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Integrals - Previous Year Questions Solved Examples (Class 12 Mathematics)

Integration is the reverse of differentiation and calculates areas under curves. These questions test mastery of integration techniques and applications.

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TL;DR: Integration is the reverse of differentiation and calculates areas under curves. These questions test mastery of integration techniques and applicatio…

Written & reviewed by the Syllab.in Academic Team (CBSE/NCERT subject experts) · Updated Jul 23, 2026

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Integration is the reverse of differentiation and calculates areas under curves. These questions test mastery of integration techniques and applications.

Integrals - Previous Year Questions — Solved Numerical Examples (Step by Step)

Example 1: Integrate (3x² + 2x + 1) with respect to x. [2 marks]

Solution: ∫(3x² + 2x + 1)dx = 3∫x²dx + 2∫xdx + ∫1dx = 3(x³/3) + 2(x²/2) + x + C = x³ + x² + x + C, where C is constant of integration.

Example 2: Find ∫x × e^x dx using integration by parts. [3 marks]

Solution: Using integration by parts: ∫u dv = uv - ∫v du. Let u = x, dv = e^x dx. Then du = dx, v = e^x. ∫x e^x dx = x e^x - ∫e^x dx = x e^x - e^x + C = e^x(x - 1) + C.

Example 3: Evaluate the definite integral ∫[0 to π] sin(x) dx. [2 marks]

Solution: ∫sin(x)dx = -cos(x) + C. Evaluating from 0 to π: [-cos(x)]₀^π = -cos(π) - (-cos(0)) = -(-1) + 1 = 1 + 1 = 2.

Example 4: Use substitution to find ∫(2x + 3) / (x² + 3x + 1) dx. [3 marks]

Solution: Let u = x² + 3x + 1. Then du = (2x + 3)dx. ∫(2x + 3)/(x² + 3x + 1) dx = ∫(1/u) du = ln|u| + C = ln|x² + 3x + 1| + C.

Example 5: Find ∫1/(x² + 4) dx. [3 marks]

Solution: This is standard form ∫1/(x² + a²) dx = (1/a) tan⁻¹(x/a) + C. Here a² = 4, so a = 2. ∫1/(x² + 4) dx = (1/2) tan⁻¹(x/2) + C.

Example 6: Find the area under the curve y = x² between x = 0 and x = 2. [3 marks]

Solution: Area = ∫[0 to 2] x² dx = [x³/3]₀² = (2³/3) - (0³/3) = 8/3 square units.

Example 7: Evaluate ∫e^(2x) × sin(x) dx. [4 marks]

Solution: Using integration by parts twice. Let I = ∫e^(2x) sin(x) dx. u = sin(x), dv = e^(2x) dx. Then du = cos(x) dx, v = e^(2x)/2. I = (e^(2x) sin(x))/2 - (1/2)∫e^(2x) cos(x) dx. Repeat for the second integral with u = cos(x), dv = e^(2x) dx. After algebraic manipulation: I = e^(2x)(2sin(x) - cos(x))/5 + C.

Tips

  • Always add constant C to indefinite integrals; it's essential and expected.
  • Integration by parts: choose u as LIATE priority (Log, Inverse, Algebraic, Trig, Exponential).
  • Substitution works when derivative of inner function appears in the integrand.
  • For definite integrals, apply limits after finding antiderivative: [F(x)] from a to b = F(b) - F(a).

Frequently Asked Questions

What is the difference between definite and indefinite integrals?

Indefinite integral has constant C and represents family of functions. Definite integral has limits and gives numerical value representing area.

Why is the constant of integration important in indefinite integrals?

Because differentiation of constant is zero, multiple functions differing only in constant have same derivative. The constant represents all possible solutions.

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