Area Under Simple Curves – Application of Integrals

01 · Overview

The Four Fundamental Cases

Area under a simple curve means the area between the curve and a reference line (axis). There are four standard cases based on which axis or line forms the boundary.

Case A

Curve & x-axis

Area between \(y = f(x)\) and the x-axis, between \(x = a\) and \(x = b\).

Case B

Curve & y-axis

Area between \(x = g(y)\) and the y-axis, between \(y = c\) and \(y = d\).

Case C

Curve & Vertical Line

Area bounded by the curve, two vertical lines \(x = a\), \(x = b\), and the x-axis.

Case D

Curve & Horizontal Line

Area bounded by curve and a horizontal line (use appropriate limits).

Case A

Area Bounded by Curve and x-axis

This is the most common case. The region lies between the curve \(y = f(x)\) above and the x-axis below, from \(x = a\) to \(x = b\).

Area between Curve and x-axis

\[ A = \int_a^b y\,dx = \int_a^b f(x)\,dx \]

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Critical Rule — Sign of Area

If \(f(x) \geq 0\) on \([a,b]\) → area \(= \int_a^b f(x)\,dx\) (positive).
If \(f(x) \leq 0\) on \([a,b]\) → area \(= \left|\int_a^b f(x)\,dx\right|\) (take absolute value).
If curve crosses x-axis → split at crossing point and add absolute values.

Example 1 — Standard Board Question

Find the area bounded by \(y = x^2\), x-axis, \(x=0\), \(x=3\).

\(A = \displaystyle\int_0^3 x^2\,dx = \left[\dfrac{x^3}{3}\right]_0^3 = \dfrac{27}{3} - 0 = \mathbf{9}\) sq. units

Example 2 — Parabola (JEE Level)

Area bounded by \(y^2 = 4x\) and x-axis from \(x=0\) to \(x=4\).

Express: \(y = 2\sqrt{x}\) (upper half). \(A = \displaystyle\int_0^4 2\sqrt{x}\,dx = 2\cdot\dfrac{2}{3}[x^{3/2}]_0^4 = \dfrac{4}{3}\cdot 8 = \mathbf{\dfrac{32}{3}}\) sq. units

Case B

Area Bounded by Curve and y-axis

When the curve is given as \(x = g(y)\), or when the region is naturally bounded on the left and right by the y-axis and curve, integrate with respect to \(y\).

Area between Curve and y-axis

\[ A = \int_c^d x\,dy = \int_c^d g(y)\,dy \]

Example 3

Find area bounded by \(x = y^2\), y-axis, \(y=0\) to \(y=2\).

\(A = \displaystyle\int_0^2 y^2\,dy = \left[\dfrac{y^3}{3}\right]_0^2 = \dfrac{8}{3}\) sq. units

Case C

Area Bounded by Curve and Vertical Lines

The region is enclosed between the curve \(y=f(x)\), the x-axis, and two vertical lines \(x = a\) and \(x = b\). This is the direct application of definite integrals.

Area — Curve + Vertical Lines + x-axis

\[ A = \int_a^b f(x)\,dx \]

Example 4 — Board 5-Mark Type

Find area enclosed between \(y = \sin x\), x-axis, \(x=0\) and \(x=\pi\).

\(A = \displaystyle\int_0^\pi \sin x\,dx = [-\cos x]_0^\pi = (-\cos\pi)-(-\cos 0) = 1+1 = \mathbf{2}\) sq. units

Note: \(\sin x \geq 0\) on \([0,\pi]\), so no sign issue here.

Example 5 — Sign Issue

Find area enclosed between \(y = \sin x\), x-axis, from \(x=0\) to \(x = 2\pi\).

Curve is positive on \([0,\pi]\) and negative on \([\pi,2\pi]\).

\(A = \displaystyle\int_0^\pi \sin x\,dx + \left|\int_\pi^{2\pi} \sin x\,dx\right| = 2 + |-(-2)| = \mathbf{4}\) sq. units

Never compute \(\int_0^{2\pi}\sin x\,dx = 0\) and say area is 0!

Quick Reference

Standard Area Results — Memorise These!

Shape / Curve	Bounded By	Area Formula
Parabola \(y^2=4ax\)	Latus Rectum	\(\dfrac{8a^2}{3}\)
Circle \(x^2+y^2=r^2\)	Full circle	\(\pi r^2\)
Ellipse \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\)	Full ellipse	\(\pi ab\)
\(y=x^n\)	x-axis, \(x=0\) to \(x=a\)	\(\dfrac{a^{n+1}}{n+1}\)
\(y = \sin x\)	x-axis, \(0\) to \(\pi\)	\(2\)
\(y = \cos x\)	x-axis, \(-\pi/2\) to \(\pi/2\)	\(2\)

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Board Exam — Must-Know

Sketch the curve and shade the bounded region before starting.
For parabola \(y^2=4ax\), use the upper half \(y=2\sqrt{ax}\) to integrate w.r.t. x.
Circle area: use quarter circle \(\int_0^r\sqrt{r^2-x^2}\,dx = \dfrac{\pi r^2}{4}\), then multiply by 4.
Always mention correct limits with justification in your solution.

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JEE Main / Advanced

Area of one loop of \(r = a\sin\theta\) (polar) = \(\dfrac{\pi a^2}{4}\) — might appear in advanced.
Area between \(y=x^2\) and \(y=\sqrt{x}\) is \(\dfrac{1}{3}\) — a JEE classic.
When curve is given parametrically: \(A = \int y\,\dfrac{dx}{dt}\,dt\) over correct limits.
For negative region: area = \(\left|\int_a^b f(x)\,dx\right|\), not just the raw value.