CBSE Class 12 · Chapter 8 · Calculus · 2026-27

Application of Integrals Class 12: MCQ and case study questions

15 multiple-choice questions and 2 case-based questions on the current (rationalised) syllabus, in the style of the board paper's Sections A and E. Try each one first; the answer, a one-line reason and a worked solution open on a tap.

15 MCQs (1 mark each)
2 case studies (4 marks each)
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Multiple-choice questions

Tap an option to check it; the answer and a worked solution open. Section A of the board paper has 18 MCQs of 1 mark each, spread over all chapters.

Q1

·1 mark·Multiple choiceArea under simple curves

The area of the region bounded by the curve $y = 3 x^{2}$ , the $x$ -axis and the ordinates $x = 0$ and $x = 2$ is

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Answer: (a)

8

sq units

Why: ∫₀² 3x² dx = [x³]₀².

$\int_{0}^{2} 3 x^{2} d x = {[x^{3}]}_{0}^{2} = 8$ sq units.

Q2

·1 mark·Multiple choiceArea under simple curves

Find the area of the triangular region under the line $y = 2 x$ , above the $x$ -axis, from $x = 0$ to $x = 5$ .

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Answer: (b)

25

sq units

Why: ∫₀⁵ 2x dx = 25 (a triangle with base 5 and height 10).

$\int_{0}^{5} 2 x d x = {[x^{2}]}_{0}^{5} = 25$ sq units (check: $\frac{1}{2} \times 5 \times 10$ ).

Q3

·1 mark·Multiple choiceAreas of circles and ellipses

The area enclosed by the circle $x^{2} + y^{2} = 49$ is

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Answer: (c)

49 π

sq units

Why: 4∫₀⁷ √(49 − x²) dx = 4 × 49π/4.

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

Q4

·1 mark·Multiple choiceAreas of circles and ellipses

The ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{4} = 1$ ( $a > 0$ ) encloses an area of $10 π$ square units. Then $a$ is

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Answer: (d)

5

Why: Area = 4 × (2/a)∫₀ᵃ √(a² − x²) dx = 2πa.

$y = \frac{2}{a} \sqrt{a^{2} - x^{2}}$ , so area $= 4 \cdot \frac{2}{a} \int_{0}^{a} \sqrt{a^{2} - x^{2}} d x = \frac{8}{a} \cdot \frac{π a^{2}}{4} = 2 π a = 10 π \Rightarrow a = 5$ .

Q5

·1 mark·Multiple choiceArea under simple curves

The area of the region bounded by the parabola $y^{2} = 9 x$ and the line $x = 4$ is

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Answer: (a)

32

sq units

Why: The region is symmetric about the x-axis: 2∫₀⁴ 3√x dx.

Area $= 2 \int_{0}^{4} 3 \sqrt{x} d x = 6 {[\frac{2}{3} x^{3 / 2}]}_{0}^{4} = 6 \times \frac{16}{3} = 32$ sq units.

Q6

·1 mark·Multiple choiceArea under simple curves

The area of the region bounded by $y = \sqrt{x}$ , the $x$ -axis and the line $x = 4$ is

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Answer: (b)

\frac{16}{3}

sq units

Why: ∫₀⁴ x^1/2 dx = (2/3) × 8.

$\int_{0}^{4} x^{1 / 2} d x = {[\frac{2}{3} x^{3 / 2}]}_{0}^{4} = \frac{2}{3} \times 8 = \frac{16}{3}$ .

Q7

·1 mark·Multiple choiceAreas of circles and ellipses

The area of the region in the first quadrant bounded by the circle $x^{2} + y^{2} = 4$ , the $x$ -axis and the lines $x = 0$ and $x = 1$ is

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Answer: (a)

\frac{\sqrt{3}}{2} + \frac{π}{3}

Why: ∫₀¹ √(4 − x²) dx = [x/2 √(4 − x²) + 2 sin⁻¹(x/2)]₀¹.

${[\frac{x}{2} \sqrt{4 - x^{2}} + 2 \sin^{- 1} \frac{x}{2}]}_{0}^{1} = \frac{1}{2} \sqrt{3} + 2 \cdot \frac{π}{6} = \frac{\sqrt{3}}{2} + \frac{π}{3}$ sq units.

Q8

·1 mark·Multiple choiceArea under simple curves

What is the total area between the curve $y = x^{3}$ and the $x$ -axis from $x = - 1$ to $x = 1$ ?

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Answer: (b)

\frac{1}{2}

sq units

Why: The part below the axis counts positively: 2 × ∫₀¹ x³ dx.

$\int_{- 1}^{1} x^{3} d x = 0$ , but area is never negative: area $= | \int_{- 1}^{0} x^{3} d x | + \int_{0}^{1} x^{3} d x = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$ .

Q9

·1 mark·Multiple choiceArea under simple curves

The area of the region bounded by $y = \cos x$ , the $x$ -axis and the lines $x = 0$ and $x = \frac{π}{2}$ is

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Answer: (c)

1

sq unit

Why: ∫₀^π/2 cos x dx = [sin x].

$\int_{0}^{π / 2} \cos x d x = {[\sin x]}_{0}^{π / 2} = 1$ sq unit.

Q10

·1 mark·Multiple choiceArea under simple curves

Find the area under the curve $y = e^{x}$ , above the $x$ -axis, from $x = 0$ to $x = 1$ .

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Answer: (d)

(e - 1)

sq units

Why: ∫₀¹ eˣ dx = e − 1.

$\int_{0}^{1} e^{x} d x = {[e^{x}]}_{0}^{1} = e - 1$ sq units.

Q11

·1 mark·Multiple choiceArea under simple curves

The area of the region between the curve $y = 4 - x^{2}$ and the $x$ -axis is

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Answer: (a)

\frac{32}{3}

sq units

Why: The curve meets the axis at x = ±2: ∫₋₂² (4 − x²) dx.

$4 - x^{2} = 0 \Rightarrow x = \pm 2$ . Area $= \int_{- 2}^{2} (4 - x^{2}) d x = {[4 x - \frac{x^{3}}{3}]}_{- 2}^{2} = \frac{16}{3} + \frac{16}{3} = \frac{32}{3}$ .

Q12

·1 mark·Multiple choiceArea under simple curves

The area of the region bounded by $y = \frac{1}{x}$ , the $x$ -axis and the lines $x = 1$ and $x = e^{2}$ is

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Answer: (b)

2

sq units

Why: ∫₁^e² dx/x = log e² = 2.

$\int_{1}^{e^{2}} \frac{d x}{x} = {[\log_{e} x]}_{1}^{e^{2}} = 2 - 0 = 2$ sq units.

Q13

·1 mark·Multiple choiceArea under simple curves

The area of the region bounded by the curve $x = y^{2}$ , the $y$ -axis and the lines $y = 0$ and $y = 3$ is

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Answer: (c)

9

sq units

Why: Integrate along y: ∫₀³ y² dy.

Horizontal strips: area $= \int_{0}^{3} x d y = \int_{0}^{3} y^{2} d y = {[\frac{y^{3}}{3}]}_{0}^{3} = 9$ sq units.

Q14

·1 mark·Multiple choiceAreas of circles and ellipses

The area of the part of the ellipse $\frac{x^{2}}{4} + \frac{y^{2}}{25} = 1$ that lies in the first quadrant is

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Answer: (d)

\frac{5 π}{2}

sq units

Why: (5/2)∫₀² √(4 − x²) dx = (5/2)(π).

$y = \frac{5}{2} \sqrt{4 - x^{2}}$ . $\frac{5}{2} \int_{0}^{2} \sqrt{4 - x^{2}} d x = \frac{5}{2} \times π = \frac{5 π}{2}$ (a quarter of $π a b = 10 π$ ).

Q15

·1 mark·Multiple choiceArea under simple curves

The line $x = a$ divides the region bounded by $y = x^{2}$ , the $x$ -axis and $x = 2$ into two parts of equal area. Then $a$ is

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Answer: (a)

\sqrt[3]{4}

Why: ∫₀ᵃ x² dx = half of 8/3, so a³/3 = 4/3.

Total area $= \frac{8}{3}$ . Need $\frac{a^{3}}{3} = \frac{4}{3} \Rightarrow a^{3} = 4 \Rightarrow a = \sqrt[3]{4}$ .

Case study questions

Section E of the board paper has three case-based questions of 4 marks: a real-life passage, then parts of 1, 1 and 2 marks, with a choice (OR) on the 2-mark part. All Class 12 case study questions by chapter.

Case study 1: Elliptical pond (4 marks)

A park has an ornamental pond whose edge, in a coordinate system with the centre of the pond at the origin and units in metres, is the ellipse $\frac{x^{2}}{25} + \frac{y^{2}}{9} = 1$ . The park engineer uses integration to find areas for tiling and lighting.

(i) Write an integral for the area of the part of the pond in the first quadrant. [1 mark]

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Answer:

\frac{3}{5} \int_{0}^{5} \sqrt{25 - x^{2}} d x

y = \frac{3}{5} \sqrt{25 - x^{2}}

, so the area is

\frac{3}{5} \int_{0}^{5} \sqrt{25 - x^{2}} d x

. A1

(ii) Find the total area of the pond. [1 mark]

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Answer:

15 π m^{2}

4 \times \frac{3}{5} \times \frac{25 π}{4} = 15 π m^{2}

. A1

(iii) Find the area of the part of the pond with

0 \leq x \leq \frac{5}{2}

and

y \geq 0

. [2 marks]

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Answer:

(\frac{15 \sqrt{3}}{8} + \frac{5 π}{4}) m^{2}

\frac{3}{5} {[\frac{x}{2} \sqrt{25 - x^{2}} + \frac{25}{2} \sin^{- 1} \frac{x}{5}]}_{0}^{5 / 2}

= \frac{3}{5} (\frac{25 \sqrt{3}}{8} + \frac{25 π}{12}) = \frac{15 \sqrt{3}}{8} + \frac{5 π}{4} m^{2}

. A1

OR Lights are placed around the part of the pond with

x \geq \frac{5}{2}

. Find the area of this part. [2 marks]

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Answer:

(5 π - \frac{15 \sqrt{3}}{4}) m^{2}

Area

= 2 \cdot \frac{3}{5} \int_{5 / 2}^{5} \sqrt{25 - x^{2}} d x = \frac{6}{5} {[\frac{x}{2} \sqrt{25 - x^{2}} + \frac{25}{2} \sin^{- 1} \frac{x}{5}]}_{5 / 2}^{5}

= \frac{6}{5} (\frac{25 π}{4} - \frac{25 \sqrt{3}}{8} - \frac{25 π}{12}) = 5 π - \frac{15 \sqrt{3}}{4} m^{2}

. A1

Case study 2: Curved flower bed (4 marks)

A landscape designer plans a flower bed bounded by a straight path along the $x$ -axis, a straight hedge along the line $x = 9$ , and a curved edging along $y = \sqrt{x}$ (all lengths in metres, starting from the corner at the origin).

(i) How wide is the bed along the hedge (the value of

y

x = 9

)? [1 mark]

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Answer:

3

y = \sqrt{9} = 3

m. A1

(ii) Find the area of the flower bed. [1 mark]

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Answer:

18 m^{2}

\int_{0}^{9} \sqrt{x} d x = {[\frac{2}{3} x^{3 / 2}]}_{0}^{9} = \frac{2}{3} \times 27 = 18 m^{2}

. A1

(iii) Roses are planted in the part of the bed between

x = 1

and

x = 4

. Find the area planted with roses. [2 marks]

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Answer:

\frac{14}{3} m^{2}

\int_{1}^{4} \sqrt{x} d x = {[\frac{2}{3} x^{3 / 2}]}_{1}^{4}

= \frac{2}{3} (8 - 1) = \frac{14}{3} m^{2}

. A1

OR Only the part of the bed between

x = 4

and

x = 9

is turfed, at ₹

30

per square metre. Find its area and the cost. [2 marks]

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Answer:

\frac{38}{3} m^{2}

; ₹

380

\int_{4}^{9} \sqrt{x} d x = {[\frac{2}{3} x^{3 / 2}]}_{4}^{9} = \frac{2}{3} (27 - 8) = \frac{38}{3} m^{2}

M1; cost

= \frac{38}{3} \times 30 = ₹ 380

. A1

Next steps for Application of Integrals

This free set is separate from the chapter's question bank. On the Application of Integrals chapter page the revision notes and the first step of the Route to 95 are free for everyone. CBSE Pro adds all 47 questions in the chapter bank (short and long answers, assertion–reason and more case studies) with their full step-marking schemes, the Route to 95 checkpoints with your progress saved, Skill Builders, the full common-mistakes library and AI marking of your written working.

Application of Integrals: important questions What Pro includes

Original questions written by CBSE Math Revision for the CBSE 2026-27 syllabus; not taken from NCERT, NCERT Exemplar or CBSE papers. Every answer was checked twice, the second time by an independent reviewer. CBSE Math Revision is independent and not affiliated with CBSE or NCERT. Spotted a slip? Tell us and it goes in the corrections log.