Class 12 Maths Notes

Application of Integrals Class 12 Notes

Complete, exam-ready notes on applying definite integrals: the area under a curve, the area between two curves, and areas of regions bounded by parabolas, lines and circles — written for CBSE boards and JEE revision.

Class12SubjectMathematicsCoversCBSE · JEE

Written byDeep Narayan· Science & Mathematics EducatorReviewed byPushpanjali

What is this chapter about in one line?

A definite integral adds up the strips of area under a curve, so it gives the exact area bounded by curves and lines.

Area Under a Curve

Definite integral as area

For $y = f(x) \geq 0$ on $[a, b]$ , the area between the curve and the x-axis is $\int_a^b f(x)\,dx$ . If the curve dips below the axis, integrate the absolute value or split at the roots.

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

Area with sign handled

Area Between Two Curves

To find the area between f and g, first find where the curves intersect — those points are the limits of integration.
The strip height is the difference of the functions at each x.
Region splits into pieces whenever the curves cross inside the interval — integrate each piece separately.

A = \int_a^b \left(f(x) - g(x)\right) dx,\qquad f(x) \geq g(x)\ \text{on } [a,b]

Area between curves

Standard Bounded Regions

Area between a parabola
$y^2 = 4ax$
and its latus rectum
$x = a$
is
$\frac{8a^2}{3}$
.
Area bounded by
$y = x^2$
and the line
$y = x$
is found from their intersection points
$(0,0)$
and
$(1,1)$
.
A full ellipse
$x^2/a^2 + y^2/b^2 = 1$
has area
$\pi ab$
; a circle has area
$\pi r^2$
.

Use symmetry

For symmetric regions like ellipses and circles, compute the area in the first quadrant and multiply by the number of quadrants instead of one long integration.

Choosing the Variable

Integrate along x by vertical strips when each region is bounded above and below by single curves.
Integrate along y by horizontal strips when the curves are easier written as x = g(y).
For
$y^2 = 4ax$
type curves, horizontal strips often avoid square roots.

Solved Examples

Example: Find the area between $y = x^2$ and $y = x$ .

Solution: Intersections at x = 0 and x = 1, with $x \geq x^2$ on [0,1]. Area $= \int_0^1 (x - x^2)\,dx = \left[\frac{x^2}{2} - \frac{x^3}{3}\right]_0^1 = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}$ .

Example: Find the area of the ellipse $x^2/9 + y^2/4 = 1$ .

Solution: $a = 3$ , $b = 2$ , so the area is $\pi ab = \pi(3)(2) = 6\pi$ — four times one quadrant computed by a single integration.

Revision

Key formulas at a glance

Memorise these before attempting numericals — most exam questions hinge on one of them.

Area under a curve

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

Area between curves

A = \int_a^b (f(x) - g(x))\,dx

Parabola and latus rectum

\text{Area} = \frac{8a^2}{3}

Ellipse area

\pi ab

Circle area

\pi r^2

Exam tips

How this chapter is asked

Where this topic appears in CBSE, JEE Main and NEET papers.

Find the intersection points first — they set the limits of integration.
Always integrate top curve minus bottom curve.
Split the integral wherever two curves cross inside the interval.
Use symmetry for circles and ellipses to halve or quarter the work.
Check the answer is positive — swap the order if f(x) < g(x).

FAQ

Common questions

How do you find the area between two curves?

Set the two functions equal to find their intersection points (the limits), then integrate the difference of the upper and lower curves between those limits.

Why must the area under a curve be positive?

Area is a magnitude, so you take the absolute value of the integrand, or split the integral at the roots, so that strips below the x-axis contribute positive area too.

When should I integrate with respect to y instead of x?

When the region is bounded by curves better written as x = g(y) — like parabolas y² = 4ax — or when horizontal strips simplify the integrand and avoid square roots.

What is the area of the region bounded by a parabola and its latus rectum?

For y² = 4ax bounded by x = a, the area is 8a²/3 — half on each side of the axis, computed by one integration and doubled.

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