MathsCBSE Class 12 · Chapter 8

Application of Integrals Class 12: Notes, Methods and Solved Questions

Q: Is the area between two curves in the 2026-27 syllabus?

No. NCERT removed the section on area between two curves, and the syllabus limits the chapter to area under simple curves. The 2026-27 sample paper does include regions bounded by a circle or a parabola and a straight line, so practise those simple cases as well.

Q: Which curves can be asked?

Lines, circles, parabolas and ellipses in standard form only, such as x squared plus y squared equals a squared, y squared equals 4ax and x squared over a squared plus y squared over b squared equals 1. Simple trigonometric or polynomial curves also appear as definite-integral area questions.

Q: Is a sketch compulsory?

Marking schemes usually give one mark for a correct figure in 3- and 5-mark area questions. Draw a neat rough sketch even if the question does not ask for one.

Q: How important is this chapter?

It is part of Calculus, which carries 35 of the 80 marks. In our question bank it has 21 questions, and nine of them are 5-mark long answers, so it is a reliable source of a full long question.

Q: Should I use vertical or horizontal strips?

Use vertical strips when the region is closed by vertical lines and y is easy to write in terms of x. Use horizontal strips when the region is closed by horizontal lines, such as a parabola x squared equals 4ay cut by y equals k.

Q: Why is my area answer negative?

Part of the curve is below the axis. Area is always positive, so split the integral where the curve crosses the axis and add the absolute values.

Updated 14 October 2026 · Fact-checked

Application of Integrals is Chapter 8 of CBSE Class 12 Maths. It uses definite integrals to find the area of a region bounded by a simple curve, an axis and two ordinates: lines, circles, parabolas and ellipses in standard form, using vertical strips, horizontal strips and symmetry, and taking care with parts below the axis.

What Application of Integrals is about

This short chapter turns the definite integral into an area calculator. Think of the region under a curve y = f(x) as made of very thin vertical strips, each of height y and width dx. Adding up their areas gives

$A = \int_{a}^{b} y d x = \int_{a}^{b} f (x) d x$

If the curve is easier to write as x = g(y), use horizontal strips instead: $A = \int_{c}^{d} x d y$ . Where the curve lies below the x-axis, the integral is negative, so you take its absolute value and add the pieces separately.

The syllabus limits the curves to lines, circles, parabolas and ellipses in standard form ( $x^{2} + y^{2} = a^{2}$ , $y^{2} = 4 a x$ , $x^{2} = 4 a y$ , $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ ). Symmetry does a lot of work: the area of a circle is 4 times its first-quadrant part, which gives $π a^{2}$ , and the ellipse gives $π ab$ . Always sketch the region first; marking schemes give a mark for the figure.

Calculus carries 35 of the 80 marks. In our question bank of eight board papers (2023 to 2026) and four sample papers, this chapter has 21 questions, nine of them 5-mark long answers, and appears in all eight board papers.

Weightage: This chapter is part of Unit III, Calculus, which carries 35 of the 80 theory marks together with Continuity and Differentiability, Application of Derivatives, Integrals and Differential Equations.Source: CBSE Mathematics syllabus 2026-27, Unit III Calculus

In real board papers: asked in 8 of the 8 board papers we hold (2023 to 2026), on average 5.1 marks per paper. See the questions

Key concepts

Area under a curve using vertical strips

NCERT section 8.2

For a curve y = f(x) lying above the x-axis between x = a and x = b, a thin vertical strip at x has height y and width dx, so its area is $d A = y d x$ . Adding all strips:

$A = \int_{a}^{b} y d x = \int_{a}^{b} f (x) d x$

Steps: sketch the curve and shade the region; write y in terms of x (taking the positive root where the region is above the axis); fix the limits from the ordinates; integrate.

Worked example

Find the area of the region bounded by y = x², the x-axis and the lines x = 1 and x = 3. [2 marks]

Try it first, then show the solutionSolution

$A = \int_{1}^{3} x^{2} d x = [\frac{x ^{3}}{3}]_{1}^{3}$
$= 9 - \frac{1}{3} = \frac{26}{3}$

Answer: $\frac{26}{3}$ square units

Area using horizontal strips

NCERT section 8.2

For a curve x = g(y) to the right of the y-axis between y = c and y = d, use horizontal strips of length x and thickness dy:

$A = \int_{c}^{d} x d y = \int_{c}^{d} g (y) d y$

Choose horizontal strips when the region is bounded by the y-axis and lines y = c, y = d, or when x is a simpler function of y (as with $x = y^{2}$ or $x^{2} = 4 a y$ cut by a horizontal line).

Worked example

Find the area bounded by the curve $x = y^{2}$ , the y-axis and the lines y = 1 and y = 2. [2 marks]

Try it first, then show the solutionSolution

$A = \int_{1}^{2} y^{2} d y = [\frac{y ^{3}}{3}]_{1}^{2}$
$= \frac{8}{3} - \frac{1}{3} = \frac{7}{3}$

Answer: $\frac{7}{3}$ square units

Curves below the axis and curves that cross it

NCERT section 8.2

If f(x) < 0 on [a, b], then $\int_{a}^{b} f (x) d x$ is negative, but area is never negative: the area is $\int_{a}^{b} f (x) d x$ .

If the curve crosses the x-axis at x = c between a and b, split the integral there and add the absolute values:

$A = \int_{a}^{c} f (x) d x + \int_{c}^{b} f (x) d x$

Integrating straight across would let the negative part cancel the positive part and give the wrong area.

Worked example

Find the area bounded by $y = x^{2} - 4$ , the x-axis and the lines x = 0 and x = 3. [3 marks]

Try it first, then show the solutionSolution

The curve cuts the x-axis at x = 2: below the axis on (0, 2), above on (2, 3).
$\int_{0}^{2} (x^{2} - 4) d x = \frac{8}{3} - 8 = \frac{16}{3}$ ; $\int_{2}^{3} (x^{2} - 4) d x = (9 - 12) - (\frac{8}{3} - 8) = \frac{7}{3}$
Total $= \frac{16}{3} + \frac{7}{3} = \frac{23}{3}$

Answer: $\frac{23}{3}$ square units

Area under straight lines and modulus graphs

NCERT section 8.2

A line y = mx + c with ordinates x = a, x = b gives a trapezium or triangle, and the integral agrees with the geometric formula, which is a good check.

For a modulus graph such as y = |x − k|, find the corner x = k, write the function without the modulus on each side (k − x on the left, x − k on the right) and split the integral at the corner. Questions often say "using integration", so show the integral even when a triangle formula would do.

Worked example

Using integration, find the area bounded by the line y = 2x + 1, the x-axis and the ordinates x = 1 and x = 3. [2 marks]

Try it first, then show the solutionSolution

The line is above the x-axis on [1, 3]: $A = \int_{1}^{3} (2 x + 1) d x = [x^{2} + x]_{1}^{3}$
= 12 − 2 = 10 (check: trapezium with parallel sides 3 and 7, width 2: ½(3 + 7)(2) = 10)

Answer: 10 square units

Area of a circle and parts of a circle

NCERT section 8.2

For $x^{2} + y^{2} = a^{2}$ , the upper half is $y = a^{2} - x^{2}$ . The circle is symmetric about both axes, so

Area $= 4 \int_{0}^{a} a^{2} - x^{2} d x = 4 [\frac{x}{2} a^{2} - x^{2} + \frac{a ^{2}}{2} sin^{- 1} \frac{x}{a}]_{0}^{a} = 4 \cdot \frac{π a ^{2}}{4} = π a^{2}$

For part of a circle (between two ordinates, or a segment cut off by x = k), use the same antiderivative with the new limits, and double it if the region is above and below the axis.

Worked example

Find the area of the region in the first quadrant enclosed by the circle $x^{2} + y^{2} = 16$ , the y-axis and the line x = 2. [3 marks]

Try it first, then show the solutionSolution

$A = \int_{0}^{2} 16 - x^{2} d x = [\frac{x}{2} 16 - x^{2} + 8 sin^{- 1} \frac{x}{4}]_{0}^{2}$
$= \frac{2}{2} 12 + 8 \cdot \frac{π}{6} = 23 + \frac{4 π}{3}$

Answer: $23 + \frac{4 π}{3}$ square units (about 7.65)

Area involving a parabola

NCERT section 8.2

Standard parabolas: $y^{2} = 4 a x$ (opening right, symmetric about the x-axis) and $x^{2} = 4 a y$ (opening up, symmetric about the y-axis).

For $y^{2} = 4 a x$ , the upper branch is $y = 2 a x$ . The region bounded by the parabola and the line x = k (a vertical line) has area $2 \int_{0}^{k} 2 a x d x$ (double for the part below the axis).

The line x = a through the focus is the latus rectum; the region it cuts off has area $\frac{8 a ^{2}}{3}$ . For $x^{2} = 4 a y$ cut by a horizontal line y = k, use horizontal strips: $2 \int_{0}^{k} 2 a y d y$ .

Worked example

Find the area of the region bounded by the parabola $y^{2} = 8 x$ and its latus rectum. [3 marks]

Try it first, then show the solutionSolution

$y^{2} = 8 x = 4 (2) x$ , so a = 2 and the latus rectum is x = 2.
$A = 2 \int_{0}^{2} 8 x d x = 2 \cdot 22 \cdot [\frac{2}{3} x^{3/2}]_{0}^{2} = 42 \cdot \frac{2}{3} \cdot 22$
$= \frac{32}{3}$ (agrees with $\frac{8 a ^{2}}{3}$ for a = 2)

Answer: $\frac{32}{3}$ square units

Area of an ellipse and parts of an ellipse

NCERT section 8.2

For $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ , the upper half is $y = \frac{b}{a} a^{2} - x^{2}$ . By symmetry,

Area $= 4 \int_{0}^{a} \frac{b}{a} a^{2} - x^{2} d x = \frac{4 b}{a} \cdot \frac{π a ^{2}}{4} = π ab$

If the equation is given as $A x^{2} + B y^{2} = C$ , divide by C first to read off a and b. For the part between x = −k and x = k, use $4 \int_{0}^{k} \frac{b}{a} a^{2} - x^{2} d x$ .

Worked example

Using integration, find the area enclosed by the ellipse $\frac{x ^{2}}{25} + \frac{y ^{2}}{9} = 1$ . [3 marks]

Try it first, then show the solutionSolution

$y = \frac{3}{5} 25 - x^{2}$ in the first quadrant; area $= 4 \int_{0}^{5} \frac{3}{5} 25 - x^{2} d x$
$= \frac{12}{5} [\frac{x}{2} 25 - x^{2} + \frac{25}{2} sin^{- 1} \frac{x}{5}]_{0}^{5} = \frac{12}{5} \cdot \frac{25 π}{4}$
$= 15 π$

Answer: $15 π$ square units

Exam note: a standard curve cut by a slanting line

NCERT section 8.2

NCERT removed the separate section on the region between a curve and a line, and the syllabus wording is "area under simple curves". However, CBSE's own 2026-27 sample paper still asks for such regions (Q34 offers a circle cut by a line, or the parabola $y^{2} = 2 x$ cut by a line), and the 2023 and 2025 board papers in our question bank also included curve-and-line regions, so it is worth knowing the method for these simple cases:

1. Sketch both and find the points where they meet (solve simultaneously).
2. Between those x values, area $= \int_{a}^{b} (y_{upper} - y_{lower}) d x$ .

Keep to a standard circle, parabola or ellipse and a straight line. Regions between two curves (such as two circles) are not asked.

Worked example

Find the area of the region bounded by the parabola $y^{2} = 4 x$ and the line y = x. [3 marks]

Try it first, then show the solutionSolution

Meeting points: $x^{2} = 4 x \Rightarrow x = 0, 4$ , i.e. (0, 0) and (4, 4). On (0, 4) the parabola $y = 2 x$ is above the line.
$A = \int_{0}^{4} (2 x - x) d x = [\frac{4}{3} x^{3/2} - \frac{x ^{2}}{2}]_{0}^{4}$
$= \frac{32}{3} - 8 = \frac{8}{3}$

Answer: $\frac{8}{3}$ square units

Maths in real life: Application of Integrals

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Area of an ellipse

Suppose a school builds an elliptical sports ground whose boundary, in metres with the centre as origin, is x²/2500 + y²/900 = 1, and needs to order turf for it.

Suppose the ground is the ellipse with a = 50 m and b = 30 m. How much turf is needed?

Try it first, then show the solutionSolution

Area $= π ab = π (50) (30) = 1500 π$
$\approx 4712$

Answer: 1500π ≈ 4712 m² of turf.

Memory hook: Ellipse area: π times the two half-axes.

Area under a parabola (horizontal strips)

Imagine a glass window shaped like the region between the parabola x² = 4y and the line y = 1, with lengths in metres.

Suppose the window is the region bounded by x² = 4y and y = 1. Find its area.

Try it first, then show the solutionSolution

Horizontal strips: $A = 2 \int_{0}^{1} 2 y d y$
$= 4 \cdot \frac{2}{3} = \frac{8}{3}$

Answer: 8/3 ≈ 2.67 m² of glass.

Memory hook: A parabolic segment is two-thirds of the rectangle around it (here 4 × 1).

Part of a circle

Suppose a semicircular lawn has boundary x² + y² = 100 above the x-axis (metres) and a gardener wants to reseed only the central strip between x = −5 and x = 5.

Find the area of that central strip.

Try it first, then show the solutionSolution

$A = 2 \int_{0}^{5} 100 - x^{2} d x = 2 [\frac{x}{2} 100 - x^{2} + 50 sin^{- 1} \frac{x}{10}]_{0}^{5}$
$= 2 [\frac{5}{2} 75 + 50 \cdot \frac{π}{6}] = 253 + \frac{50 π}{3}$

Answer: 25√3 + 50π/3 ≈ 95.7 m².

Memory hook: sin⁻¹ of (half the radius over the radius) is π/6.

Topics in this chapter

Application of Integrals formula sheet

Vertical strips: $A = \int_{a}^{b} y d x = \int_{a}^{b} f (x) d x$
Horizontal strips: $A = \int_{c}^{d} x d y = \int_{c}^{d} g (y) d y$
Curve below the axis: $A = \int_{a}^{b} f (x) d x$
Curve crossing the axis at c: $A = \int_{a}^{c} f (x) d x + \int_{c}^{b} f (x) d x$
Key antiderivative: $\int a^{2} - x^{2} d x = \frac{x}{2} a^{2} - x^{2} + \frac{a ^{2}}{2} sin^{- 1} \frac{x}{a} + C$
Circle: $x^{2} + y^{2} = a^{2} : A = 4 \int_{0}^{a} a^{2} - x^{2} d x = π a^{2}$
Ellipse: $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1 : A = 4 \int_{0}^{a} \frac{b}{a} a^{2} - x^{2} d x = π ab$
Parabola and its latus rectum: $y^{2} = 4 a x, x = a : A = 2 \int_{0}^{a} 2 a x d x = \frac{8 a ^{2}}{3}$
Between a curve and a line: $A = \int_{a}^{b} (y_{upper} - y_{lower}) d x$

How the board asks Application of Integrals

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Area under a line or modulus graph	1-mark MCQs and 2-3 mark questions such as "sketch y = \|x + k\| and find the area between x = p and x = q".	Sketch, split at the corner or where the line crosses the axis, integrate each piece.
Area of a circle or ellipse (whole or part)	5-mark "using integration, find the area enclosed by 4x² + y² = 36" or "between the lines x = −2 and x = 2".	Rewrite in standard form, use symmetry (4 × first quadrant), apply the √(a² − x²) antiderivative.
Area involving a parabola	Region bounded by y² = 4ax and x = k, or x² = 4ay and y = k; MCQs on the latus-rectum area.	Use symmetry about the axis of the parabola; choose horizontal strips when the bounding line is horizontal.
Trigonometric or polynomial curves crossing the axis	Area under y = sin x, cos x, x³ or a quadratic over an interval where the sign changes.	Find the zeros, split, take absolute values.
Region cut by a slanting line (sample-paper style)	A 5-mark question on a circle or parabola and a line through the origin or the intercepts.	Find meeting points; integrate upper minus lower; sketch to decide which is upper.

Tricks to remember Application of Integrals

Short memory aids. Say each one aloud once, then use it on a question the same day.

Sketch first, always
No figure, no strip: draw the curve, shade the region, mark the limits before writing any integral.
Use it when: Every area question (the figure itself earns a mark).
Four quarters
Circles and ellipses are symmetric about both axes: find the first-quadrant area and multiply by 4.
Use it when: Whole circle or ellipse.
Area is never negative
Below the axis, take the modulus; across the axis, split and add.
Use it when: sin x, cos x, cubics, lines crossing the axis.
Line through the region? Choose the strip
Vertical boundary lines → vertical strips (dx). Horizontal boundary lines → horizontal strips (dy).
Use it when: Parabolas like x² = 4ay cut by y = k.
Two-thirds rule
A parabolic segment cut by a line perpendicular to its axis is two-thirds of the bounding rectangle.
Use it when: Quick check on parabola answers.

Application of Integrals: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Area of an ellipse · 1 mark · MCQBoard level

The area enclosed by the ellipse $\frac{x ^{2}}{36} + \frac{y ^{2}}{4} = 1$ is:
(a) $6 π$
(b) $12 π$
(c) $24 π$
(d) $36 π$

Try it first, then show the solutionSolution

a = 6, b = 2, so the area is $π ab = 12 π$ .

Answer: (b) $12 π$

Example 2: Area versus integral · 1 mark · Assertion-reasonBoard level

Assertion (A): The area bounded by y = sin x and the x-axis from x = 0 to x = 2π is 4 square units.
Reason (R): $\int_{0}^{2 π} sin x d x = 0$ .
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.

Try it first, then show the solutionSolution

A: $\int_{0}^{π} sin x d x = 2$ and $\int_{π}^{2 π} sin x d x = 2$ , total 4: true. R is also true, but a zero integral does not explain an area of 4.

Answer: (b)

Example 3: Parabola between two ordinates · 2 marks · Very short answerHarder board level

Find the area of the region bounded by $y^{2} = 9 x$ , the x-axis and the lines x = 2 and x = 4, in the first quadrant. [2 marks]

Try it first, then show the solutionSolution

$y = 3 x$ : $A = \int_{2}^{4} 3 x d x = 2 [x^{3/2}]_{2}^{4}$
$= 2 (8 - 22) = 16 - 42$

Answer: $16 - 42$ square units (about 10.34)

Example 4: A modulus graph · 3 marks · Short answerHarder board level

Sketch y = |x − 1| and, using integration, find the area bounded by it, the x-axis and the lines x = −2 and x = 3. [3 marks]

Try it first, then show the solutionSolution

The graph is a V with its corner at (1, 0): y = 1 − x for x ≤ 1, y = x − 1 for x ≥ 1.
$\int_{- 2}^{1} (1 - x) d x = [x - \frac{x ^{2}}{2}]_{- 2}^{1} = \frac{1}{2} - (- 4) = \frac{9}{2}$
$\int_{1}^{3} (x - 1) d x = [\frac{x ^{2}}{2} - x]_{1}^{3} = \frac{3}{2} + \frac{1}{2} = 2$ ; total $= \frac{13}{2}$

Answer: $\frac{13}{2}$ square units

Example 5: A trigonometric curve crossing the axis · 3 marks · Short answerHarder board level

Find the area bounded by y = sin 2x, the x-axis and the lines x = 0 and x = π. [3 marks]

Try it first, then show the solutionSolution

sin 2x ≥ 0 on $[0, \frac{π}{2}]$ and ≤ 0 on $[\frac{π}{2}, π]$ .
$\int_{0}^{π /2} sin 2 x d x = [- \frac{c o s 2 x}{2}]_{0}^{π /2} = 1$ ; $\int_{π /2}^{π} sin 2 x d x = 1$
Area = 2

Answer: 2 square units

Example 6: A strip of an ellipse · 5 marks · Long answerHarder board level

Using integration, find the area of the region enclosed by the ellipse $9 x^{2} + 4 y^{2} = 36$ and lying between the lines x = −1 and x = 1. [5 marks]

Try it first, then show the solutionSolution

Standard form: $\frac{x ^{2}}{4} + \frac{y ^{2}}{9} = 1$ , so a = 2, b = 3 and $y = \frac{3}{2} 4 - x^{2}$ in the upper half.
By symmetry about both axes: $A = 4 \int_{0}^{1} \frac{3}{2} 4 - x^{2} d x = 6 \int_{0}^{1} 4 - x^{2} d x$
$= 6 [\frac{x}{2} 4 - x^{2} + 2 sin^{- 1} \frac{x}{2}]_{0}^{1}$
$= 6 [\frac{3}{2} + 2 \cdot \frac{π}{6}]$
$= 33 + 2 π$

Answer: $33 + 2 π$ square units (about 11.48)

Example 7: Smaller segment of a circle · 5 marks · Long answerHOTS / competency

Find the area of the smaller region cut off from the circle $x^{2} + y^{2} = 9$ by the line $x = \frac{3}{2}$ . [5 marks]

Try it first, then show the solutionSolution

The smaller region lies between $x = \frac{3}{2}$ and x = 3, symmetric about the x-axis: $A = 2 \int_{3/2}^{3} 9 - x^{2} d x$ .
$= 2 [\frac{x}{2} 9 - x^{2} + \frac{9}{2} sin^{- 1} \frac{x}{3}]_{3/2}^{3}$
Upper limit: $0 + \frac{9}{2} \cdot \frac{π}{2} = \frac{9 π}{4}$ . Lower limit: $\frac{3}{4} \cdot \frac{3 3}{2} + \frac{9}{2} \cdot \frac{π}{6} = \frac{9 3}{8} + \frac{3 π}{4}$ .
$A = 2 (\frac{9 π}{4} - \frac{3 π}{4} - \frac{9 3}{8}) = 3 π - \frac{9 3}{4}$

Answer: $3 π - \frac{9 3}{4}$ square units (about 5.53)

Example 8: Cutting off a fraction of a parabolic region · 3 marks · Short answerHOTS / competency

The region bounded by the parabola $y^{2} = 4 x$ and the line x = 4 is cut by the line x = k (0 < k < 4). Find k so that the part to the left of x = k is one-eighth of the whole region. [3 marks]

Try it first, then show the solutionSolution

The region is symmetric about the x-axis, with $y = 2 x$ on top. Area to the left of x = k: $2 \int_{0}^{k} 2 x d x = \frac{8}{3} k^{3/2}$ .
Whole region (k = 4): $\frac{8}{3} \cdot 4^{3/2} = \frac{64}{3}$ . One-eighth of it is $\frac{8}{3}$ .
$\frac{8}{3} k^{3/2} = \frac{8}{3} \Rightarrow k^{3/2} = 1 \Rightarrow k = 1$

Answer: k = 1

Case-study practice (Section E style)

Read the situation, then answer every part. In the board paper this is a 4-mark question split into 1 + 1 + 2 marks.

An elliptical garden

Suppose a park designer lays out an elliptical garden whose boundary, in metres with the centre as origin, is $\frac{x ^{2}}{400} + \frac{y ^{2}}{100} = 1$ . A straight footpath is to be laid along the major axis.

(i) · 1 mark
Write the lengths of the semi-major and semi-minor axes.
Answer
Answer: a = 20 m, b = 10 m
$a^{2} = 400$ , $b^{2} = 100$ .
(ii) · 1 mark
Find the total area of the garden.
Answer
Answer: $200 π \approx 628$ m²
$π ab = π (20) (10)$ .
(iii) · 2 marks
Using integration, find the area of the part of the garden between the lines x = −10 and x = 10.
Answer
Answer: $1003 + \frac{200 π}{3} \approx 382.6$ m²
$y = \frac{1}{2} 400 - x^{2}$ ; area $= 4 \int_{0}^{10} \frac{1}{2} 400 - x^{2} d x = 2 [\frac{x}{2} 400 - x^{2} + 200 sin^{- 1} \frac{x}{20}]_{0}^{10} = 2 (5300 + \frac{200 π}{6})$ .

A dish antenna cross-section

Suppose the cross-section of a dish antenna is the region bounded by the parabola $x^{2} = 8 y$ and the line y = 2 (metres).

(i) · 1 mark
Where does the line y = 2 meet the parabola?
Answer
Answer: At (−4, 2) and (4, 2)
$x^{2} = 16$ .
(ii) · 1 mark
Write the area as an integral using horizontal strips.
Answer
Answer: $A = 2 \int_{0}^{2} 8 y d y$
For each y the strip runs from $x = - 8 y$ to $8 y$ .
(iii) · 2 marks
Evaluate the area and compare it with the 8 m × 2 m rectangle around it.
Answer
Answer: $\frac{32}{3}$ m², two-thirds of the rectangle's 16 m²
$2 \cdot 22 \cdot \frac{2}{3} y^{3/2}_{0}^{2} = 2 \cdot 22 \cdot \frac{2}{3} \cdot 22 = \frac{32}{3}$ .

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Using integration, find the area bounded by y = 3x², the x-axis and the lines x = 1 and x = 2. [2 marks] (2 marks)

Step	Marks
Curve above the x-axis on [1, 2]: $A = \int_{1}^{2} 3 x^{2} d x$	1
$= [x^{3}]_{1}^{2} = 8 - 1 = 7$ square units	1

Find the area bounded by the curve y = cos x, the x-axis and the lines x = 0 and $x = \frac{3 π}{2}$ . [3 marks] (3 marks)

Step	Marks
Sketch: cos x ≥ 0 on $[0, \frac{π}{2}]$ and ≤ 0 on $[\frac{π}{2}, \frac{3 π}{2}]$ .	1
$\int_{0}^{π /2} cos x d x = 1$ ; $\int_{π /2}^{3 π /2} cos x d x = ∣ - 1 - 1∣ = 2$	1
Area = 1 + 2 = 3 square units	1

Using integration, show that the area enclosed by the ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ is $π ab$ . [5 marks] (5 marks)

Step	Marks
Sketch the ellipse with vertices (±a, 0), (0, ±b); it is symmetric about both axes.	1
In the first quadrant, $y = \frac{b}{a} a^{2} - x^{2}$ ; area $= 4 \int_{0}^{a} \frac{b}{a} a^{2} - x^{2} d x$	1
$= \frac{4 b}{a} [\frac{x}{2} a^{2} - x^{2} + \frac{a ^{2}}{2} sin^{- 1} \frac{x}{a}]_{0}^{a}$	1
$= \frac{4 b}{a} [(0 + \frac{a ^{2}}{2} \cdot \frac{π}{2}) - 0]$	1
$= \frac{4 b}{a} \cdot \frac{π a ^{2}}{4} = π ab$ square units	1

Where students lose marks in Application of Integrals

The slips examiners see most often, and what to write instead.

Mistake: No sketch of the region
Going straight to the integral.
Do this: Draw a neat rough sketch with the curve, the lines and the shaded region.
Usually costs: 1 mark in most 3- and 5-mark questions.
Mistake: Integrating across a sign change
Letting the part below the axis cancel the part above.
Wrong working
Area under y = sin x from 0 to 2π $= \int_{0}^{2 π} sin x d x = 0$
Right working
$\int_{0}^{π} sin x d x + \int_{π}^{2 π} sin x d x = 2 + 2 = 4$
Do this: Find the zeros, split and add absolute values.
Usually costs: 2-3 marks.
Mistake: Forgetting the factor from symmetry
Computing only the first-quadrant integral for a whole circle or ellipse.
Do this: Write "by symmetry, area = 4 × (area in the first quadrant)" explicitly.
Usually costs: 1-2 marks.
Mistake: Wrong limits for an ellipse
Using 0 to b with $y = \frac{b}{a} a^{2} - x^{2}$ , mixing the two axes.
Do this: With vertical strips (dx), the x-limits go from 0 to a, the semi-axis along x.
Usually costs: Most of the marks.
Mistake: Leaving out units
Writing just "8/3".
Do this: Write "square units" (or m² in a context question).
Usually costs: ½ mark.

NCERT exercise guide: Application of Integrals

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Exercise 8.1 (area under simple curves)

What it tests: Areas of ellipses, parts of circles in the first quadrant, and a parabola with the y-axis and a horizontal line.

How to approach it: Sketch, use symmetry, apply the √(a² − x²) antiderivative or horizontal strips.

Same kind as Exercise 8.1

Find the area of the region bounded by the ellipse $\frac{x ^{2}}{49} + \frac{y ^{2}}{16} = 1$ .

Show the solutionSolution

Area $= 4 \int_{0}^{7} \frac{4}{7} 49 - x^{2} d x = \frac{16}{7} [\frac{x}{2} 49 - x^{2} + \frac{49}{2} sin^{- 1} \frac{x}{7}]_{0}^{7}$
$= \frac{16}{7} \cdot \frac{49 π}{4} = 28 π$

Answer: $28 π$ square units

Miscellaneous Exercise on Chapter 8

What it tests: Areas under y = x², x⁴, a line with a modulus, sin x over a full period, x³ across the origin and x|x|.

How to approach it: Split wherever the curve crosses the x-axis; add absolute values.

Same kind as the Miscellaneous Exercise

Find the area bounded by $y = x^{3}$ , the x-axis and the ordinates x = −1 and x = 3.

Show the solutionSolution

$\int_{- 1}^{0} x^{3} d x = \frac{1}{4}$ ; $\int_{0}^{3} x^{3} d x = \frac{81}{4}$
Area $= \frac{1}{4} + \frac{81}{4} = \frac{41}{2}$

Answer: $\frac{41}{2}$ square units

Application of Integrals: previous year questions

Asked in 8 of the 8 board papers we hold (2023 to 2026), on average 5.1 marks per paper; 13 questions in all, counting OR alternatives. In 4 of 4 sample papers.

Counted from the real papers on our previous year papers page, not estimated. Average marks are over every board paper we hold, with an OR pair counted once. Chapter tags are ours.

Solved board questions from Application of Integrals

Real questions from recent papers. Try each one first, then open our step-by-step solution with marks per step.

2026, set 65/1/1, Q10 · 1 mark · MCQ

The area bounded by the curve $y = x ∣ x ∣$ , $x$ -axis and the ordinates $x = - 1$ and $x = 1$ is given by

(A)0
(B) $\frac{1}{3}$
(C) $\frac{2}{3}$
(D) $\frac{4}{3}$

Try it first, then show our solutionSolution

$y = x ∣ x ∣$ is $- x^{2}$ for $x < 0$ , $x^{2}$ for $x \geq 0$ ; the region is symmetric.
Area $= 2 \int_{0}^{1} x^{2} d x = 2 \cdot \frac{1}{3} = \frac{2}{3}$ . Option (C).
1 mark

Answer: $\frac{2}{3}$ , option (C)

Where the marks are

Takes absolute area on both sides, $2 \int_{0}^{1} x^{2} d x = \frac{2}{3}$ (1 mark)

See it in the full 2026 board paper, set 65/1/1

2026, set 65/1/1, Q38 · 4 marks · Case-basedCompetency-based

Roundabouts are often made on busy roads to ease the traffic and avoid red lights.

One such round-about is made such that equation representing its boundary is given by $C_{1} : x^{2} + y^{2} = 64$ .

There is a circular pond with a fountain in the middle of the roundabout whose equation is given by $C_{2} : x^{2} + y^{2} = 4$ .

Based on the given information, answer the following questions :

(i)
Represent the given equations $C_{1}$ and $C_{2}$ with the help of a diagram.
1 mark
(ii)
Express y as a function of $x$ , ( $y = f (x)$ ), for both $C_{1}$ and $C_{2}$ .
1 mark
(iii)(a)
Using integration find the area of region covered by the roundabout.
2 marks
OR
(iii)(b)
Using integration, find the area of region covered by circular pond.
2 marks

Figure in the paper: Aerial picture of a roundabout with a circular pond and fountain at its centre, surrounded by trees and roads. See it in the official paper.

Try it first, then show our solutionSolution

(i) Diagram: draw the $x$ - and $y$ -axes with origin O.
- Draw $C_{1}$ , a circle centred at O with radius 8, passing through $(\pm 8, 0)$ and $(0, \pm 8)$ .
- Draw $C_{2}$ , a smaller concentric circle with radius 2, passing through $(\pm 2, 0)$ and $(0, \pm 2)$ .
- Label both circles.
1 mark
(ii) For $C_{1}$ : $y = 64 - x^{2}$ (upper half) or $y = - 64 - x^{2}$ (lower half), with $- 8 \leq x \leq 8$ .
½ mark
(ii) For $C_{2}$ : $y = 4 - x^{2}$ or $y = - 4 - x^{2}$ , with $- 2 \leq x \leq 2$ .
½ mark
(iii)(a) The circle is symmetric about both axes, so the area is 4 times the area in the first quadrant:
Area $= 4 \int_{0}^{8} 64 - x^{2} d x$ .
1 mark
(iii)(a) Use $\int a^{2} - x^{2} d x = \frac{x}{2} a^{2} - x^{2} + \frac{a ^{2}}{2} sin^{- 1} \frac{x}{a}$ :
$= 4 [\frac{x}{2} 64 - x^{2} + 32 sin^{- 1} \frac{x}{8}]_{0}^{8} = 4 [0 + 32 \cdot \frac{π}{2} - 0] = 64 π$ sq. units.
1 mark
(iii)(b) By symmetry, Area of pond $= 4 \int_{0}^{2} 4 - x^{2} d x$ .
1 mark
(iii)(b) $= 4 [\frac{x}{2} 4 - x^{2} + 2 sin^{- 1} \frac{x}{2}]_{0}^{2} = 4 [0 + 2 \cdot \frac{π}{2} - 0] = 4 π$ sq. units.
1 mark

Answer: (ii) $C_{1}$ : $y = \pm 64 - x^{2}$ ; $C_{2}$ : $y = \pm 4 - x^{2}$ ; (iii)(a) $64 π$ sq. units; (iii)(b) $4 π$ sq. units

Where the marks are

(i) Draws two concentric circles centred at O with radii 8 and 2, correctly labelled (1 mark)
(ii) Writes $y = \pm 64 - x^{2}$ for $C_{1}$ (½ mark)
(ii) Writes $y = \pm 4 - x^{2}$ for $C_{2}$ (½ mark)
(iii)(a) Sets up Area $= 4 \int_{0}^{8} 64 - x^{2} d x$ (1 mark)
(iii)(a) Evaluates to $64 π$ sq. units (1 mark)
(iii)(b) Sets up Area $= 4 \int_{0}^{2} 4 - x^{2} d x$ (1 mark)
(iii)(b) Evaluates to $4 π$ sq. units (1 mark)

See it in the full 2026 board paper, set 65/1/1

2026, set 65/5/1, Q34 · 5 marks · Long answer

Sketch the curve ${(x, y) : 100 x^{2} + 25 y^{2} = 2500}$ and find the area of the region enclosed by it, using integration.

Try it first, then show our solutionSolution

Divide by 2500: $\frac{x ^{2}}{25} + \frac{y ^{2}}{100} = 1$ . This is an ellipse centred at the origin with major axis along the y-axis. It meets the x-axis at $(\pm 5, 0)$ and the y-axis at $(0, \pm 10)$ .
Sketch: draw a vertical ellipse through $(5, 0), (- 5, 0), (0, 10), (0, - 10)$ and shade the region inside it.
1½ marks
In the first quadrant $y = \frac{10}{5} 25 - x^{2} = 2 25 - x^{2}$ . The ellipse is symmetric about both axes, so
Required area $= 4 \int_{0}^{5} 2 25 - x^{2} d x = 8 \int_{0}^{5} 25 - x^{2} d x$ .
1½ marks
Using $\int a^{2} - x^{2} d x = \frac{x}{2} a^{2} - x^{2} + \frac{a ^{2}}{2} sin^{- 1} \frac{x}{a}$ :
$= 8 [\frac{x}{2} 25 - x^{2} + \frac{25}{2} sin^{- 1} \frac{x}{5}]_{0}^{5}$
1 mark
$= 8 [(0 + \frac{25}{2} \cdot \frac{π}{2}) - (0 + 0)] = 8 \cdot \frac{25 π}{4} = 50 π$ sq. units.
1 mark

Answer: Area $= 50 π$ sq. units

Where the marks are

Writes the ellipse in standard form $\frac{x ^{2}}{25} + \frac{y ^{2}}{100} = 1$ and draws a correct sketch (1½ marks)
Sets up the area as $4 \int_{0}^{5} 2 25 - x^{2} d x$ (1½ marks)
Integrates correctly using the standard formula (1 mark)
Obtains the area $50 π$ sq. units (1 mark)

See it in the full 2026 board paper, set 65/5/1

Every Application of Integrals question, by year

For 2027

Sample paper 2026-27: Q34(A) (5 marks, solved), Q34(B) (5 marks, solved)

2026

Set 65/1/1: Q10 (1 mark, solved), Q38 (4 marks, solved)

Set 65/5/1: Q12 (1 mark, solved), Q34 (5 marks, solved)

Sample paper 2025-26: Q23(B) (2 marks, solved), Q28(A) (3 marks, solved), Q28(B) (3 marks, solved)

2025

Set 65/1/1: Q17 (1 mark, solved), Q31 (3 marks, solved)

Set 65/7/1: Q33 (5 marks, solved)

Sample paper 2024-25: Q18 (1 mark, solved), Q32 (5 marks, solved)

2024

Set 65/1/1: Q34 (5 marks, solved)

Set 65/5/1: Q13 (1 mark, solved), Q33 (5 marks, solved)

Sample paper 2023-24: Q32 (5 marks, solved)

2023

Set 65/1/1: Q35 (5 marks, solved)

Set 65/5/1: Q23 (2 marks, solved), Q29 (3 marks, solved)

Practice questions on Application of Integrals

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The area bounded by y = 2x, the x-axis and the lines x = 0 and x = 3 is:
(a) 6
(b) 9
(c) 12
(d) 18
Hint
A triangle; check by integration.
Answer
(b) 9, since $\int_{0}^{3} 2 x d x = 9$
Q2 · 1 mark · MCQBoard level
The area of the region in the first quadrant enclosed by the circle $x^{2} + y^{2} = 25$ and the axes is:
(a) $25 π$
(b) $\frac{25 π}{2}$
(c) $\frac{25 π}{4}$
(d) $5 π$
Hint
A quarter of the circle.
Answer
(c) $\frac{25 π}{4}$
Q3 · 1 mark · MCQBoard level
The area bounded by y = cos x, the x-axis and the lines x = 0 and $x = \frac{π}{2}$ is:
(a) 0
(b) 1
(c) 2
(d) $\frac{π}{2}$
Hint
$[sin x]_{0}^{π /2}$
Answer
(b) 1
Q4 · 1 mark · MCQBoard level
The area bounded by the curve $x = y^{2}$ , the y-axis and the lines y = 0 and y = 3 is:
(a) 3
(b) 6
(c) 9
(d) 27
Hint
Horizontal strips.
Answer
(c) 9, since $\int_{0}^{3} y^{2} d y = 9$
Q5 · 2 marks · Very short answerHarder board level
Using integration, find the area bounded by $y = x^{2} + 2$ , the x-axis and the lines x = −1 and x = 2.
Hint
The curve is always above the x-axis.
Answer
$[\frac{x ^{3}}{3} + 2 x]_{- 1}^{2} = (\frac{8}{3} + 4) - (- \frac{1}{3} - 2) = 9$ square units.
Q6 · 3 marks · Short answerHarder board level
Using integration, find the area enclosed by the ellipse $4 x^{2} + 9 y^{2} = 36$ .
Hint
Standard form $\frac{x ^{2}}{9} + \frac{y ^{2}}{4} = 1$ .
Answer
$4 \int_{0}^{3} \frac{2}{3} 9 - x^{2} d x = \frac{8}{3} \cdot \frac{9 π}{4} = 6 π$ square units.
Q7 · 3 marks · Short answerHarder board level
Find the area bounded by the line y = 2x − 3, the x-axis and the lines x = 0 and x = 3.
Hint
The line crosses the axis at $x = \frac{3}{2}$ .
Answer
$\int_{0}^{3/2} (2 x - 3) d x + \int_{3/2}^{3} (2 x - 3) d x = \frac{9}{4} + \frac{9}{4} = \frac{9}{2}$ square units.
Q8 · 3 marks · Short answerHarder board level
Find the area of the region bounded by $y^{2} = 4 x$ , the lines x = 1 and x = 4 and the x-axis, in the first quadrant.
Hint
$y = 2 x$ .
Answer
$\int_{1}^{4} 2 x d x = \frac{4}{3} [x^{3/2}]_{1}^{4} = \frac{4}{3} (8 - 1) = \frac{28}{3}$ square units.
Q9 · 5 marks · Long answerHarder board level
Sketch the curve $25 x^{2} + 16 y^{2} = 400$ and, using integration, find the area enclosed by it.
Hint
$\frac{x ^{2}}{16} + \frac{y ^{2}}{25} = 1$ : the major axis is along the y-axis.
Answer
a = 4 (along x), b = 5 (along y). Area $= 4 \int_{0}^{4} \frac{5}{4} 16 - x^{2} d x = 5 [\frac{x}{2} 16 - x^{2} + 8 sin^{- 1} \frac{x}{4}]_{0}^{4} = 5 \cdot 4 π = 20 π$ square units.
Q10 · 3 marks · Short answerHarder board level
Find the area of the region bounded by the parabola $y = x^{2}$ and the line y = x.
Hint
They meet at x = 0 and x = 1; the line is on top between them.
Answer
$\int_{0}^{1} (x - x^{2}) d x = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}$ square units.
Q11 · 5 marks · Long answerHOTS / competency
Find the area of the region in the first quadrant enclosed by the x-axis, the line $y = 3 x$ and the circle $x^{2} + y^{2} = 16$ .
Hint
The line meets the circle at $(2, 23)$ ; split the region at x = 2.
Answer
$\int_{0}^{2} 3 x d x + \int_{2}^{4} 16 - x^{2} d x = 23 + [\frac{x}{2} 16 - x^{2} + 8 sin^{- 1} \frac{x}{4}]_{2}^{4} = 23 + (4 π - 23 - \frac{4 π}{3}) = \frac{8 π}{3}$ square units (a sector of angle $\frac{π}{3}$ of a circle of radius 4, as a check).
Q12 · 2 marks · Very short answerHOTS / competency
Explain why $\int_{- 2}^{2} x^{3} d x = 0$ , and find the area bounded by $y = x^{3}$ , the x-axis and the lines x = −2 and x = 2.
Hint
Odd function; area counts both parts as positive.
Answer
$x^{3}$ is odd, so the negative part on [−2, 0] cancels the positive part. Area $= 2 \int_{0}^{2} x^{3} d x = 2 \cdot 4 = 8$ square units.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Area under y = f(x) from x = a to x = b?Show answer
$\int_{a}^{b} f (x) d x$ (with modulus if f is negative).
2.Area between x = g(y), the y-axis and y = c to y = d?Show answer
$\int_{c}^{d} g (y) d y$
3.Area of the circle x² + y² = a²?Show answer
$π a^{2}$
4.Area of the ellipse x²/a² + y²/b² = 1?Show answer
$π ab$
5.Area cut off by the latus rectum of y² = 4ax?Show answer
$\frac{8 a ^{2}}{3}$
6.Area bounded by y = sin x and the x-axis from 0 to π?Show answer
2 square units.
7.Antiderivative of √(a² − x²)?Show answer
$\frac{x}{2} a^{2} - x^{2} + \frac{a ^{2}}{2} sin^{- 1} \frac{x}{a}$
8.What do you do when the curve is below the x-axis?Show answer
Take the absolute value of the integral.

Best order to study it

Revise definite integrals and the antiderivative of √(a² − x²) from Chapter 7.
Learn vertical strips with lines and simple polynomials; then curves below or crossing the axis.
Practise horizontal strips with x = g(y) and parabolas of the form x² = 4ay.
Derive the areas of the circle and ellipse using symmetry.
Practise parts of circles and ellipses between given ordinates, and parabola regions.
Finish with modulus graphs and sample-paper style regions cut by a line.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you sketch a standard circle, parabola or ellipse quickly from its equation?
Can you find the area under a line using integration and check it with geometry?
Can you split an integral where the curve crosses the x-axis?
Can you choose between vertical and horizontal strips?
Can you derive the area πa² of a circle by integration?
Can you derive the area πab of an ellipse, and read a and b from Ax² + By² = C?
Can you find the area of a circle or ellipse between two ordinates?
Can you find the area bounded by a parabola and its latus rectum?
Can you handle y = |x − k| by splitting at the corner?
Can you find the meeting points of a standard curve and a line and integrate upper minus lower?

Application of Integrals: frequently asked questions

Is the area between two curves in the 2026-27 syllabus?

No. NCERT removed the section on area between two curves, and the syllabus limits the chapter to area under simple curves. The 2026-27 sample paper does include regions bounded by a circle or a parabola and a straight line, so practise those simple cases as well.

Which curves can be asked?

Lines, circles, parabolas and ellipses in standard form only, such as x squared plus y squared equals a squared, y squared equals 4ax and x squared over a squared plus y squared over b squared equals 1. Simple trigonometric or polynomial curves also appear as definite-integral area questions.

Is a sketch compulsory?

Marking schemes usually give one mark for a correct figure in 3- and 5-mark area questions. Draw a neat rough sketch even if the question does not ask for one.

How important is this chapter?

It is part of Calculus, which carries 35 of the 80 marks. In our question bank it has 21 questions, and nine of them are 5-mark long answers, so it is a reliable source of a full long question.

Should I use vertical or horizontal strips?

Use vertical strips when the region is closed by vertical lines and y is easy to write in terms of x. Use horizontal strips when the region is closed by horizontal lines, such as a parabola x squared equals 4ay cut by y equals k.

Why is my area answer negative?

Part of the curve is below the axis. Area is always positive, so split the integral where the curve crosses the axis and add the absolute values.

Based on: CBSE Mathematics (041) syllabus 2026-27; NCERT Mathematics Part II, Textbook for Class XII, Chapter 8 (Reprint 2026-27); CBSE Mathematics sample question paper 2026-27