Integrals — Class 12 Mathematics NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 12 Mathematics chapter "Integrals" — 8 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 12 Mathematics chapter "Integrals" — 8 important questions with detailed answers for CBSE board exam prepa…
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NCERT Solutions for Class 12 Mathematics — Integrals. Step-by-step answers to all 8 textbook questions from this chapter, written for CBSE board preparation and free to use.
Integrals — All 8 Questions Solved
Q1. Find ∫ 5x⁴ dx.
We need to find the antiderivative of 5x⁴.
Using the power rule for integration: ∫ xⁿ dx = xⁿ⁺¹/(n+1) + C (where n ≠ -1)
∫ 5x⁴ dx = 5 ∫ x⁴ dx
= 5 × (x⁴⁺¹/(4+1)) + C
= 5 × (x⁵/5) + C
= x⁵ + C
VERIFICATION:
d/dx(x⁵ + C) = 5x⁴ ✓
CONCLUSION: ∫ 5x⁴ dx = x⁵ + C
Q2. Evaluate ∫ (3x² + 2x - 1) dx.
We integrate each term separately using the power rule.
∫ (3x² + 2x - 1) dx = ∫ 3x² dx + ∫ 2x dx - ∫ 1 dx
Term 1: ∫ 3x² dx = 3 × (x²⁺¹/(2+1)) = 3 × (x³/3) = x³
Term 2: ∫ 2x dx = 2 × (x¹⁺¹/(1+1)) = 2 × (x²/2) = x²
Term 3: ∫ 1 dx = x
Combining:
∫ (3x² + 2x - 1) dx = x³ + x² - x + C
VERIFICATION:
d/dx(x³ + x² - x + C) = 3x² + 2x - 1 ✓
CONCLUSION: ∫ (3x² + 2x - 1) dx = x³ + x² - x + C
Q3. Evaluate the definite integral ∫₀¹ 2x dx.
We need to evaluate the definite integral from 0 to 1 of 2x.
Step 1: Find the antiderivative
∫ 2x dx = 2 × (x²/2) = x² + C
Step 2: Apply the Fundamental Theorem of Calculus
∫₀¹ 2x dx = [x²]₀¹ = (1)² - (0)² = 1 - 0 = 1
GEOMETRIC INTERPRETATION:
The function y = 2x is a straight line through the origin.
The area under this line from x = 0 to x = 1 forms a triangle with base 1 and height 2.
Area = (1/2) × 1 × 2 = 1 ✓
CONCLUSION: ∫₀¹ 2x dx = 1
Q4. Find ∫ e^x dx.
We need to find the antiderivative of the exponential function e^x.
The exponential function has the special property that it is its own derivative:
de^x/dx = e^x
Therefore, the antiderivative is:
∫ e^x dx = e^x + C
VERIFICATION:
d/dx(e^x + C) = e^x ✓
CONCLUSION: ∫ e^x dx = e^x + C
This is a fundamental formula in calculus.
Q5. Evaluate ∫₀^(π/2) sin(x) dx.
We need to evaluate the definite integral of sin(x) from 0 to π/2.
Step 1: Find the antiderivative
We know that d/dx(-cos(x)) = sin(x)
So ∫ sin(x) dx = -cos(x) + C
Step 2: Apply the Fundamental Theorem of Calculus
∫₀^(π/2) sin(x) dx = [-cos(x)]₀^(π/2)
= (-cos(π/2)) - (-cos(0))
= (-0) - (-1)
= 0 + 1
= 1
GEOMETRIC INTERPRETATION:
The integral represents the area under the sine curve from x = 0 to x = π/2, which is indeed 1.
CONCLUSION: ∫₀^(π/2) sin(x) dx = 1
Q6. Find ∫ (1/x) dx for x > 0.
We need to find the antiderivative of 1/x.
Note: This is a special case where the power rule doesn't apply directly because n = -1.
We use the logarithmic integration rule:
∫ (1/x) dx = ln|x| + C
For x > 0, the absolute value can be dropped:
∫ (1/x) dx = ln(x) + C
VERIFICATION:
d/dx(ln(x)) = 1/x ✓
CONCLUSION: ∫ (1/x) dx = ln(x) + C (for x > 0)
∫ (1/x) dx = ln|x| + C (for x ≠ 0, general form)
Q7. Evaluate ∫ (4x³ - 3x² + 2) dx from x = 1 to x = 2.
We evaluate the definite integral ∫₁² (4x³ - 3x² + 2) dx.
Step 1: Find the antiderivative
∫ (4x³ - 3x² + 2) dx = (4x⁴/4) - (3x³/3) + 2x + C
= x⁴ - x³ + 2x + C
Step 2: Apply the Fundamental Theorem of Calculus
∫₁² (4x³ - 3x² + 2) dx = [x⁴ - x³ + 2x]₁²
Step 3: Evaluate at the bounds
At x = 2: (2)⁴ - (2)³ + 2(2) = 16 - 8 + 4 = 12
At x = 1: (1)⁴ - (1)³ + 2(1) = 1 - 1 + 2 = 2
Step 4: Calculate the difference
∫₁² (4x³ - 3x² + 2) dx = 12 - 2 = 10
CONCLUSION: ∫₁² (4x³ - 3x² + 2) dx = 10
Q8. Find ∫ cos(x) dx and evaluate ∫₀^(π/4) cos(x) dx.
Part A: Finding the antiderivative
We know that d/dx(sin(x)) = cos(x)
Therefore: ∫ cos(x) dx = sin(x) + C
VERIFICATION:
d/dx(sin(x)) = cos(x) ✓
Part B: Evaluating the definite integral
∫₀^(π/4) cos(x) dx = [sin(x)]₀^(π/4)
= sin(π/4) - sin(0)
= (√2/2) - 0
= √2/2
Alternative form: √2/2 = 1/√2 ≈ 0.707
GEOMETRIC INTERPRETATION:
The integral represents the area under the cosine curve from x = 0 to x = π/4.
CONCLUSION: ∫ cos(x) dx = sin(x) + C
∫₀^(π/4) cos(x) dx = √2/2
Integrals — Practice MCQs with Answers
Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.
MCQ 1. Evaluate ∫ (x² + 2x + 1) dx.
- A. x³/3 + x² + x + C
- B. x³ + x² + x + C
- C. 2x + 2 + C
- D. x³/3 + x² + C
Answer: A. x³/3 + x² + x + C — Apply the power rule for integration, ∫ xⁿ dx = (xⁿ⁺¹)/(n+1) + C, term by term to each polynomial term.
MCQ 2. Evaluate ∫ (sin x + cos x) dx.
- A. cos x - sin x + C
- B. -cos x + sin x + C
- C. sin x - cos x + C
- D. -sin x + cos x + C
Answer: B. -cos x + sin x + C — The integral of sin x is -cos x and the integral of cos x is sin x. Integrate each term separately.
MCQ 3. Evaluate ∫ x sin(x²) dx.
- A. -½ cos(x²) + C
- B. ½ cos(x²) + C
- C. x² cos(x²) + C
- D. -cos(x²) + C
Answer: A. -½ cos(x²) + C — Use substitution u = x², so du = 2x dx. The integral becomes ½ ∫ sin u du, which evaluates to -½ cos u + C.
MCQ 4. Evaluate ∫ x eˣ dx.
- A. xeˣ - eˣ + C
- B. xeˣ + eˣ + C
- C. eˣ + C
- D. x² eˣ + C
Answer: A. xeˣ - eˣ + C — Use integration by parts formula ∫ u dv = uv - ∫ v du. Let u=x and dv=eˣ dx.
MCQ 5. Evaluate ∫ 1/(x² + 4) dx.
- A. ½ tan⁻¹(x/2) + C
- B. 2 tan⁻¹(x/2) + C
- C. ¼ tan⁻¹(x/2) + C
- D. tan⁻¹(x/2) + C
Answer: A. ½ tan⁻¹(x/2) + C — This is a standard integral of the form ∫ 1/(x² + a²) dx = (1/a) tan⁻¹(x/a) + C. Here a=2.
MCQ 6. Evaluate ∫ (sin x)/√(4 - cos² x) dx.
- A. sin⁻¹(cos x / 2) + C
- B. -sin⁻¹(cos x / 2) + C
- C. cos⁻¹(cos x / 2) + C
- D. ½ sin⁻¹(cos x / 2) + C
Answer: B. -sin⁻¹(cos x / 2) + C — Let t = cos x, so dt = -sin x dx. The integral becomes ∫ (-dt)/√(2² - t²) which is -sin⁻¹(t/2) + C.
MCQ 7. Evaluate ∫₀¹ (2x + 1) dx.
- A. 1
- B. 2
- C. 3
- D. 0
Answer: B. 2 — The indefinite integral is x² + x. Evaluate this from 0 to 1: (1²+1) - (0²+0) = 2.
MCQ 8. Evaluate ∫₀^(π/2) (√sin x) / (√sin x + √cos x) dx.
- A. π
- B. π/2
- C. π/4
- D. 0
Answer: C. π/4 — Use the property ∫₀ᵃ f(x) dx = ∫₀ᵃ f(a-x) dx. Adding the original integral and its transformed form yields 2I = ∫₀^(π/2) 1 dx, leading to I = π/4.
MCQ 9. Evaluate ∫ 1/(3x+5) dx.
- A. ln|3x+5| + C
- B. ⅓ ln|3x+5| + C
- C. 3 ln|3x+5| + C
- D. -⅓(3x+5)⁻² + C
Answer: B. ⅓ ln|3x+5| + C — Use substitution u = 3x+5, so du = 3 dx. The integral becomes ⅓ ∫ 1/u du, which is ⅓ ln|u| + C.
MCQ 10. Evaluate ∫₁ᵉ ln x dx.
- A. 1
- B. e-1
- C. e
- D. 0
Answer: A. 1 — Using integration by parts, ∫ ln x dx = x ln x - x + C. Evaluating from 1 to e gives (e ln e - e) - (1 ln 1 - 1) = (e - e) - (0 - 1) = 1.
MCQ 11. Evaluate ∫ 1/(x² - 1) dx.
- A. ½ ln|(x-1)/(x+1)| + C
- B. ln|(x-1)/(x+1)| + C
- C. ½ ln|x²-1| + C
- D. tan⁻¹ x + C
Answer: A. ½ ln|(x-1)/(x+1)| + C — Use partial fractions to decompose 1/(x²-1) as ½/(x-1) - ½/(x+1). Integrate each term to get ½ (ln|x-1| - ln|x+1|) + C.
MCQ 12. Evaluate ∫ tan² x dx.
- A. tan x - x + C
- B. sec² x - x + C
- C. 2 tan x sec² x + C
- D. tan³ x / 3 + C
Answer: A. tan x - x + C — Use the trigonometric identity tan² x = sec² x - 1. Then integrate (sec² x - 1) which is tan x - x + C.
MCQ 13. Evaluate ∫ e³ˣ dx.
- A. e³ˣ + C
- B. 3e³ˣ + C
- C. ⅓e³ˣ + C
- D. e³ˣ/x + C
Answer: C. ⅓e³ˣ + C — The integral of e^(ax) is (1/a)e^(ax) + C. Here a=3.
MCQ 14. Evaluate ∫₀^(π/2) cos³ x sin x dx.
- A. ½
- B. ⅓
- C. ¼
- D. ⅕
Answer: C. ¼ — Let u = cos x, so du = -sin x dx. Change limits: x=0 => u=1, x=π/2 => u=0. The integral becomes -∫₁⁰ u³ du = ∫₀¹ u³ du = [u⁴/4]₀¹ = ¼.
MCQ 15. Evaluate ∫ eˣ sin x dx.
- A. ½ eˣ (sin x - cos x) + C
- B. ½ eˣ (sin x + cos x) + C
- C. eˣ (sin x - cos x) + C
- D. eˣ (sin x + cos x) + C
Answer: A. ½ eˣ (sin x - cos x) + C — Use integration by parts twice, denoting the integral as I. The equation I = eˣ sin x - eˣ cos x - I will be formed, solving for I.
MCQ 16. Evaluate ∫ 1/√(9 - x²) dx.
- A. sin⁻¹(x/3) + C
- B. cos⁻¹(x/3) + C
- C. ⅓ sin⁻¹(x/3) + C
- D. √(9-x²) + C
Answer: A. sin⁻¹(x/3) + C — This is a standard integral of the form ∫ 1/√(a² - x²) dx = sin⁻¹(x/a) + C. Here a=3.
MCQ 17. Evaluate ∫ 1/x³ dx.
- A. -1/(2x²) + C
- B. 1/(2x²) + C
- C. -2x² + C
- D. ln|x³| + C
Answer: A. -1/(2x²) + C — Rewrite as ∫ x⁻³ dx and apply the power rule for integration: (x⁻³⁺¹)/(-3+1) + C = x⁻²/-2 + C = -1/(2x²) + C.
MCQ 18. Evaluate ∫₋₁¹ (x³ + x) dx.
- A. 0
- B. 1
- C. 2
- D. 4
Answer: A. 0 — The integrand f(x) = x³ + x is an odd function because f(-x) = -f(x). For an odd function over a symmetric interval [-a, a], the definite integral is 0.
MCQ 19. Evaluate ∫ 1/(x² + 2x + 2) dx.
- A. tan⁻¹(x+1) + C
- B. ½ tan⁻¹(x+1) + C
- C. ln(x² + 2x + 2) + C
- D. 1/√2 tan⁻¹((x+1)/√2) + C
Answer: A. tan⁻¹(x+1) + C — Complete the square in the denominator: x² + 2x + 2 = (x+1)² + 1. Then use the standard form ∫ 1/(u² + a²) du = (1/a) tan⁻¹(u/a) + C.
MCQ 20. Evaluate ∫ 1/(sin² x cos² x) dx.
- A. tan x - cot x + C
- B. tan x + cot x + C
- C. sec x - cosec x + C
- D. -tan x + cot x + C
Answer: A. tan x - cot x + C — Rewrite the integrand as (sin² x + cos² x)/(sin² x cos² x) = sec² x + cosec² x. The integral of this is tan x - cot x + C.
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.