Application of Integrals: Class 12 Maths · Chapter 8

Areas under curves, one thin strip at a time

Mathematics Part 2 · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published 12 Oct 2026 · Updated 12 Oct 2026

50 min
1 simulation
12 practice questions
14 MCQs

What you will learn in Application of Integrals

Explain why a region with a curved edge needs integration, and set up its area as a sum of thin strips
Write the area between a curve y = f(x) and the x-axis, over an interval from a to b, as a definite integral and evaluate it
Use horizontal strips to find the area between a curve x = g(y) and the y-axis, from y = c to y = d
Handle curves that lie below or cross the x-axis by splitting at the zeros and adding absolute values
Use symmetry to find the areas enclosed by circles and ellipses in standard form, recovering πr² and πab
Check every computed area against a quick geometric estimate

Original companion text aligned to NCERT Mathematics Class 12, Chapter 8, pp. 292–299. Not affiliated with or endorsed by NCERT.

8.1 Introduction

A farmer near Nashik owns a plot with a road along its front, two straight fences at the sides and a stream curling along the back. She wants its area to order the right amount of fertiliser. A tape measure gives her the three straight sides at once; the stream is the problem.

School geometry handles shapes made of straight pieces: cut a polygon into triangles, rectangles or trapezia and add their areas. The circle is the one curved shape with a ready formula. For a boundary like the stream, or the arc of a parabola, no formula exists, and every polygon we fit is a little too big in some places and a little too small in others (Fig. 8.1).

3/3Sixteen thin strips hug the curve far better; thinner strips do better still.

Fig. 8.1A curved edge has no area formula, but thin strips can approach it as closely as we like.

A region sits on a straight base, with straight vertical sides and a wavy curved top. In the first step it is shaded and labelled: no formula fits. In the second step a dashed trapezium is drawn under the chord joining the two top corners; near the hump the curve is above the chord, a gap, and near the dip the trapezium pokes above the curve, extra area. In the third step the region is cut into sixteen thin rectangles whose tops almost follow the curve, labelled: thinner is closer.

The previous chapter built the tool we need. A definite integral is the limit of sums of thin rectangles, and the fundamental theorem lets us evaluate it with an antiderivative. Here we turn that tool round: given a region, we slice it into strips, write the area of one strip, and integrate.

The curves in this chapter are the friendly ones, each in its standard form: parabolas, ellipses, circles and straight lines. Alongside every integral we make a rough geometric estimate, so that a slip in the algebra shows up at once.

Quick check

1 of 2

Which of these areas can a school geometry formula give without integration?

A field shaped like a trapeziumA semicircular stageThe region under

y = x^{2}

from 0 to 1A lake with a wavy shore

Select all that apply

Show answer

Answer

(A) A field shaped like a trapezium
(B) A semicircular stage

A trapezium and a semicircle (half of

π r^{2}

) have formulas. The parabola and the wavy shore are curved edges with no ready formula; integration gives the first as

\frac{1}{3}

8.2 Area under Simple Curves

Throughout this section the region is bounded by a curve, one coordinate axis and two straight lines parallel to the other axis. The vertical lines $x = a$ and $x = b$ are called ordinates, and the size of the region they close off is the area under the curve. We first slice the region vertically, then horizontally.

Vertical strips

DerivationArea as a sum of vertical strips

Let $f$ be continuous with $f (x) \geq 0$ on $[a, b]$ . Cut the region under $y = f (x)$ into vertical strips. Over a short stretch of width $d x$ at $x$ the curve hardly changes height, so the strip is almost a rectangle of height $y = f (x)$ .

Its area is the elementary area $d A = y d x$ . Adding the strips from $x = a$ to $x = b$ and letting their width shrink to zero turns the sum into a definite integral.

\begin{aligned} A & = \int_{a}^{b} y d x \\ = \int_{a}^{b} f (x) d x \end{aligned}

(8.1)

where

$A$: the area of the region under $y = f (x)$ from $x = a$ to $x = b$
$d x$: the width of one strip

2/3A strip of width $d x$ and height $y$ has area $d A = y d x$ .

Fig. 8.2Area under a curve: one strip has area

y d x

; the integral adds them all.NCERT Fig. 8.1 · p. 292

Axes with a curve rising gently from left to right, labelled y equals f of x. In the first step the region between the curve and the x-axis, from a to b, is shaded. In the second step one thin vertical strip is highlighted: its width is marked dx and its height y, and a label reads dA equals y dx. In the third step the region is filled with twelve thin rectangles, and the label reads A equals the integral from a to b of y dx.

Definition

Elementary area

The area $d A$ of one thin strip of a region: $y d x$ for a vertical strip of width $d x$ , or $x d y$ for a horizontal strip of thickness $d y$ .

ExampleUnder $y = x^{2}$ , the strip at $x = 3$ of width $0.01$ has area about $9 \times 0.01 = 0.09$ .

The working routine is short. Sketch the region and read off the limits, write the strip height as a function of $x$ , integrate, and then compare the answer with a rectangle or trapezium you can see in the sketch.

Example

Under a parabola, with a sanity check

How much area lies between the curve $y = x^{2} + 1$ and the $x$ -axis, from $x = 0$ to $x = 2$ ?

Try it yourself first. Then reveal the solution one step at a time.

Horizontal strips

Sometimes the boundary is easier to write as $x = g (y)$ , and the region lies between this curve, the $y$ -axis and the horizontal lines $y = c$ and $y = d$ (Fig. 8.3). Then slice across. A horizontal strip of thickness $d y$ at height $y$ reaches from the $y$ -axis to the curve, so its length is $x = g (y)$ and its area is $x d y$ .

\begin{aligned} A & = \int_{c}^{d} x d y \\ = \int_{c}^{d} g (y) d y \end{aligned}

(8.2)

where

$A$: the area of the region between $x = g (y)$ and the $y$ -axis, from $y = c$ to $y = d$

2/3A horizontal strip has length $x = g (y)$ and thickness $d y$ .

Fig. 8.3With

x

given in terms of

y

, horizontal strips of length

x

and thickness

d y

do the job.NCERT Fig. 8.2 · p. 293

Axes with a curve x equals g of y that moves further from the y-axis as y increases. In the first step the region between the y-axis and the curve, from height c to height d, is shaded. In the second step one thin horizontal strip is highlighted, its thickness marked dy and its length x, with the label dA equals x dy. In the third step ten horizontal rectangles fill the region and the label reads A equals the integral from c to d of x dy.

Which way to slice is your choice. Choose the direction in which one formula describes every strip of the region; the next example shows how much work a good choice saves.

Example

A sideways parabola

Find the area of the region enclosed between the parabola $x = 4 - y^{2}$ and the $y$ -axis.

Try it yourself first. Then reveal the solution one step at a time.

3/3It fills two-thirds of the box, so the area is $\frac{32}{3}$ .

Fig. 8.4The parabolic region fills exactly two-thirds of its bounding box:

\frac{2}{3} \times 16 = \frac{32}{3}

Axes with a parabola opening to the left, its tip at (4, 0) and crossing the y-axis at 2 and minus 2. In the first step the region between the parabola and the y-axis is shaded and one horizontal strip of thickness dy is highlighted, with the integral from minus 2 to 2 of 4 minus y squared dy. In the second step a dashed 4 by 4 box surrounds the region, with area 16. In the third step a note says the region fills two-thirds of the box, so its area is 32 over 3.

Quick check

1 of 3

What area do the line

y = 2 x

, the

x

-axis and the line

x = 3

enclose?

sq. units

Show answer

Answer 9 sq. units

Integrating gives

{[x^{2}]}_{0}^{3} = 9

square units, the same as the triangle with base 3 and height 6.

When the curve dips below the x-axis

Formula (8.1) assumed $f (x) \geq 0$ . If the curve lies below the $x$ -axis on $[a, b]$ , every strip "height" $y$ is negative, so each $y d x$ is negative and so is the integral. Area is never negative: in that case the area is the absolute value of the integral.

3/3The area is $A_{1} + ∣ A_{2} ∣$ , not $A_{1} + A_{2}$ .

Fig. 8.5Below the axis the integral is negative; the area adds the sizes of the pieces.NCERT Figs 8.3, 8.4 · p. 293

Axes with a curve that starts above the x-axis at a, rises to a hump, falls through the axis at c and dips to a trough before the line x equals b. In the first step only the hump between a and c is shaded with a plus sign, labelled A1 greater than 0. In the second step the dip between c and b is also shaded, hatched with a minus sign, labelled A2 less than 0: below the axis the integral is negative. In the third step the label reads Area equals A1 plus the absolute value of A2.

Often the curve crosses the axis inside $[a, b]$ , at some point $c$ (Fig. 8.5). A single integral from $a$ to $b$ then lets the piece below cancel part of the piece above and returns a net value, not the area. Split at $c$ instead.

A = ∣ A_{1} ∣ + ∣ A_{2} ∣

(8.3)

where

$A_{1}$: $\int_{a}^{c} f (x) d x$ , the integral over the first piece
$A_{2}$: $\int_{c}^{b} f (x) d x$ , the integral over the second piece
$c$: the point where the curve crosses the $x$ -axis

Example

A region under the axis

The curve $y = x^{2} - 2 x$ dips below the $x$ -axis and comes back up. Find the area of the region it encloses with the axis.

Try it yourself first. Then reveal the solution one step at a time.

InteractiveArea explorer

An interactive graph with seven curves from this chapter: x squared plus 1, 2x minus 3, x squared minus 2x, sine 2x, x cubed, x times the absolute value of x, and the sideways parabola x equals 4 minus y squared. Two limits are shown as vertical lines (horizontal lines for the sideways parabola) with round handles that can be dragged, and two sliders set them too. The region between the curve and the axis is shaded plain where it is above the axis and hatched where it is below. A status panel gives the signed integral, the area, and the size of each piece between zeros of the curve. Buttons draw 4, 8, 16 or 40 strips and show their total area and how far it is from the true area. Nothing moves until the learner acts.

Fig. 8.6Pick a curve, then drag the limits (or use the sliders) and compare the signed integral with the area. Turn on strips to watch their total approach the area.

Quick check

1 of 3

From

x = 0

x = 6

, what total area lies between the line

y = x - 2

and the

x

-axis?

sq. units

Show answer

Answer 10 sq. units

The line crosses at

x = 2

. Below the axis, the piece from 0 to 2 has area 2 (its integral is

- 2

); above it, the piece from 2 to 6 has area 8. The area is 10. One integral from 0 to 6 would give only 6.

Areas of circles and ellipses

A circle or an ellipse is not the graph of one function $y = f (x)$ : each $x$ between the extremes gives two values of $y$ , one above the axis and one below. Symmetry solves this. In standard form both curves are symmetric about both axes, so the enclosed area is four times the area of the piece in the first quadrant, where we may take $y \geq 0$ .

Example 1NCERT Example 1 · p. 294

A round table top

A round table top has radius 6 dm. With its centre at the origin, its rim is $x^{2} + y^{2} = 36$ . Find its area by integration, first with vertical strips and then with horizontal strips.

Try it yourself first. Then reveal the solution one step at a time.

2/4A vertical strip has height $y = \sqrt{36 - x^{2}}$ .

Fig. 8.7One quarter of the circle, sliced either way, then multiplied by 4.NCERT Figs 8.5, 8.6 · p. 294

A circle of radius 6 centred at the origin O, through A(6, 0) and B(0, 6). In the first step the quarter OAB is shaded and the equation x squared plus y squared equals 36 is shown. In the second step a vertical strip of width dx and height y equals the square root of 36 minus x squared is highlighted in the quarter. In the third step a horizontal strip of thickness dy and length x equals the square root of 36 minus y squared is highlighted instead. In the last step all four quarters are shaded, with four times 9 pi equals 36 pi.

SimulationTwo points for every x

A coordinate plane from minus 8 to 8 on both axes with a fixed origin O and a draggable point P that starts at (3.6, 4.8), six units from O. The distance OP is drawn and measured, and the mirror image of P in the x-axis is shown, so the learner sees that each x on the circle has one point above the axis and one below.

Fig. 8.8P starts on the rim

x^{2} + y^{2} = 36

of the table. Its mirror image in the

x

-axis has the same

x

and is on the rim too.

Try this

Drag P to $(4.8, 3.6)$ , $(0, 6)$ and $(6, 0)$ . Does $O P$ stay 6?
Watch the mirror point. Both satisfy $x^{2} + y^{2} = 36$ ; which one gives the strip height in the first quadrant?
Put P at $(3, 5)$ . Is it inside or outside the rim? Compare $3^{2} + 5^{2}$ with 36.

Try this

Count your way to π

You’ll need

1 cm graph paper
a compass
a sharp pencil

Predict first

A quarter circle of radius 6 cm is drawn on 1 cm squares. Roughly how many squares' worth of area does it cover?

about 18about 28about 36

Put the compass point on a grid corner and draw a quarter circle of radius 6 cm in one quadrant.
Count the squares that lie wholly inside the arc.
Count the squares the arc cuts through.
Estimate the area as the whole squares plus half the cut squares, and compare it with $9 π \approx 28.3$ .

Table 8.1

What you count	Your value
Whole squares inside
Squares cut by the arc
Estimate: whole + half of cut

Your entries are saved automatically.

Example 2NCERT Example 2 · p. 295

An elliptical lawn

A lawn in a park is an ellipse 10 m long and 6 m wide. With its centre at the origin and its long axis along the $x$ -axis, its edge is $\frac{x^{2}}{25} + \frac{y^{2}}{9} = 1$ . Find its area by integration, with vertical strips and then with horizontal strips.

Try it yourself first. Then reveal the solution one step at a time.

2/4Multiply every height by $\frac{3}{5}$ : the point at height 4 drops to 2.4.

Fig. 8.9Squashing every vertical strip by

\frac{3}{5}

turns the circle into the ellipse and its area

25 π

into

15 π

.NCERT Figs 8.7, 8.8 · p. 295

Axes with a dashed circle of radius 5 centred at the origin. In the first step only the circle, x squared plus y squared equals 25, is shown. In the second step the ellipse with semi-axes 5 and 3 is drawn inside it, and arrows at x equals 3 and minus 3 move the circle's points at height 4 down to height 2.4: every height is multiplied by three-fifths. In the third step the first-quadrant quarter of the ellipse is shaded with a vertical strip whose height is three-fifths of the square root of 25 minus x squared. In the last step the whole ellipse is shaded and the areas are compared: circle 25 pi, ellipse 15 pi.

The figure explains the answer $π a b$ . Squash the circle $x^{2} + y^{2} = a^{2}$ vertically by the factor $\frac{b}{a}$ . Each vertical strip keeps its width while its height is multiplied by $\frac{b}{a}$ , so the total area becomes $\frac{b}{a} \times π a^{2} = π a b$ . The numbers $a$ and $b$ are the semi-axes of the ellipse.

Quick check

1 of 3

The area enclosed by the ellipse

\frac{x^{2}}{16} + \frac{y^{2}}{49} = 1

11 π

784 π

65 π

28 π

Show answer

Answer (D) $28 π$

The semi-axes are

a = 4

and

b = 7

, so the area is

π a b = 28 π

. Multiplying 16 by 49 gives

784 π

; adding gives

11 π

65 π

Practice · 5 questions

Practice 8.1: questions with answers

Short answerMediumNCERT Exercise 8.1, Q1
Use integration to find the area enclosed by the ellipse $\frac{x^{2}}{36} + \frac{y^{2}}{16} = 1$ .
Show solution
Answer $24 π \approx 75.4$ square units
In the first quadrant $y = \frac{2}{3} \sqrt{36 - x^{2}}$ , and the ellipse is symmetric about both axes.
So the area is $\frac{8}{3}$ times the integral $\int_{0}^{6} \sqrt{36 - x^{2}} d x$ .
That integral is the quarter-circle area for radius 6, namely $9 π$ . The standard formula gives the same: at $x = 6$ the bracket is $18 \cdot \frac{π}{2}$ , and at $x = 0$ it is 0.
So $A = \frac{8}{3} \times 9 π = 24 π$ . Check: $π a b = π \cdot 6 \cdot 4$ .
Short answerMediumNCERT Exercise 8.1, Q2
Find, by integration, the area enclosed by the curve $4 x^{2} + y^{2} = 36$ .
Show solution
Answer $18 π \approx 56.5$ square units
Dividing by 36 gives $\frac{x^{2}}{9} + \frac{y^{2}}{36} = 1$ : an ellipse with semi-axes 3 along $x$ and 6 along $y$ .
Horizontal strips suit this tall ellipse. In the first quadrant $x = \frac{1}{2} \sqrt{36 - y^{2}}$ .
So the area is twice $\int_{0}^{6} \sqrt{36 - y^{2}} d y$ . That integral is a quarter-circle area, $9 π$ , so $A = 18 π$ .
Check: $π a b = π \cdot 3 \cdot 6 = 18 π$ (about 56.5), between the rhombus (36) and the rectangle (72) on the axes.
Quick answerEasyNCERT Exercise 8.1, Q3
The part of the second quadrant that lies inside the circle $x^{2} + y^{2} = 9$ has area
$\frac{3 π}{4}$ $\frac{9 π}{4}$ $\frac{9 π}{2}$ $9 π$
Show answer
Answer (B) $\frac{9 π}{4}$
The region is a quarter of a circle of radius 3, so its area is $\frac{9 π}{4}$ (as an integral, $\int_{- 3}^{0} \sqrt{9 - x^{2}} d x$ over the second quadrant). Half the circle would be $\frac{9 π}{2}$ , and $\frac{3 π}{4}$ uses $r$ in place of $r^{2}$ .
Quick answerMediumNCERT Exercise 8.1, Q4
The parabola $y^{2} = 9 x$ , the $y$ -axis and the line $y = 6$ enclose a region of area
7224168
Show answer
Answer (D) 8
Horizontal strips give $\int_{0}^{6} \frac{y^{2}}{9} d y$ , which works out to ${[\frac{y^{3}}{27}]}_{0}^{6} = 8$ square units. Dropping the $\frac{1}{9}$ gives 72; 24 is the whole 4 by 6 rectangle; 16 is the region on the other side of the curve, next to the $x$ -axis ( $16 + 8 = 24$ ).
Quick answerEasy
Find the area enclosed by the ellipse $\frac{x^{2}}{49} + \frac{y^{2}}{4} = 1$ , to two decimal places.
sq. units
Show answer
Answer 43.98 sq. units
The semi-axes are 7 and 2, so the area is $14 π \approx 43.98$ .

Miscellaneous Examples

Both examples below cross the $x$ -axis between their limits, so the routine is the same each time: find the zeros, split, integrate each piece, add the sizes, and check with a shape you can measure.

Example 3NCERT Example 3 · p. 296

A line that crosses the axis

Between $x = 0$ and $x = 4$ , find the total area enclosed by the line $y = 2 x - 3$ and the $x$ -axis.

Try it yourself first. Then reveal the solution one step at a time.

3/3Area $= \frac{9}{4} + \frac{25}{4} = \frac{17}{2}$ .

Fig. 8.10Two triangles:

\frac{9}{4}

below the axis and

\frac{25}{4}

above it.NCERT Fig. 8.9 · p. 297

Axes with the line y equals 2x minus 3 crossing the x-axis at three-halves; a dashed vertical line marks x equals 4, where the line is at height 5. In the first step only the line and the limits are shown. In the second step the triangle below the axis, from 0 to three-halves, is hatched with a minus sign and the triangle above the axis, from three-halves to 4, is shaded with a plus sign. In the third step the triangles are labelled nine-quarters and twenty-five quarters, and a note gives the area as their sum, seventeen halves.

Example 4NCERT Example 4 · p. 297

A wave that changes sign

Find the total area enclosed between the wave $y = \sin 2 x$ and the $x$ -axis, from $x = 0$ to $x = \frac{3 π}{4}$ .

Try it yourself first. Then reveal the solution one step at a time.

3/3Area $= 1 + \frac{1}{2} = \frac{3}{2}$ .

Fig. 8.11A hump of area 1 and half a dip of area

\frac{1}{2}

: the area is

\frac{3}{2}

.NCERT Fig. 8.10 · p. 297

Axes with one full wave of y equals sine 2x from 0 to pi, rising to 1 at pi over 4 and falling to minus 1 at 3 pi over 4, where a dashed vertical line marks the right limit. In the first step only the curve and the limit are shown. In the second step the hump from 0 to pi over 2 is shaded with a plus sign and the part of the dip from pi over 2 to 3 pi over 4 is hatched with a minus sign. In the third step the pieces are labelled 1 and one-half and the area is given as three-halves.

Try both examples in the area explorer: move the right-hand limit and watch the signed integral and the area separate as soon as a piece slips below the axis.

Quick check

1 of 3

Match each region, between the curve and the

x

-axis over the interval, with its area.

Tap an item on the left, then its match on the right.

Show answer

Answer

$y = x$ on $[- 1, 1]$ : $1$
$y = \sin x$ on $[0, π]$ : $2$
$y = x^{2}$ on $[- 1, 2]$ : $3$
$y = x^{3}$ on $[- 1, 1]$ : $\frac{1}{2}$

y = x

: two triangles of

\frac{1}{2}

\sin x

[- \cos x]_{0}^{π} = 2

x^{2}

stays above the axis:

\frac{1}{3} + \frac{8}{3} = 3

x^{3}

: two pieces of

\frac{1}{4}

From historyNCERT Historical Note · p. 298

From exhaustion to ∫

Greek geometers found curved areas by the method of exhaustion: fill the region with polygons of known area, then keep adding smaller ones until what is left over is as small as you please. Eudoxus made the method rigorous in the fourth century BCE, and Archimedes used it for the circle and the parabola.

In Kerala, Madhava of Sangamagrama (around 1400) and his followers added up many tiny pieces to reach infinite series for π, reasoning close in spirit to integration.

In the 1660s Isaac Newton found areas by running his rule for rates backwards. Independently, Gottfried Leibniz wrote an area as a sum of infinitely thin strips and chose the sign ∫, a stretched S for summa (sum), publishing his calculus in the journal Acta Eruditorum in the 1680s. The name integral came from the Bernoulli brothers in the 1690s. Both men saw that finding areas undoes finding rates.

Their "infinitely small" strips worried later mathematicians. In the 1820s Augustin-Louis Cauchy defined the integral as a limit of sums, the definition you met in the previous chapter.

Recap

Application of Integrals: key points and summary

Areas with curved edges have no geometry formula; we slice them into thin strips and integrate (Section 8.1).
A vertical strip of width $d x$ under $y = f (x)$ has area $d A = y d x$ (Derivation).
If $f \geq 0$ on $[a, b]$ , the area under $y = f (x)$ is $\int_{a}^{b} f (x) d x$ ((8.1)).
If $g \geq 0$ on $[c, d]$ , the area between $x = g (y)$ and the $y$ -axis is $\int_{c}^{d} g (y) d y$ ((8.2)).
Below the $x$ -axis the integral is negative; the area is its absolute value (When the curve dips below the x-axis).
When the curve crosses the axis, split at the zeros and add the absolute values of the pieces ((8.3)).
Symmetry about both axes lets us find a circle's or an ellipse's area as four times a first-quadrant integral (Areas of circles and ellipses).
A circle of radius $r$ encloses $π r^{2}$ (Example 1); the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ encloses $π a b$ (Example 2).
Always compare an answer with a rectangle, triangle or trapezium you can see in the sketch (Example).

Flash cards · 7

Formulas to remember

1 / 7

Miscellaneous Exercise on Chapter 8

Find your NCERT exercise question

Our practice questions are written for ncert.ai and follow the NCERT exercises. Find the one that matches the question in your textbook; its answer is with it.

In your NCERT textbook	Practice question here
NCERT Exercise 8.1, Q1	Practice 8.1 · Q1
NCERT Exercise 8.1, Q2	Practice 8.1 · Q2
NCERT Exercise 8.1, Q3	Practice 8.1 · Q3
NCERT Exercise 8.1, Q4	Practice 8.1 · Q4
NCERT Miscellaneous Exercise 8, Q1	Miscellaneous exercise · Q1
NCERT Miscellaneous Exercise 8, Q2	Miscellaneous exercise · Q2
NCERT Miscellaneous Exercise 8, Q3	Miscellaneous exercise · Q3
NCERT Miscellaneous Exercise 8, Q4	Miscellaneous exercise · Q4
NCERT Miscellaneous Exercise 8, Q5	Miscellaneous exercise · Q5

Quick check: MCQs on Application of Integrals

8.1 Introduction2 questions
Horizontal strips3 questions
When the curve dips below the x-axis3 questions
Areas of circles and ellipses3 questions
Miscellaneous Examples3 questions

Glossary · 10

Key terms (English–Hindi)

area under a curveवक्र के अंतर्गत क्षेत्रफलvakra ke antargat kshetraphal: The area of the region between a curve, the $x$ -axis and two ordinates; it equals $\int_{a}^{b} f (x) d x$ when $f \geq 0$ .First seen in the chapter →

definite integralनिश्चित समाकलनnishchit samaakalan: The limit of sums of thin rectangles, $\int_{a}^{b} f (x) d x$ , evaluated as $F (b) - F (a)$ with an antiderivative $F$ .First seen in the chapter →

elementary areaप्रारंभिक क्षेत्रफलpraarambhik kshetraphal: The area of one thin strip of a region: $y d x$ for a vertical strip, $x d y$ for a horizontal one.First seen in the chapter →
ellipseदीर्घवृत्तdeerghvritt: The curve $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ in standard form, a circle stretched or squashed along one axis; it encloses $π a b$ .First seen in the chapter →

method of exhaustionनिःशेषण विधिnihsheshan vidhi: The Greek way of finding curved areas by filling a region with ever smaller polygons until the leftover is negligible.First seen in the chapter →

ordinateकोटिkoti: Here, a vertical line such as $x = a$ that closes off a region under a curve at one side.First seen in the chapter →

parabolaपरवलयparvalay: A curve such as $y = x^{2}$ or $y^{2} = 4 a x$ ; Archimedes showed a parabolic region fills $\frac{2}{3}$ of its bounding box.First seen in the chapter →

quadrantचतुर्थांशchaturthaansh: One of the four regions into which the axes divide the plane; in the first quadrant $x \geq 0$ and $y \geq 0$ .First seen in the chapter →

semi-axisअर्ध-अक्षardh-aksh: Half of an axis of an ellipse: the distances $a$ and $b$ from the centre to the curve along the two axes.First seen in the chapter →
symmetryसममितिsammiti: A curve is symmetric about an axis when reflecting it in that axis gives the same curve; then the two sides have equal areas.First seen in the chapter →

Application of Integrals: Class 12 Maths · Chapter 8

What you will learn in Application of Integrals

8.1 Introduction

8.2 Area under Simple Curves

Vertical strips

Horizontal strips

When the curve dips below the x-axis

Areas of circles and ellipses

Practice 8.1: questions with answers

Miscellaneous Examples

Application of Integrals: key points and summary

Miscellaneous Exercise on Chapter 8

Practice questions with answers (NCERT “Miscellaneous exercise”)

Find your NCERT exercise question

Quick check: MCQs on Application of Integrals

Key terms (English–Hindi)

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