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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)
  • No calculator needed
  • Free, no sign-in

Multiple-choice questions

Choose one option. 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

\(\displaystyle\int (3x^2 + 2x)\,dx\) equals

  1. (a)\(x^3 + x^2 + C\)
  2. (b)\(6x + 2 + C\)
  3. (c)\(3x^3 + 2x^2 + C\)
  4. (d)\(x^3 + 2x^2 + C\)
Show answer
Answer: (a) \(x^3 + x^2 + C\)

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

\(\int 3x^2\,dx = x^3\), \(\int 2x\,dx = x^2\): the answer is \(x^3 + x^2 + C\).

Q2

·1 mark·Multiple choiceIndefinite integrals: standard forms

\(\displaystyle\int e^{3x}\,dx\) equals

  1. (a)\(3e^{3x} + C\)
  2. (b)\(\dfrac{e^{3x}}{3} + C\)
  3. (c)\(e^{3x} + C\)
  4. (d)\(\dfrac{e^{3x + 1}}{3x + 1} + C\)
Show answer
Answer: (b) \(\dfrac{e^{3x}}{3} + C\)

Why: d/dx (e3x/3) = e3x.

\(\dfrac{d}{dx}\left(\dfrac{e^{3x}}{3}\right) = e^{3x}\), so \(\int e^{3x}\,dx = \dfrac{e^{3x}}{3} + C\).

Q3

·1 mark·Multiple choiceIndefinite integrals: standard forms

\(\displaystyle\int \sec^2(2x)\,dx\) equals

  1. (a)\(2\tan(2x) + C\)
  2. (b)\(\tan(2x) + C\)
  3. (c)\(\dfrac{\tan(2x)}{2} + C\)
  4. (d)\(\dfrac{\sec^3(2x)}{3} + C\)
Show answer
Answer: (c) \(\dfrac{\tan(2x)}{2} + C\)

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

\(\dfrac{d}{dx}\tan(2x) = 2\sec^2(2x)\), hence \(\int\sec^2(2x)\,dx = \dfrac12\tan(2x) + C\).

Q4

·1 mark·Multiple choiceIndefinite integrals: standard forms

For \(x \gt 4\), an antiderivative of \(\dfrac{1}{x^2 - 16}\) is

  1. (a)\(\dfrac14\log\left|\dfrac{x - 4}{x + 4}\right|\)
  2. (b)\(\dfrac18\log\left|\dfrac{x + 4}{x - 4}\right|\)
  3. (c)\(\dfrac14\tan^{-1}\dfrac{x}{4}\)
  4. (d)\(\dfrac18\log\left|\dfrac{x - 4}{x + 4}\right|\)
Show answer
Answer: (d) \(\dfrac18\log\left|\dfrac{x - 4}{x + 4}\right|\)

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

\(\dfrac{1}{x^2 - 16} = \dfrac18\left(\dfrac{1}{x - 4} - \dfrac{1}{x + 4}\right)\), so \(\displaystyle\int\dfrac{dx}{x^2 - 16} = \dfrac18\log\left|\dfrac{x - 4}{x + 4}\right| + C\). (Here \(\log\) means \(\log_e\).)

Q5

·1 mark·Multiple choiceIndefinite integrals: standard forms

\(\displaystyle\int \dfrac{dx}{\sqrt{16 - x^2}}\) (for \(|x| \lt 4\)) equals

  1. (a)\(\sin^{-1}\dfrac{x}{4} + C\)
  2. (b)\(\dfrac14\sin^{-1}\dfrac{x}{4} + C\)
  3. (c)\(4\sin^{-1}\dfrac{x}{4} + C\)
  4. (d)\(\sin^{-1}\dfrac{x}{16} + C\)
Show answer
Answer: (a) \(\sin^{-1}\dfrac{x}{4} + C\)

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

\(\displaystyle\int\dfrac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\dfrac{x}{a} + C\), \(a = 4\).

Q6

·1 mark·Multiple choiceIntegration by parts

Integrating by parts, \(\displaystyle\int xe^{x}\,dx\) is found to be

  1. (a)\(xe^x + C\)
  2. (b)\(e^x(x - 1) + C\)
  3. (c)\(e^x(x + 1) + C\)
  4. (d)\(\dfrac{x^2}{2}e^x + C\)
Show answer
Answer: (b) \(e^x(x - 1) + C\)

Why: By parts with u = x, dv = ex dx: xex − ∫ex dx.

\(\int xe^x\,dx = xe^x - \int e^x\,dx = xe^x - e^x + C = e^x(x - 1) + C\).

Q7

·1 mark·Multiple choiceIntegration by substitution

\(\displaystyle\int \dfrac{2x}{x^2 + 1}\,dx\) equals (Here \(\log\) means \(\log_e\).)

  1. (a)\(\tan^{-1}x + C\)
  2. (b)\(2\log(x^2 + 1) + C\)
  3. (c)\(\log(x^2 + 1) + C\)
  4. (d)\(\dfrac{1}{x^2 + 1} + C\)
Show answer
Answer: (c) \(\log(x^2 + 1) + C\)

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

Put \(t = x^2 + 1\), \(dt = 2x\,dx\): \(\int\dfrac{dt}{t} = \log|t| + C = \log(x^2 + 1) + C\).

Q8

·1 mark·Multiple choiceIntegration by partial fractions

\(\displaystyle\int \dfrac{dx}{x(x + 1)}\) equals (Here \(\log\) means \(\log_e\).)

  1. (a)\(\log|x(x + 1)| + C\)
  2. (b)\(\log\left|\dfrac{x}{x + 1}\right| + C\)
  3. (c)\(\log\left|\dfrac{x + 1}{x}\right| + C\)
  4. (d)\(\dfrac{1}{x} - \dfrac{1}{x + 1} + C\)
Show answer
Answer: (b) \(\log\left|\dfrac{x}{x + 1}\right| + C\)

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

\(\dfrac{1}{x(x + 1)} = \dfrac1x - \dfrac{1}{x + 1}\), so the integral is \(\log|x| - \log|x + 1| + C = \log\left|\dfrac{x}{x + 1}\right| + C\).

Q9

·1 mark·Multiple choiceDefinite integrals and their properties

\(\displaystyle\int_0^1 (2x + 1)\,dx\) equals

  1. (a)\(1\)
  2. (b)\(3\)
  3. (c)\(\dfrac32\)
  4. (d)\(2\)
Show answer
Answer: (d) \(2\)

Why: [x² + x] from 0 to 1.

\(\left[x^2 + x\right]_0^1 = (1 + 1) - 0 = 2\).

Q10

·1 mark·Multiple choiceDefinite integrals and their properties

\(\displaystyle\int_0^{\pi/2}\cos x\,dx\) equals

  1. (a)\(0\)
  2. (b)\(1\)
  3. (c)\(-1\)
  4. (d)\(\dfrac{\pi}{2}\)
Show answer
Answer: (b) \(1\)

Why: [sin x] from 0 to π/2.

\(\left[\sin x\right]_0^{\pi/2} = 1 - 0 = 1\).

Q11

·1 mark·Multiple choiceDefinite integrals and their properties

\(\displaystyle\int_{-2}^{2} x^3\,dx\) equals

  1. (a)\(0\)
  2. (b)\(8\)
  3. (c)\(16\)
  4. (d)\(4\)
Show answer
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 \(\int_{-a}^{a} f(x)\,dx = 0\). Directly: \(\left[\tfrac{x^4}{4}\right]_{-2}^{2} = 4 - 4 = 0\).

Q12

·1 mark·Multiple choiceDefinite integrals and their properties

\(\displaystyle\int_0^{\pi/2}\sin^2 x\,dx\) equals

  1. (a)\(\dfrac{\pi}{2}\)
  2. (b)\(\dfrac12\)
  3. (c)\(\dfrac{\pi}{4}\)
  4. (d)\(\pi\)
Show answer
Answer: (c) \(\dfrac{\pi}{4}\)

Why: sin² x = (1 − cos 2x)/2.

\(\int_0^{\pi/2}\dfrac{1 - \cos 2x}{2}\,dx = \left[\dfrac{x}{2} - \dfrac{\sin 2x}{4}\right]_0^{\pi/2} = \dfrac{\pi}{4}\).

Q13

·1 mark·Multiple choiceDefinite integrals and their properties

\(\displaystyle\int_1^{e}\dfrac{1}{x}\,dx\) equals

  1. (a)\(e\)
  2. (b)\(1\)
  3. (c)\(0\)
  4. (d)\(e - 1\)
Show answer
Answer: (b) \(1\)

Why: [log x] from 1 to e = 1 − 0.

\(\left[\log_e x\right]_1^{e} = \log_e e - \log_e 1 = 1 - 0 = 1\).

Q14

·1 mark·Multiple choiceDefinite integrals and their properties

\(\displaystyle\int_0^3 |x - 1|\,dx\) equals

  1. (a)\(\dfrac32\)
  2. (b)\(2\)
  3. (c)\(\dfrac72\)
  4. (d)\(\dfrac52\)
Show answer
Answer: (d) \(\dfrac52\)

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

\(\int_0^1 (1 - x)\,dx + \int_1^3 (x - 1)\,dx = \dfrac12 + 2 = \dfrac52\).

Q15

·1 mark·Multiple choiceIntegration by parts

For \(x \gt 0\), \(\displaystyle\int e^{x}\left(\dfrac1x - \dfrac{1}{x^2}\right)dx\) equals

  1. (a)\(\dfrac{e^x}{x^2} + C\)
  2. (b)\(e^x\log x + C\)
  3. (c)\(\dfrac{e^x}{x} + C\)
  4. (d)\(-\dfrac{e^x}{x} + C\)
Show answer
Answer: (c) \(\dfrac{e^x}{x} + C\)

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

With \(f(x) = \dfrac1x\), \(f'(x) = -\dfrac{1}{x^2}\): \(\int e^x(f + f')\,dx = e^xf(x) + C = \dfrac{e^x}{x} + C\).

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.

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 \(v(t) = 6t - t^2\) m/s for \(0 \le t \le 6\).

(i) Find her velocity at \(t = 2\) s. [1 mark]
Show answer
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: \(s(t) = 3t^2 - \dfrac{t^3}{3}\)
\(s(t) = \int(6t - t^2)\,dt = 3t^2 - \dfrac{t^3}{3}\) (constant \(0\) as \(s(0) = 0\)). A1
(iii) Find the distance she covers in the first \(6\) seconds. [2 marks]
Show answer
Answer: \(36\) m
\(\int_0^6(6t - t^2)\,dt = \left[3t^2 - \dfrac{t^3}{3}\right]_0^6\) M1 \(= 108 - 72 = 36\) m. A1
OR Find the distance she covers between \(t = 3\) s and \(t = 6\) s. [2 marks]
Show answer
Answer: \(18\) m
\(\int_3^6(6t - t^2)\,dt = \left[3t^2 - \dfrac{t^3}{3}\right]_3^6\) M1 \(= 36 - 18 = 18\) m. A1

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

A small company making school bags finds that its marginal revenue is \(MR(x) = 20 - 2x\) (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: \(R(x) = 20x - x^2\)
\(R(x) = \int(20 - 2x)\,dx = 20x - x^2 + C\), 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 \(400\) to \(600\) bags. [2 marks]
Show answer
Answer: ₹\(20\) thousand
\(\int_4^6(20 - 2x)\,dx = \left[20x - x^2\right]_4^6\) M1 \(= 84 - 64 = 20\), i.e. ₹\(20{,}000\). A1
OR Evaluate \(\displaystyle\int_0^{10}(20 - 2x)\,dx\) and say what it represents. [2 marks]
Show answer
Answer: \(100\): revenue from \(1000\) bags
\(\left[20x - x^2\right]_0^{10} = 200 - 100 = 100\) 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 Essentials adds all 43 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 and the full common-mistakes library. There is no AI marking on CBSE Math Revision: you check your work against the marking scheme.

Integrals: notes and questionsNCERT solutions, Chapter 7What Essentials 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 re-checked by computer algebra and 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.