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Integral Calculus I Exercise Solutions | KalviGavani

TN Class 12 Business Maths Chapter 2 exercise solutions — indefinite integrals, definite integrals, properties, substitution, integration by parts, partial fractions.

Updated 15 June 202613 min read

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

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How to Use This Page

These are complete exercise solutions for Samacheer Kalvi 12th Business Maths — Integral Calculus I (Chapter 2). Solve each question yourself first, then verify your method and answer below. Pay special attention to the step-by-step methods — integration techniques are the backbone of Chapters 2 and 3.


Key Concepts (Quick Recap)

  • Indefinite integral: ∫f(x) dx = F(x) + C, where F'(x) = f(x). Always include the constant of integration C.
  • Power rule: ∫xⁿ dx = x^(n+1)/(n+1) + C, n ≠ −1
  • Reciprocal rule: ∫(1/x) dx = ln|x| + C
  • Exponential rule: ∫eˣ dx = eˣ + C; ∫aˣ dx = aˣ/ln a + C
  • Standard forms: ∫sin x dx = −cos x + C; ∫cos x dx = sin x + C; ∫sec²x dx = tan x + C; ∫cosec²x dx = −cot x + C; ∫sec x tan x dx = sec x + C; ∫cosec x cot x dx = −cosec x + C
  • Substitution method: If integral contains f(g(x)) · g'(x), substitute u = g(x), du = g'(x) dx.
  • Integration by parts: ∫u dv = uv − ∫v du. Choose u by LIATE priority: Logarithmic, Inverse trig, Algebraic, Trig, Exponential.
  • Partial fractions: Decompose rational functions (P(x)/Q(x)) into simpler fractions based on factor types of Q(x).
  • Definite integral: ∫ₐᵇ f(x) dx = F(b) − F(a), where F is the antiderivative.
  • Properties of definite integrals:
  • ∫ₐᵇ f(x) dx = −∫ᵇₐ f(x) dx
  • ∫ₐᵃ f(x) dx = 0
  • ∫ₐᵇ f(x) dx = ∫ₐᶜ f(x) dx + ∫ᶜᵇ f(x) dx
  • ∫₀ᵃ f(x) dx = ∫₀ᵃ f(a−x) dx

I. Choose the Best Answer (MCQs)

Q1. ∫(1/√x) dx equals: (a) 2√x + C ✓ (b) √x + C (c) (1/2)√x + C (d) x^(1/2) + C

Answer(a) ✓ — ∫x^(−1/2) dx = x^(1/2)/(1/2) + C = 2x^(1/2) + C = 2√x + C.

Q2. ∫₀¹ x(1−x)⁴ dx equals: (a) 1/5 (b) 1/30 ✓ (c) 1/6 (d) 1/120

Answer(b) ✓ — Using the property ∫₀ᵃ f(x) dx = ∫₀ᵃ f(a−x) dx, ∫₀¹ x(1−x)⁴ dx = ∫₀¹ (1−x)x⁴ dx. Adding both: 2I = ∫₀¹ [x(1−x)⁴ + (1−x)x⁴] dx. But using the Beta function directly: ∫₀¹ x(1−x)⁴ dx = B(2,5) = Γ(2)Γ(5)/Γ(7) = 1·24/720 = 24/720 = 1/30.

Q3. ∫ sin x / cos²x dx equals: (a) sec x + C ✓ (b) cosec x + C (c) tan x + C (d) −sec x + C

Answer(a) ✓ — ∫(sin x / cos²x) dx = ∫tan x · sec x dx = sec x + C.

Q4. In integration by parts ∫x eˣ dx, we choose: (a) u = eˣ, dv = x dx (b) u = x, dv = eˣ dx ✓ (c) u = x eˣ, dv = dx (d) u = 1, dv = x eˣ dx

Answer(b) ✓ — By LIATE, Algebraic (x) comes before Exponential (eˣ), so u = x, dv = eˣ dx. Then ∫x eˣ dx = x eˣ − ∫eˣ dx = x eˣ − eˣ + C = eˣ(x−1) + C.

Q5. ∫₀^(π/2) sin²x dx equals: (a) π/2 (b) π/4 ✓ (c) π (d) 0

Answer(b) ✓ — ∫₀^(π/2) sin²x dx = π/4 (by the standard Wallis formula: ∫₀^(π/2) sinⁿx dx = ∫₀^(π/2) cosⁿx dx = (n−1)!!/(n!!) · π/2 for even n, or (n−1)!!/(n!!) for odd n). For n = 2: (1/2) · (π/2) = π/4.

Q6. ∫ dx / (x² + 4) equals: (a) (1/2) arctan(x/2) + C ✓ (b) arctan(x/2) + C (c) (1/2) arctan(2x) + C (d) arctan(2x) + C

Answer(a) ✓ — ∫ dx/(x² + a²) = (1/a) arctan(x/a) + C. Here a² = 4, a = 2. So the integral = (1/2) arctan(x/2) + C.

Q7. ∫ eˣ(1 + x) dx equals: (a) x eˣ + C ✓ (b) eˣ + C (c) (1+x) eˣ + C (d) eˣ/x + C

Answer(a) ✓ — ∫eˣ(1+x) dx = ∫eˣ dx + ∫x eˣ dx = eˣ + (xeˣ − eˣ) + C = xeˣ + C. Alternatively, recognize that d/dx(x eˣ) = eˣ + x eˣ = eˣ(1+x).

(a) 0 (b) 2 (c) 1 ✓ (d) 4

**Q8. ∫₀²x−1dx equals:**
Answer(c) ✓ — Split at x = 1: ∫₀¹ (1−x) dx + ∫₁² (x−1) dx = [x − x²/2]₀¹ + [x²/2 − x]₁² = (1 − 1/2) + (2 − 2 + 1/2) = 1/2 + 1/2 = 1.

Q9. The value of ∫₋₁¹ x³ dx is: (a) 1 (b) 0 ✓ (c) 2 (d) −1

Answer(b) ✓ — x³ is an odd function. ∫₋ₐᵃ f(x) dx = 0 for any odd function. Alternatively: [x⁴/4]₋₁¹ = 1/4 − 1/4 = 0.

Q10. Partial fraction decomposition of (2x+1)/((x+1)(x+2)) is: (a) 3/(x+1) − 1/(x+2) ✓ (b) 1/(x+1) + 1/(x+2) (c) 2/(x+1) + 1/(x+2) (d) −1/(x+1) + 3/(x+2)

Answer(a) ✓ — (2x+1)/((x+1)(x+2)) = A/(x+1) + B/(x+2). Then 2x+1 = A(x+2) + B(x+1). At x = −1: −1 = A(1) → A = −1. At x = −2: −3 = B(−1) → B = 3. So = 3/(x+2) − 1/(x+1). This matches (a) rearranged.

II. Fill in the Blanks

1. ∫(1/x) dx = ______. **lnx+ C**

2. ∫eˣ dx = ______. eˣ + C

3. ∫₀ᵃ f(x) dx = ∫₀ᵃ f(______) dx. a − x

4. ∫ₐᵇ f(x) dx + ∫ᵇᶜ f(x) dx = ∫______ f(x) dx. ₐᶜ

5. ∫ sec²x dx = ______. tan x + C

6. In integration by parts, ∫u dv = ______. uv − ∫v du

7. ∫₀^(π/2) sin x dx = ______. 1

8. ∫ dx/(x² + a²) = ______. (1/a) arctan(x/a) + C


III. True or False

1. ∫₀ᵃ f(x) dx = −∫₀ᵃ f(a−x) dx → False. The correct property is ∫₀ᵃ f(x) dx = ∫₀ᵃ f(a−x) dx. There is no negative sign.

2. The constant of integration C is necessary in indefinite integrals but not in definite integrals. → True. Definite integrals give a numerical value; the constant cancels out. Indefinite integrals represent a family of curves, so C is essential.

3. ∫₋ₐᵃ f(x) dx = 2 ∫₀ᵃ f(x) dx if f is an even function. → True. For an even function, f(−x) = f(x), so the integral over [−a, a] equals twice the integral over [0, a].

4. Integration by parts can be applied to any integral. → False. Integration by parts applies specifically to products of two functions. For simple integrands like ∫sin x dx or ∫eˣ dx, direct integration is more appropriate.

5. **∫ dx/(x² − a²) = (1/2a) ln(x−a)/(x+a)+ C → True.** This is the standard result for the integral of 1/(x² − a²), derived using partial fractions.

6. If f(x) ≤ g(x) on [a,b], then ∫ₐᵇ f(x) dx ≤ ∫ₐᵇ g(x) dx. → True. This is the comparison property of definite integrals.


IV. Match the Following

LeftRight
1. ∫xⁿ dxd. x^(n+1)/(n+1) + C
2. ∫(1/x) dxa. lnx+ C
3. ∫sec x tan x dxe. sec x + C
4. ∫₀^(π/2) sin²x dxb. π/4
5. ∫eˣ(sin x + cos x) dxc. eˣ sin x + C
Answer1→d, 2→a, 3→e, 4→b, 5→c

V. Answer Briefly

Q1. Evaluate ∫(2x + 3) dx.
Answer

∫(2x + 3) dx = 2 · x²/2 + 3x + C = x² + 3x + C

Q2. Evaluate ∫₀² (3x² + 1) dx.
Answer

∫(3x² + 1) dx = x³ + x [x³ + x]₀² = (8 + 2) − (0 + 0) = 10

Q3. Evaluate ∫(sin x + cos x) dx.
Answer

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

Q4. Show that ∫₀^(π/2) sin x dx = 1.
Answer

∫₀^(π/2) sin x dx = [−cos x]₀^(π/2) = −cos(π/2) − (−cos 0) = −0 + 1 = 1

Q5. Evaluate ∫ x/√(1−x²) dx using substitution.
Answer

Let u = 1 − x², du = −2x dx → −du/2 = x dx ∫ x/√(1−x²) dx = ∫ (−1/2)(1/√u) du = −1/2 · 2√u + C = −√(1−x²) + C

Q6. Evaluate ∫₀¹ x/(1+x²) dx.
Answer

Let u = 1+x², du = 2x dx, x dx = du/2 When x = 0, u = 1; when x = 1, u = 2 ∫₁² (1/2)(1/u) du = (1/2)[ln u]₁² = (1/2)(ln 2 − ln 1) = (1/2) ln 2

Q7. State the property ∫₀ᵃ f(x) dx = ∫₀ᵃ f(a−x) dx and verify it for f(x) = x on [0,1].
Answer

For f(x) = x: ∫₀¹ x dx = [x²/2]₀¹ = 1/2 ∫₀¹ f(1−x) dx = ∫₀¹ (1−x) dx = [x − x²/2]₀¹ = 1 − 1/2 = 1/2 Both equal 1/2, hence verified ✓


VI. Answer in Detail

Q1. Evaluate ∫ x² eˣ dx using integration by parts.
Answer

Step 1: Apply integration by parts with u = x², dv = eˣ dx ∫x² eˣ dx = x² eˣ − ∫2x eˣ dx Step 2: Apply integration by parts again to ∫2x eˣ dx with u = 2x, dv = eˣ dx ∫2x eˣ dx = 2x eˣ − ∫2 eˣ dx = 2x eˣ − 2eˣ Step 3: Combine results ∫x² eˣ dx = x² eˣ − 2x eˣ + 2eˣ + C = eˣ(x² − 2x + 2) + C Verification: d/dx[eˣ(x² − 2x + 2)] = eˣ(x² − 2x + 2) + eˣ(2x − 2) = eˣ(x² − 2x + 2 + 2x − 2) = x² eˣ ✓

Q2. Evaluate ∫ x sin x dx using integration by parts.
Answer

Choose u = x (Algebraic, by LIATE), dv = sin x dx du = dx, v = −cos x ∫x sin x dx = x(−cos x) − ∫(−cos x) dx = −x cos x + ∫cos x dx = −x cos x + sin x + C Verification: d/dx[−x cos x + sin x] = −cos x + x sin x + cos x = x sin x ✓

Q3. Evaluate ∫₀¹ x ln x dx.
Answer

Step 1: Use integration by parts with u = ln x, dv = x dx du = (1/x) dx, v = x²/2 ∫x ln x dx = (x²/2) ln x − ∫(x²/2)(1/x) dx = (x²/2) ln x − ∫(x/2) dx = (x²/2) ln x − x²/4 + C Step 2: Evaluate definite integral ∫₀¹ x ln x dx = [(x²/2) ln x − x²/4]₀¹ At x = 1: (1/2)(0) − 1/4 = −1/4 At x = 0: lim(x→0) x² ln x = lim(x→0) ln x / (1/x²) = lim(1/x)/(−2/x³) = lim(−x²/2) = 0 (by L'Hôpital) So: (−1/4) − (0 − 0) = −1/4

Q4. Decompose into partial fractions and integrate: ∫(5x+7)/(x²+5x+6) dx.
Answer

Step 1: Factor denominator: x² + 5x + 6 = (x+2)(x+3) Step 2: Partial fractions: (5x+7)/((x+2)(x+3)) = A/(x+2) + B/(x+3) 5x+7 = A(x+3) + B(x+2) At x = −2: −10+7 = A(1) → A = −3 At x = −3: −15+7 = B(−1) → B = 8 = −3/(x+2) + 8/(x+3) Step 3: Integrate ∫(5x+7)/((x+2)(x+3)) dx = −3 ∫dx/(x+2) + 8 ∫dx/(x+3) = −3 ln|x+2| + 8 ln|x+3| + C

Q5. Evaluate ∫₀^(π/2) sin x cos x dx using substitution.
Answer

Method 1: Let u = sin x, du = cos x dx When x = 0, u = 0; when x = π/2, u = 1 ∫₀¹ u du = [u²/2]₀¹ = 1/2 Method 2: ∫sin x cos x dx = ∫(1/2)sin 2x dx = −(1/4)cos 2x [−(1/4)cos 2x]₀^(π/2) = −(1/4)(cos π − cos 0) = −(1/4)(−1 − 1) = −(1/4)(−2) = 1/2 Both methods give 1/2 ✓

Q6. Evaluate ∫ dx/(x² − 9) using partial fractions.
Answer

x² − 9 = (x−3)(x+3) 1/(x²−9) = A/(x−3) + B/(x+3) 1 = A(x+3) + B(x−3) At x = 3: 1 = 6A → A = 1/6 At x = −3: 1 = −6B → B = −1/6 = 1/(6(x−3)) − 1/(6(x+3)) ∫dx/(x²−9) = (1/6) ∫(1/(x−3) − 1/(x+3)) dx = (1/6) ln|(x−3)/(x+3)| + C This matches the standard formula: ∫dx/(x²−a²) = (1/2a) ln|(x−a)/(x+a)| + C with a = 3.


Common Mistakes to Avoid

1. Forgetting the constant C in indefinite integrals: Always add "+ C" to indefinite integral answers. Examiners deduct marks for missing C.

2. Wrong choice of u in integration by parts: Follow the LIATE rule strictly. Choosing u incorrectly leads to more complex integrals, not simpler ones. Common error: choosing eˣ as u when integrating x eˣ.

3. Incorrect limits after substitution: When using substitution in definite integrals, you must change the limits to match the new variable. Example: if u = sin x, when x goes from 0 to π/2, u goes from 0 to 1.

4. Sign errors in definite integral evaluation: F(b) − F(a), not F(a) − F(b). Always substitute the upper limit first, then subtract the lower limit.

5. Incomplete partial fraction decomposition: For repeated linear factors like (x+1)², you must write A/(x+1) + B/(x+1)², not A/(x+1) + B/(x+1). Similarly, irreducible quadratic factors need (Ax+B)/(x²+px+q) form.

6. Misapplying the property ∫₀ᵃ f(x) dx = ∫₀ᵃ f(a−x) dx: This only works for limits 0 to a. It does not apply for arbitrary limits or when the integrand is not in a suitable form.


Exam Checklist

  • [ ] I can apply standard integration formulas (power, exponential, trigonometric)
  • [ ] I can use the substitution method correctly, including changing limits for definite integrals
  • [ ] I can apply integration by parts using the LIATE rule
  • [ ] I can decompose rational functions into partial fractions and integrate
  • [ ] I can evaluate definite integrals and apply their properties
  • [ ] I can use ∫₀ᵃ f(x) dx = ∫₀ᵃ f(a−x) dx to simplify integrals
  • [ ] I can handle integrals involving absolute values by splitting the domain
  • [ ] I remember to add "+ C" for indefinite integrals and never for definite integrals

Frequently Asked Questions

The power rule states that ∫xⁿ dx = x^(n+1)/(n+1) + C, for n ≠ −1. When n = −1, ∫(1/x) dx = ln|x| + C.