Integrals — Class 12 MCQs with Answers

Integrals — important questions & MCQs with answers (Class 12 Mathematics)

6 solved questions from this chapter's difficulty levels, each with its answer and explanation. Tap an option to check yourself.

0 of 6 answered

1. Q1 of 6Easy

   ∫ x² dx =

   Ax³ + CBx³/3 + CC2x + CD3x² + C
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   Answer: x³/3 + C

   Explanation: Power rule: ∫xⁿ dx = xⁿ⁺¹/(n+1) + C. Here n=2, so x³/3 + C — the tempting mistake is skipping the ÷3, which gives just x³.

2. Q2 of 6Easy

   ∫ xⁿ dx (n ≠ −1) =

   Axⁿ + CBxⁿ⁺¹Cxⁿ⁺¹/(n + 1) + CDnxⁿ⁻¹
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   Answer: xⁿ⁺¹/(n + 1) + C

   Explanation: Power rule: raise the exponent by 1 and divide by the new exponent, xⁿ⁺¹/(n+1) + C. Forgetting the ÷(n+1) — just writing xⁿ⁺¹ — is the most common slip.

3. Q3 of 6Medium

   ∫ x cos x dx =

   Acos x − x sin xBx sin xCx sin x + cos x + CDsin x
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   Answer: x sin x + cos x + C

   Explanation: Integration by parts with u = x, dv = cos x dx gives uv − ∫v du = x sin x − ∫sin x dx = x sin x + cos x + C. Forgetting the second term, ∫sin x dx, is the usual gap.

4. Q4 of 6Medium

   Assertion (A): ∫ 2x cos(x²) dx = sin(x²) + C. Reason (R): Putting x² = t gives 2x dx = dt, so the integral becomes ∫ cos t dt.

   ABoth A and R are true and R is the correct explanation of ABBoth A and R are true but R is not the correct explanation of ACA is true but R is falseDA is false but R is true
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   Answer: Both A and R are true and R is the correct explanation of A

   Explanation: R is true: with t = x², dt = 2x dx. A follows from R directly, since ∫ cos t dt = sin t + C = sin(x²) + C. R is the step that gives A.

5. Q5 of 6Hard

   ∫_0^(π/4) tan x dx =

   A1Bln 2C(1/2) ln 2Dπ/4
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   Answer: (1/2) ln 2

   Explanation: Antiderivative of tan x is −ln|cos x|; evaluating from 0 to π/4 gives −ln(1/√2) − (−ln 1) = ln√2 = (1/2)ln 2. Forgetting cos(π/4) = 1/√2 is the usual slip.

6. Q6 of 6Hard

   Assertion (A): ∫₀^(π/2) sin x / (sin x + cos x) dx = π/4. Reason (R): ∫₀ᵃ f(x) dx = ∫₀ᵃ f(a − x) dx.

   ABoth A and R are true and R is the correct explanation of ABBoth A and R are true but R is not the correct explanation of ACA is true but R is falseDA is false but R is true
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   Answer: Both A and R are true and R is the correct explanation of A

   Explanation: R is property P4, true. Let I be the integral. By R with a = π/2, sin x becomes cos x and cos x becomes sin x, so I = ∫₀^(π/2) cos x / (cos x + sin x) dx. Adding, 2I = ∫₀^(π/2) 1 dx = π/2, so I = π/4. R gives A.

Integrals notes: key concepts (Class 12 Mathematics)

Integration is introduced as the inverse of differentiation. The chapter builds standard integrals, substitution, integration by parts (ILATE) and partial fractions, integrals yielding inverse-trig forms, then the definite integral through the Fundamental Theorem and its properties — symmetry and the king's rule.

Indefinite integral

The antiderivative family ∫f(x)dx = F(x) + C, where F′ = f and C is an arbitrary constant of integration.

Standard integrals

∫xⁿdx = x^(n+1)/(n+1)+C (n≠−1), ∫eˣdx = eˣ+C, ∫(1/x)dx = ln|x|+C, plus the basic trig integrals.

Integration by substitution

Put u = g(x), du = g′(x)dx to convert the integral into a simpler one in u, e.g. ∫2x cos(x²)dx = sin(x²)+C.

Integration by parts

∫u dv = uv − ∫v du; pick u by the ILATE order (Inverse, Log, Algebraic, Trig, Exponential).

The eˣ product rule

∫eˣ[f(x) + f′(x)]dx = eˣ f(x) + C — a shortcut whenever the integrand is eˣ times a function plus its derivative.

Partial fractions

Split a proper rational function, e.g. 1/((x−1)(x+2)) = A/(x−1) + B/(x+2), then integrate each piece to logs.

Integrals giving inverse trig

∫dx/(1+x²) = tan⁻¹x + C and ∫dx/√(1−x²) = sin⁻¹x + C; recognise the 1+x² and 1−x² patterns.

Trigonometric integrals

Use half-angle identities: ∫sin²x dx and ∫cos²x dx come from sin²x = (1−cos2x)/2, cos²x = (1+cos2x)/2.

Definite integral & FTC

∫ₐᵇ f(x)dx = F(b) − F(a), the signed area under f; the constant of integration cancels.

Properties of definite integrals

∫ₐᵃ f = 0, ∫ₐᵇ f = −∫ᵦᵃ f, and ∫ₐᶜ = ∫ₐᵇ + ∫ᵦᶜ split at any interior point b.

Symmetry property

Over [−a, a], ∫f = 2∫₀ᵃ f when f is even and ∫f = 0 when f is odd — test f(−x) before integrating.

King's property

∫ₐᵇ f(x)dx = ∫ₐᵇ f(a+b−x)dx; adding the integral to its reflected copy collapses many hard definite integrals.