Chapter 8: Application of Integrals

The definite integral has widespread applications in computing the area of bounded regions.

Area Under Simple Curves

Let a curve be represented by $y = f(x)$. The area of the region bounded by the curve $y = f(x)$, $x$-axis and the ordinates $x = a$ and $x = b$ is given by the definite integral: $$ \text{Area} = \int_a^b y dx = \int_a^b f(x) dx $$ Similarly, the area of the region bounded by a curve $x = g(y)$, $y$-axis, $y = c$ and $y = d$ is: $$ \text{Area} = \int_c^d x dy = \int_c^d g(y) dy $$

Area between Two Curves

The area of a region bounded by two curves $y = f(x)$ and $y = g(x)$ (where $f(x) \ge g(x)$ in $[a, b]$), and the lines $x = a$ and $x = b$, is given by: $$ \text{Area} = \int_a^b [f(x) - g(x)] dx $$

Competency-Based Questions

1. [CBSE 2021] Find the area of the region bounded by the curve $y = x^2$ and the line $y = 4$.

Solution:

The curve is $x^2 = y$ which is a parabola opening upwards with vertex at the origin.

The line is $y = 4$ which is parallel to the $x$-axis.

The intersection points of $y = x^2$ and $y = 4$ are $x^2 = 4 \implies x = \pm 2$.

The region is symmetric about the $y$-axis.

So, Required Area = $2 \times (\text{Area bounded by curve in first quadrant})$

Area $= 2 \times \int_0^2 (\text{upper curve} - \text{lower curve}) dx$

Upper curve is $y = 4$, lower curve is $y = x^2$.

Area $= 2 \times \int_0^2 (4 - x^2) dx$

Area $= 2 \times \left[ 4x - \frac{x^3}{3} \right]_0^2$

Area $= 2 \times \left[ 4(2) - \frac{2^3}{3} - (0) \right]$

Area $= 2 \times \left[ 8 - \frac{8}{3} \right] = 2 \times \left[ \frac{16}{3} \right]$

Area $= \frac{32}{3} \text{ sq. units}$.

Alternatively, using integration w.r.t $y$:

Area $= \int_0^4 2x dy = \int_0^4 2\sqrt{y} dy = 2 \left[ \frac{y^{3/2}}{3/2} \right]_0^4 = \frac{4}{3} [4^{3/2}] = \frac{4}{3} (8) = \frac{32}{3} \text{ sq. units}$.

2. [Sample Paper 2024] Find the area of the region bounded by $y = \cos x$, $y = \sin x$, the $y$-axis and $x \in [0, \pi/4]$.

Solution:

In the interval $[0, \pi/4]$, $\cos x \ge \sin x$.

Therefore, upper curve $f(x) = \cos x$ and lower curve $g(x) = \sin x$.

Area $= \int_0^{\pi/4} (\cos x - \sin x) dx$

$= \left[ \sin x - (-\cos x) \right]_0^{\pi/4} = \left[ \sin x + \cos x \right]_0^{\pi/4}$

$= (\sin(\pi/4) + \cos(\pi/4)) - (\sin(0) + \cos(0))$

$= \left(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}\right) - (0 + 1)$

$= \frac{2}{\sqrt{2}} - 1 = \sqrt{2} - 1 \text{ sq. units}$.

3. [CBSE 2018] Find the area of the smaller part of the circle $x^2 + y^2 = a^2$ cut off by the line $x = \frac{a}{\sqrt{2}}$.

Solution:

The circle $x^2 + y^2 = a^2$ has its center at $(0,0)$ and radius $a$.

The line $x = \frac{a}{\sqrt{2}}$ is perpendicular to the $x$-axis.

The bounded region is symmetric w.r.t positive $x$-axis.

So, Required Area $= 2 \times \int_{a/\sqrt{2}}^a y dx$

For the circle, $y^2 = a^2 - x^2 \implies y = \sqrt{a^2 - x^2}$ (taking positive root for 1st quadrant).

Area $= 2 \int_{a/\sqrt{2}}^a \sqrt{a^2 - x^2} dx$

We know the standard integral: $\int \sqrt{a^2 - x^2} dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right)$

Area $= 2 \left[ \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right) \right]_{a/\sqrt{2}}^a$

$= 2 \left( \left[ 0 + \frac{a^2}{2}\sin^{-1}(1) \right] - \left[ \frac{a}{2\sqrt{2}}\sqrt{a^2 - \frac{a^2}{2}} + \frac{a^2}{2}\sin^{-1}\left(\frac{1}{\sqrt{2}}\right) \right] \right)$

$= 2 \left( \frac{a^2}{2}\cdot\frac{\pi}{2} - \frac{a}{2\sqrt{2}}\cdot\frac{a}{\sqrt{2}} - \frac{a^2}{2}\cdot\frac{\pi}{4} \right)$

$= 2 \left( \frac{\pi a^2}{4} - \frac{a^2}{4} - \frac{\pi a^2}{8} \right)$

$= 2 \left( \frac{\pi a^2}{8} - \frac{a^2}{4} \right) = \frac{\pi a^2}{4} - \frac{a^2}{2} = \frac{a^2}{2} \left[ \frac{\pi}{2} - 1 \right] \text{ sq. units}$.