Class 9 Mathematics Chapter 7 Integrals Important Questions with Solutions

The CBSE Class 9 board exam allocates approximately 8–12 marks to Chapter 7 (Integrals) across multiple question types. The 2024-25 rationalized curriculum emphasizes: (1) conceptual understanding of integration as the inverse of differentiation, (2) computation of indefinite integrals using standard formulas, (3) evaluation of definite integrals with limits, and (4) application of the Fundamental Theorem of Calculus. Unlike pure algebra chapters, Integrals require both symbolic manipulation and geometric intuition (area under curves). Board examiners frequently ask: "Find ∫(3x² + 2x) dx", "Evaluate ∫₀² 4x³ dx", or conceptual questions like "How does the Fundamental Theorem connect derivatives and integrals?" These questions test whether students can shift from localized rate-of-change thinking (derivatives) to global accumulation thinking (integrals)—a cognitive leap critical for physics, economics, and engineering streams.

Single-mark multiple-choice questions focus on recall of formulas, quick antiderivatives, and basic properties. Here are 5 representative board-pattern questions: **Q1.** ∫ 5 dx equals: (A) 5x + C (B) x + C (C) 5 + C (D) 5x **Answer:** (A) 5x + C. Explanation: The antiderivative of a constant 5 is 5x plus the constant of integration C. **Q2.** ∫ x⁵ dx equals: (A) x⁶/6 + C (B) 6x⁶ + C (C) x⁶ + C (D) x⁴/4 + C **Answer:** (A) x⁶/6 + C. Using the power rule: ∫ xⁿ dx = xⁿ⁺¹/(n+1) + C, where n = 5, so ∫ x⁵ dx = x⁶/6 + C. **Q3.** ∫₀¹ 2x dx equals: (A) 0 (B) 1 (C) 2 (D) 1/2 **Answer:** (B) 1. Using the Fundamental Theorem: ∫₀¹ 2x dx = [x²]₀¹ = 1² − 0² = 1. **Q4.** Which of the following is the antiderivative of cos(x)? (A) −sin(x) + C (B) sin(x) + C (C) −cos(x) + C (D) cos(x) + C **Answer:** (B) sin(x) + C. The derivative of sin(x) is cos(x), so ∫ cos(x) dx = sin(x) + C. **Q5.** ∫ (3x + 2) dx equals: (A) 3x² + 2x + C (B) 3x²/2 + 2x + C (C) x³ + 2x + C (D) 3x² + C **Answer:** (B) 3x²/2 + 2x + C. Integrating term-by-term: ∫ 3x dx + ∫ 2 dx = 3x²/2 + 2x + C.

Two-mark questions require clear reasoning, correct use of antiderivative formulas, and substitution skills. Here are 5 key examples: **Q1.** Find ∫ (4x³ + 3x² + 1) dx. **Answer:** Using the power rule and linearity of integration: ∫ 4x³ dx = 4 · x⁴/4 = x⁴ ∫ 3x² dx = 3 · x³/3 = x³ ∫ 1 dx = x Therefore, ∫ (4x³ + 3x² + 1) dx = **x⁴ + x³ + x + C**. **Q2.** Evaluate ∫₁² 3x² dx. **Answer:** Using the Fundamental Theorem of Calculus: ∫₁² 3x² dx = [x³]₁² = 2³ − 1³ = 8 − 1 = **7**. **Q3.** Find ∫ (√x + 1/x) dx. **Answer:** Rewrite as ∫ (x^(1/2) + x⁻¹) dx. ∫ x^(1/2) dx = x^(3/2)/(3/2) = 2x^(3/2)/3 ∫ x⁻¹ dx = ln|x| Therefore, ∫ (√x + 1/x) dx = **2x^(3/2)/3 + ln|x| + C**. **Q4.** State the Fundamental Theorem of Calculus (Part 1). **Answer:** If F(x) is a continuous function on [a, b] and F'(x) = f(x), then ∫ₐᵇ f(x) dx = F(b) − F(a). This theorem connects the definite integral to the antiderivative. **Q5.** Evaluate ∫₀^(π/2) sin(x) dx. **Answer:** Using the Fundamental Theorem: ∫₀^(π/2) sin(x) dx = [−cos(x)]₀^(π/2) = −cos(π/2) − (−cos(0)) = 0 − (−1) = **1**.

Three-mark questions test deeper conceptual understanding, multi-step calculations, and the ability to interpret integrals geometrically. Here are 4 representative questions: **Q1.** Find ∫ (2x + 3)² dx. **Answer:** Expand (2x + 3)² = 4x² + 12x + 9. Then ∫ (4x² + 12x + 9) dx = 4 · x³/3 + 12 · x²/2 + 9x + C = **4x³/3 + 6x² + 9x + C**. **Q2.** Evaluate ∫₁³ (2x + 1) dx and interpret the result geometrically. **Answer:** ∫₁³ (2x + 1) dx = [x² + x]₁³ = (9 + 3) − (1 + 1) = 12 − 2 = **10**. Geometrically, this represents the area under the line y = 2x + 1 between x = 1 and x = 3. The area is a trapezoid with bases 3 and 7, and height 2, giving area = (1/2)(3 + 7)(2) = 10, confirming our integral result. **Q3.** Find ∫ eˣ(1 + eˣ) dx. **Answer:** Let u = 1 + eˣ, then du = eˣ dx. ∫ eˣ(1 + eˣ) dx = ∫ u du = u²/2 + C = **(1 + eˣ)²/2 + C**. Alternatively, expand: ∫ (eˣ + e²ˣ) dx = eˣ + e²ˣ/2 + C. **Q4.** Use the Fundamental Theorem of Calculus to find d/dx [∫₀ˣ t² dt]. **Answer:** By the Fundamental Theorem (Part 2), d/dx [∫₀ˣ f(t) dt] = f(x). Therefore, d/dx [∫₀ˣ t² dt] = **x²**. This shows that differentiation and integration are inverse operations.

Five-mark questions demand detailed working, multi-step reasoning, and synthesis of multiple concepts. Here are 3 comprehensive problems: **Q1.** Find ∫ (x³ − 2x² + 5x − 3) dx and verify by differentiation. **Solution:** Integrating term-by-term: ∫ x³ dx = x⁴/4 ∫ −2x² dx = −2x³/3 ∫ 5x dx = 5x²/2 ∫ −3 dx = −3x Therefore, ∫ (x³ − 2x² + 5x − 3) dx = **x⁴/4 − 2x³/3 + 5x²/2 − 3x + C**. Verification: d/dx [x⁴/4 − 2x³/3 + 5x²/2 − 3x + C] = x³ − 2x² + 5x − 3 ✓ **Q2.** Evaluate ∫₀² (x² + 2x + 1) dx and find the area under the curve from x = 0 to x = 2. **Solution:** Notice that x² + 2x + 1 = (x + 1)². ∫₀² (x + 1)² dx = [(x + 1)³/3]₀² = (3)³/3 − (1)³/3 = 27/3 − 1/3 = 26/3 ≈ 8.67 square units. This represents the area of the region bounded by y = (x + 1)², the x-axis, x = 0, and x = 2. **Q3.** Given F(x) = ∫₀ˣ (2t + 1) dt, find F(x) and F'(x). **Solution:** First, find the antiderivative: ∫ (2t + 1) dt = t² + t + C. Using the Fundamental Theorem: F(x) = [t² + t]₀ˣ = x² + x − 0 = **F(x) = x² + x**. Then, F'(x) = d/dx [x² + x] = **2x + 1**, which equals the integrand at x, confirming the Fundamental Theorem.

**Case Study:** A water tank is being filled. The rate of water inflow (in litres per minute) at time t (in minutes) is given by r(t) = 3t + 2. Find the total volume of water (in litres) accumulated in the tank from t = 0 to t = 5 minutes. **Solution Steps:** **Step 1:** Set up the integral. Total volume = ∫₀⁵ r(t) dt = ∫₀⁵ (3t + 2) dt. **Step 2:** Find the antiderivative. ∫ (3t + 2) dt = 3t²/2 + 2t + C. **Step 3:** Apply the Fundamental Theorem of Calculus. ∫₀⁵ (3t + 2) dt = [3t²/2 + 2t]₀⁵ = (3(25)/2 + 2(5)) − (0) = 75/2 + 10 = 37.5 + 10 = **47.5 litres**. **Step 4:** Interpret. The tank accumulates 47.5 litres of water over 5 minutes with the given inflow rate. **Higher-Order Thinking:** How would the answer change if the inflow rate were r(t) = 3t² + 2 instead? (Answer: ∫₀⁵ (3t² + 2) dt = [t³ + 2t]₀⁵ = 125 + 10 = 135 litres—a significantly higher volume due to the quadratic growth of the inflow rate.)

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