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Integral Calculus Exercise Solutions | KalviGavani

TN Class 11 Maths Ch11 Integral Calculus exercise solutions — indefinite integrals, definite integrals, substitution, partial fractions with worked examples.

Updated 15 June 202610 min read

Exercise Solutions Available

Step-by-step answers for all textbook questions in this chapter

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1. How to Use This Page

This page contains complete exercise solutions for Chapter 11 — Integral Calculus of the Tamil Nadu State Board Class 11 Mathematics textbook. Solve each exercise independently first, then verify your answers and methods below. All worked examples include full step-by-step numerical solutions in code blocks.


2. Key Concepts (Quick Recap)

Standard Integration Formulas

IntegralResult
∫xⁿ dxx^(n+1)/(n+1) + C, n ≠ −1
∫(1/x) dxln\x\+ C
∫eˣ dxeˣ + C
∫aˣ dxaˣ/ln a + C
∫sin x dx−cos x + C
∫cos x dxsin x + C
∫sec²x dxtan x + C
∫csc²x dx−cot x + C
∫sec x tan x dxsec x + C
∫csc x cot x dx−csc x + C
∫dx/√(1−x²)sin⁻¹x + C
∫dx/(1+x²)tan⁻¹x + C
∫1 dxx + C
∫0 dxC

Key Properties

∫kf(x) dx = k ∫f(x) dx (constant multiple) ∫[f(x) ± g(x)] dx = ∫f(x) dx ± ∫g(x) dx (sum/difference) ∫ₐᵇ f(x) dx = F(b) − F(a) (Fundamental theorem) ∫ₐᵇ f(x) dx = −∫ᵇₐ f(x) dx (reverse limits) ∫ₐᵇ f(x) dx = ∫ₐᶜ f(x) dx + ∫ᶜᵇ f(x) dx (interval splitting) ∫₀ᵃ f(x) dx = ∫₀ᵃ f(a−x) dx (reflection property)

Methods of Integration

1. Direct integration — Use standard formulas 2. Substitution — Let u = g(x), transform to u-variable 3. Partial fractions — Decompose rational functions 4. Integration by parts — ∫u dv = uv − ∫v du (for Class 12)


3. I. Choose the Best Answer (MCQs)

Q1. ∫(2x + 3) dx equals:

(a) x² + 3x + C   (b) 2x² + 3x + C   (c) x² + C   (d) 2x + C

Answer(a) x² + 3x + C
Explanation
∫(2x + 3) dx = 2(x²/2) + 3x + C = x² + 3x + C

Q2. ∫₀¹ x² dx equals:

(a) 1   (b) 1/2   (c) 1/3   (d) 1/4

Answer(c) 1/3
Explanation
∫₀¹ x² dx = [x³/3]₀¹ = 1³/3 − 0³/3 = 1/3

Q3. ∫(1/x) dx equals:

(a) x⁻¹ + C   (b) ln x + C   (c) ln\x\+ C   (d) x + C
Answer: (c) ln\x\+ C
Explanation: The power rule does not apply when n = −1. Instead, ∫(1/x) dx = ln\x\+ C. The absolute value is needed because ln is not defined for negative numbers.

Q4. If d/dx[F(x)] = f(x), then ∫f(x) dx equals:

(a) F(x) + C   (b) f(x) + C   (c) F'(x) + C   (d) f'(x) + C

Answer(a) F(x) + C
ExplanationBy definition, F is an antiderivative of f, so ∫f(x) dx = F(x) + C.

Q5. ∫sin(2x) dx equals:

(a) cos(2x) + C   (b) −cos(2x) + C   (c) −(1/2)cos(2x) + C   (d) (1/2)cos(2x) + C

Answer(c) −(1/2)cos(2x) + C
Explanation
Let u = 2x, du = 2dx, dx = du/2 ∫sin(2x) dx = ∫sin(u)(du/2) = (1/2)(−cos u) + C = −(1/2)cos(2x) + C

4. II. Fill in the Blanks

1. ∫5 dx = 5x + C.

2. ∫x⁻² dx = −x⁻¹ + C = −1/x + C.

3. ∫cos x dx = sin x + C.

4. ∫₀² 3x² dx = 8. (3·2³/3 = 8)

5. The constant C in indefinite integrals is called the constant of integration.

6. ∫eˣ dx = eˣ + C.

7. ∫(sec²x + csc²x) dx = tan x − cot x + C.

8. If ∫f(x) dx = F(x) + C, then ∫ₐᵇ f(x) dx = F(b) − F(a).

9. ∫(1 + x + x²) dx = x + x²/2 + x³/3 + C.

10. ∫sin(3x) dx = −(1/3)cos(3x) + C.

11. ∫dx/(1 + x²) = tan⁻¹x + C.

12. ∫₂³ dx = 1 (integral of 1 over interval of length 1).


5. III. True or False (with Corrections)

1. Statement: ∫x⁻¹ dx = x⁰/0 + C.

Answer: False. The power rule ∫xⁿ dx = x^(n+1)/(n+1) does not apply for n = −1. The correct result is ln\x\+ C.

2. Statement: The integral of a constant k is kx + C. Answer: True. ∫k dx = kx + C.

3. Statement: ∫ₐᵇ f(x) dx = −∫ᵇₐ f(x) dx. Answer: True. Reversing the limits changes the sign.

4. Statement: ∫sin x dx = cos x + C. Answer: False. ∫sin x dx = −cos x + C. The derivative of cos x is −sin x, so the integral of sin x is −cos x.

5. Statement: Two different functions can have the same derivative. Answer: True. They differ by a constant. If F'(x) = f(x), then (F(x) + C)' = f(x) for any constant C.


6. IV. Match the Following

Column AColumn B
1. ∫x³ dxa. ln\x\+ C
2. ∫(1/x) dxb. x + C
3. ∫5 dxc. x⁴/4 + C
4. ∫cos x dxd. −cos x + C
5. ∫sin x dxe. 5x + C
6. ∫1 dxf. sin x + C
Answer1→c, 2→a, 3→e, 4→f, 5→d, 6→b

7. V. Answer Briefly (2-Mark Questions)

Q1. Evaluate ∫(3x² + 4x − 7) dx.

∫(3x² + 4x − 7) dx = 3(x³/3) + 4(x²/2) − 7x + C = x³ + 2x² − 7x + C

Q2. Evaluate ∫₀¹ (2x + 1) dx.

∫(2x + 1) dx = x² + x + C F(1) − F(0) = (1 + 1) − (0 + 0) = 2

Q3. Evaluate ∫(sin x + cos x) dx.

∫(sin x + cos x) dx = −cos x + sin x + C

Q4. Find ∫(1/√x) dx.

1/√x = x^(−1/2) ∫x^(−1/2) dx = x^(1/2)/(1/2) + C = 2√x + C

8. VI. Answer in Detail (5-Mark Questions)

Q1. Evaluate ∫(2x + 3)/(x² + 3x + 5) dx using substitution.

Step 1: Identify the substitution Let u = x² + 3x + 5 Then du = (2x + 3) dx Step 2: Rewrite the integral ∫(2x + 3)/(x² + 3x + 5) dx = ∫du/u Step 3: Integrate = ln|u| + C Step 4: Substitute back = ln|x² + 3x + 5| + C Note: Since x² + 3x + 5 = (x + 3/2)² + 11/4 > 0 for all x, we can write: ln(x² + 3x + 5) + C

Q2. Evaluate ∫₀² (3x² − 2x + 1) dx.

Step 1: Find the antiderivative ∫(3x² − 2x + 1) dx = 3(x³/3) − 2(x²/2) + x + C = x³ − x² + x + C Step 2: Let F(x) = x³ − x² + x Step 3: Apply limits using Fundamental Theorem ∫₀² (3x² − 2x + 1) dx = F(2) − F(0) F(2) = 8 − 4 + 2 = 6 F(0) = 0 − 0 + 0 = 0 Answer = 6 − 0 = 6

Q3. Evaluate ∫sin²x dx using the identity sin²x = (1 − cos 2x)/2.

Step 1: Apply the identity sin²x = (1 − cos 2x)/2 Step 2: Integrate term by term ∫sin²x dx = ∫(1 − cos 2x)/2 dx = (1/2) ∫(1 − cos 2x) dx = (1/2) [x − (1/2)sin 2x] + C = x/2 − (sin 2x)/4 + C Step 4: Verify by differentiating d/dx[x/2 − (sin 2x)/4] = 1/2 − (2 cos 2x)/4 = 1/2 − (cos 2x)/2 = (1 − cos 2x)/2 = sin²x ✓

Q4. Evaluate ∫x√(x + 2) dx using substitution.

Step 1: Let u = x + 2, so x = u − 2 and dx = du Step 2: Substitute ∫x√(x + 2) dx = ∫(u − 2)√u du = ∫(u − 2)u^(1/2) du = ∫(u^(3/2) − 2u^(1/2)) du Step 3: Integrate each term = u^(5/2)/(5/2) − 2 · u^(3/2)/(3/2) + C = (2/5)u^(5/2) − (4/3)u^(3/2) + C Step 4: Substitute back u = x + 2 = (2/5)(x + 2)^(5/2) − (4/3)(x + 2)^(3/2) + C Step 5: Factor out (x + 2)^(3/2) to simplify = (x + 2)^(3/2) [(2/5)(x + 2) − 4/3] + C = (x + 2)^(3/2) [2(x + 2)/5 − 4/3] + C = (x + 2)^(3/2) [2x/5 + 4/5 − 4/3] + C = (x + 2)^(3/2) [2x/5 + (12 − 20)/15] + C = (x + 2)^(3/2) [2x/5 − 8/15] + C = (x + 2)^(3/2) · (6x − 8)/15 + C = (2/15)(3x − 4)(x + 2)^(3/2) + C

Q5. Evaluate ∫₀^(π/2) cos²x dx using the identity cos²x = (1 + cos 2x)/2.

Step 1: Apply identity cos²x = (1 + cos 2x)/2 Step 2: Integrate ∫₀^(π/2) cos²x dx = (1/2) ∫₀^(π/2) (1 + cos 2x) dx Step 3: Find antiderivative = (1/2) [x + (1/2)sin 2x]₀^(π/2) Step 4: Evaluate at limits = (1/2) [(π/2 + (1/2)sin π) − (0 + (1/2)sin 0)] = (1/2) [(π/2 + 0) − (0 + 0)] = (1/2)(π/2) = π/4 Answer: ∫₀^(π/2) cos²x dx = π/4

9. Common Mistakes to Avoid

1. Forgetting the constant of integration (+C): Every indefinite integral must include +C. Without it, the answer is incomplete and marks are lost.

2. Applying power rule for n = −1: ∫(1/x) dx = ln\x\+ C, NOT x⁰/0. The power rule formula ∫xⁿ dx = x^(n+1)/(n+1) + C only works for n ≠ −1.

3. Sign errors in trigonometric integrals: ∫sin x dx = −cos x + C (there IS a minus sign). ∫csc²x dx = −cot x + C. Always verify by differentiating your answer.

4. Wrong substitution choice: Choose u so that du appears in the integrand (or a constant multiple of it). If nothing simplifies after substitution, try a different u.

5. Forgetting to substitute back: After integrating in terms of u, you must convert back to x before writing the final answer (for indefinite integrals).

6. Definite integral limits: When using substitution for a definite integral, either change the limits to u-values or substitute back to x before evaluating. Don't mix x-limits with u-expressions.


10. Exam Checklist

  • [ ] Know all standard integration formulas (power, trig, exponential, logarithmic)
  • [ ] Can integrate polynomials term by term
  • [ ] Can apply substitution method: identify u, compute du, transform, integrate, back-substitute
  • [ ] Remember ∫(1/x) dx = ln\|x\| + C (exception to power rule)
  • [ ] Can evaluate definite integrals using the Fundamental Theorem
  • [ ] Know properties: ∫ₐᵇ f(x)dx = −∫ᵇₐ f(x)dx, ∫ₐᵇ = ∫ₐᶜ + ∫ᶜᵇ
  • [ ] Can use trig identities: sin²x = (1−cos2x)/2, cos²x = (1+cos2x)/2
  • [ ] Always include +C for indefinite integrals
  • [ ] Can verify answers by differentiating the antiderivative
  • [ ] Know ∫₀^(π/2) sinⁿx dx and ∫₀^(π/2) cosⁿx dx patterns for small n

Frequently Asked Questions

∫xⁿ dx = x^(n+1)/(n+1) + C, valid for n ≠ −1. When n = −1, ∫(1/x) dx = ln|x| + C.