CBSE Class 12 · Chapter 7 · Calculus · 2026-27

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 choiceIndefinite integrals: standard forms

$\int (3 x^{2} + 2 x) d x$ equals

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Answer: (a)

x^{3} + x^{2} + C

Why: Integrate term by term: xⁿ → xⁿ⁺¹/(n + 1).

$\int 3 x^{2} d x = x^{3}$ , $\int 2 x d x = x^{2}$ : the answer is $x^{3} + x^{2} + C$ .

Q2

·1 mark·Multiple choiceIndefinite integrals: standard forms

$\int e^{3 x} d x$ equals

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Answer: (b)

\frac{e^{3 x}}{3} + C

Why: d/dx (e^3x/3) = e^3x.

$\frac{d}{d x} (\frac{e^{3 x}}{3}) = e^{3 x}$ , so $\int e^{3 x} d x = \frac{e^{3 x}}{3} + C$ .

Q3

·1 mark·Multiple choiceIndefinite integrals: standard forms

$\int \sec^{2} (2 x) d x$ equals

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Answer: (c)

\frac{\tan (2 x)}{2} + C

Why: d/dx tan(2x) = 2 sec²(2x), so divide by 2.

$\frac{d}{d x} \tan (2 x) = 2 \sec^{2} (2 x)$ , hence $\int \sec^{2} (2 x) d x = \frac{1}{2} \tan (2 x) + C$ .

Q4

·1 mark·Multiple choiceIndefinite integrals: standard forms

For $x > 4$ , an antiderivative of $\frac{1}{x^{2} - 16}$ is

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Answer: (d)

\frac{1}{8} \log | \frac{x - 4}{x + 4} |

Why: Standard form ∫dx/(x² − a²) = (1/2a) log|(x − a)/(x + a)| with a = 4.

$\frac{1}{x^{2} - 16} = \frac{1}{8} (\frac{1}{x - 4} - \frac{1}{x + 4})$ , so $\int \frac{d x}{x^{2} - 16} = \frac{1}{8} \log | \frac{x - 4}{x + 4} | + C$ . (Here $\log$ means $\log_{e}$ .)

Q5

·1 mark·Multiple choiceIndefinite integrals: standard forms

$\int \frac{d x}{\sqrt{16 - x^{2}}}$ (for $| x | < 4$ ) equals

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Answer: (a)

\sin^{- 1} \frac{x}{4} + C

Why: Standard form ∫dx/√(a² − x²) = sin⁻¹(x/a), with a = 4.

$\int \frac{d x}{\sqrt{a^{2} - x^{2}}} = \sin^{- 1} \frac{x}{a} + C$ , $a = 4$ .

Q6

·1 mark·Multiple choiceIntegration by parts

Integrating by parts, $\int x e^{x} d x$ is found to be

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Answer: (b)

e^{x} (x - 1) + C

Why: By parts with u = x, dv = e^x dx: xe^x − ∫e^x dx.

$\int x e^{x} d x = x e^{x} - \int e^{x} d x = x e^{x} - e^{x} + C = e^{x} (x - 1) + C$ .

Q7

·1 mark·Multiple choiceIntegration by substitution

$\int \frac{2 x}{x^{2} + 1} d x$ equals (Here $\log$ means $\log_{e}$ .)

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Answer: (c)

\log (x^{2} + 1) + C

Why: The numerator is the derivative of the denominator: substitute t = x² + 1.

Put $t = x^{2} + 1$ , $d t = 2 x d x$ : $\int \frac{d t}{t} = \log | t | + C = \log (x^{2} + 1) + C$ .

Q8

·1 mark·Multiple choiceIntegration by partial fractions

$\int \frac{d x}{x (x + 1)}$ equals (Here $\log$ means $\log_{e}$ .)

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Answer: (b)

\log | \frac{x}{x + 1} | + C

Why: Partial fractions: 1/(x(x + 1)) = 1/x − 1/(x + 1).

$\frac{1}{x (x + 1)} = \frac{1}{x} - \frac{1}{x + 1}$ , so the integral is $\log | x | - \log | x + 1 | + C = \log | \frac{x}{x + 1} | + C$ .

Q9

·1 mark·Multiple choiceDefinite integrals and their properties

$\int_{0}^{1} (2 x + 1) d x$ equals

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Answer: (d)

2

Why: [x² + x] from 0 to 1.

${[x^{2} + x]}_{0}^{1} = (1 + 1) - 0 = 2$ .

Q10

·1 mark·Multiple choiceDefinite integrals and their properties

$\int_{0}^{π / 2} \cos x d x$ equals

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Answer: (b)

1

Why: [sin x] from 0 to π/2.

${[\sin x]}_{0}^{π / 2} = 1 - 0 = 1$ .

Q11

·1 mark·Multiple choiceDefinite integrals and their properties

$\int_{- 2}^{2} x^{3} d x$ equals

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Answer: (a)

0

Why: x³ is odd and the interval is symmetric about 0.

$f (x) = x^{3}$ is odd ( $f (- x) = - f (x)$ ), so $\int_{- a}^{a} f (x) d x = 0$ . Directly: ${[\frac{x^{4}}{4}]}_{- 2}^{2} = 4 - 4 = 0$ .

Q12

·1 mark·Multiple choiceDefinite integrals and their properties

$\int_{0}^{π / 2} \sin^{2} x d x$ equals

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Answer: (c)

\frac{π}{4}

Why: sin² x = (1 − cos 2x)/2.

$\int_{0}^{π / 2} \frac{1 - \cos 2 x}{2} d x = {[\frac{x}{2} - \frac{\sin 2 x}{4}]}_{0}^{π / 2} = \frac{π}{4}$ .

Q13

·1 mark·Multiple choiceDefinite integrals and their properties

$\int_{1}^{e} \frac{1}{x} d x$ equals

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Answer: (b)

1

Why: [log x] from 1 to e = 1 − 0.

${[\log_{e} x]}_{1}^{e} = \log_{e} e - \log_{e} 1 = 1 - 0 = 1$ .

Q14

·1 mark·Multiple choiceDefinite integrals and their properties

$\int_{0}^{3} | x - 1 | d x$ equals

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Answer: (d)

\frac{5}{2}

Why: Split at x = 1: ∫₀¹(1 − x) dx + ∫₁³(x − 1) dx = 1/2 + 2.

$\int_{0}^{1} (1 - x) d x + \int_{1}^{3} (x - 1) d x = \frac{1}{2} + 2 = \frac{5}{2}$ .

Q15

·1 mark·Multiple choiceIntegration by parts

For $x > 0$ , $\int e^{x} (\frac{1}{x} - \frac{1}{x^{2}}) d x$ equals

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Answer: (c)

\frac{e^{x}}{x} + C

Why: Form ∫e^x (f + f′) dx = e^x f + C with f(x) = 1/x.

With $f (x) = \frac{1}{x}$ , $f^{'} (x) = - \frac{1}{x^{2}}$ : $\int e^{x} (f + f^{'}) d x = e^{x} f (x) + C = \frac{e^{x}}{x} + C$ .

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: Cyclist's speed (4 marks)

On a straight training road, a cyclist starts from rest. Her velocity $t$ seconds after starting is $v (t) = 6 t - t^{2}$ m/s for $0 \leq t \leq 6$ .

(i) Find her velocity at

t = 2

s. [1 mark]

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Answer:

8

m/s

v (2) = 12 - 4 = 8

m/s. A1

(ii) Find the distance function

s (t)

with

s (0) = 0

. [1 mark]

Show answer

Answer:

s (t) = 3 t^{2} - \frac{t^{3}}{3}

s (t) = \int (6 t - t^{2}) d t = 3 t^{2} - \frac{t^{3}}{3}

(constant

0

s (0) = 0

). A1

(iii) Find the distance she covers in the first

6

seconds. [2 marks]

Show answer

Answer:

36

\int_{0}^{6} (6 t - t^{2}) d t = {[3 t^{2} - \frac{t^{3}}{3}]}_{0}^{6}

= 108 - 72 = 36

m. A1

OR Find the distance she covers between

t = 3

s and

t = 6

s. [2 marks]

Show answer

Answer:

18

\int_{3}^{6} (6 t - t^{2}) d t = {[3 t^{2} - \frac{t^{3}}{3}]}_{3}^{6}

= 36 - 18 = 18

m. A1

Case study 2: Revenue from marginal revenue (4 marks)

A small company making school bags finds that its marginal revenue is $M R (x) = 20 - 2 x$ (in thousands of rupees per hundred bags) when it sells $x$ hundred bags a month. Revenue is zero when nothing is sold.

(i) Find the revenue function

R (x)

. [1 mark]

Show answer

Answer:

R (x) = 20 x - x^{2}

R (x) = \int (20 - 2 x) d x = 20 x - x^{2} + C

, and

R (0) = 0 \Rightarrow C = 0

. A1

(ii) Find the revenue when

500

bags are sold. [1 mark]

Show answer

Answer: ₹

75

thousand

x = 5

R (5) = 100 - 25 = 75

, i.e. ₹

75,000

. A1

(iii) Find the extra revenue from increasing sales from

400

600

bags. [2 marks]

Show answer

Answer: ₹

20

thousand

\int_{4}^{6} (20 - 2 x) d x = {[20 x - x^{2}]}_{4}^{6}

= 84 - 64 = 20

, i.e. ₹

20,000

. A1

OR Evaluate

\int_{0}^{10} (20 - 2 x) d x

and say what it represents. [2 marks]

Show answer

Answer:

100

: revenue from

1000

bags

{[20 x - x^{2}]}_{0}^{10} = 200 - 100 = 100

M1: the total revenue, ₹

1, 00,000

, from selling

1000

bags. A1

Next steps for Integrals

This free set is separate from the chapter's question bank. On the Integrals chapter page the revision notes and the first step of the Route to 95 are free for everyone. CBSE Pro adds all 67 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.

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.