; (4)$

$A_4 = \left|I_4\right|$ where

$\begin{aligned} I_4 & = \int \limits_{0}^{3} y \; dx \hspace{2em} \text{where } y = 2x^2 + 10 \\\\ & = \int \limits_{0}^{3} \left(2x^2 + 10\right) \; dx \\\\ & = \dfrac{2}{3} \left[x^3\right]_{0}^{3} + 10 \left[x\right]_{0}^{3} = \dfrac{2}{3} \times 27 + 10 \times 3 = 48 \end{aligned}$

$\therefore$ $\;$ $A_4 = 48$ $\;$ sq units $\;\;\; \cdots \; (5)$

$\therefore$ $\;$ In view of equations $(2)$, $(3)$, $(4)$ and $(5)$, equation $(1)$ becomes

$A = 98 - \left(18 + \dfrac{32}{3} + 48\right) = \dfrac{64}{3}$ sq units

Applications of Definite Integration

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 $\;\;\; \cdots \; (5)$

$\begin{aligned} \text{Now, }A_4 & = A_5 - A_6 - A_1 \\\\ & = 8 - \dfrac{16}{3} - \dfrac{5}{6} \hspace{2em} \left[\text{from equations } (3), (4) \text{ and } (5)\right] \\\\ & = \dfrac{11}{6} \; \text{sq units} \end{aligned}$