Application of Integrals — Class 12 Mathematics NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 12 Mathematics chapter "Application of 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 "Application of Integrals" — 8 important questions with detailed answers for CBSE b…
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NCERT Solutions for Class 12 Mathematics — Application of Integrals. Step-by-step answers to all 8 textbook questions from this chapter, written for CBSE board preparation and free to use.
Application of Integrals — All 8 Questions Solved
Q1. Find the area enclosed by the curves y = x² and y = x.
To find the area between y = x² and y = x:
Step 1: Find points of intersection.
Set x² = x
x² - x = 0
x(x - 1) = 0
x = 0 or x = 1
Step 2: Determine which curve is above.
At x = 0.5: y = 0.5 (line) and y = 0.25 (parabola)
So y = x is above y = x²
Step 3: Set up the integral.
Area = ∫₀¹ (x - x²) dx
Step 4: Evaluate.
= [x²/2 - x³/3]₀¹
= (1/2 - 1/3) - 0
= 3/6 - 2/6
= 1/6 square units
Final Answer: 1/6 square units
Q2. Find the area of the region bounded by the curve y² = 4x and the line x = 1.
To find the area bounded by parabola y² = 4x and line x = 1:
Step 1: Identify the curve.
y² = 4x is a parabola opening rightward with vertex at origin.
Step 2: Find intersection points.
When x = 1: y² = 4(1) = 4
y = ±2
Points: (1, 2) and (1, -2)
Step 3: Set up the integral.
Using x as the variable of integration from 0 to 1:
Area = ∫₀¹ 2√(4x) dx
= ∫₀¹ 4√x dx
= 4 ∫₀¹ x^(1/2) dx
Step 4: Evaluate.
= 4 × [x^(3/2)/(3/2)]₀¹
= 4 × (2/3) × [x^(3/2)]₀¹
= (8/3) × 1
= 8/3 square units
Final Answer: 8/3 square units
Q3. Find the area of the circle x² + y² = a² using integration.
To find the area of circle x² + y² = a²:
Step 1: Use symmetry.
The circle is symmetric about both axes.
Area = 4 × (area in first quadrant)
Step 2: Express y in terms of x.
From x² + y² = a²:
y = √(a² - x²) (taking positive value for first quadrant)
Step 3: Set up the integral for first quadrant.
Area (first quadrant) = ∫₀ᵃ √(a² - x²) dx
Step 4: Evaluate using substitution.
Let x = a sin θ, dx = a cos θ dθ
When x = 0, θ = 0; when x = a, θ = π/2
= ∫₀^(π/2) √(a² - a² sin²θ) × a cos θ dθ
= ∫₀^(π/2) a cos θ × a cos θ dθ
= a² ∫₀^(π/2) cos²θ dθ
= a² ∫₀^(π/2) (1 + cos 2θ)/2 dθ
= a²/2 × [θ + sin 2θ/2]₀^(π/2)
= a²/2 × [π/2 + 0]
= πa²/4
Step 5: Multiply by 4 for total area.
Total Area = 4 × πa²/4 = πa²
Final Answer: πa² square units
Q4. Find the area bounded by y = sin x, the x-axis, and the lines x = 0 and x = π.
To find the area under sine curve from 0 to π:
Step 1: Identify the region.
y = sin x is above the x-axis for all x ∈ [0, π]
Step 2: Set up the integral.
Area = ∫₀^π sin x dx
Step 3: Evaluate.
= [-cos x]₀^π
= -cos π - (-cos 0)
= -(-1) - (-1)
= 1 + 1
= 2 square units
Final Answer: 2 square units
Q5. Find the area of the region between the parabola y = x² - 2x and the x-axis.
To find the area between parabola y = x² - 2x and the x-axis:
Step 1: Find x-intercepts.
Set y = 0:
x² - 2x = 0
x(x - 2) = 0
x = 0 or x = 2
Step 2: Check if curve is above or below x-axis.
At x = 1: y = 1 - 2 = -1
The curve is below the x-axis on [0, 2]
Step 3: Set up the integral.
Area = |∫₀² (x² - 2x) dx|
= -∫₀² (x² - 2x) dx (since function is negative)
Step 4: Evaluate.
= -[x³/3 - x²]₀²
= -[(8/3 - 4) - 0]
= -[8/3 - 12/3]
= -[-4/3]
= 4/3 square units
Final Answer: 4/3 square units
Q6. Find the volume of the solid generated by rotating the region bounded by y = √x, x = 4, x-axis about the x-axis.
To find the volume using disk method:
Step 1: Identify the setup.
Rotating y = √x from x = 0 to x = 4 about x-axis
Radius at position x: r(x) = √x
Step 2: Apply disk method formula.
V = π ∫₀⁴ [r(x)]² dx
= π ∫₀⁴ (√x)² dx
= π ∫₀⁴ x dx
Step 3: Evaluate.
= π [x²/2]₀⁴
= π × (16/2 - 0)
= 8π cubic units
Final Answer: 8π cubic units
Q7. Find the area between the curves y = e^x and y = e^(-x) from x = 0 to x = 1.
To find the area between y = e^x and y = e^(-x):
Step 1: Determine which curve is above.
At x = 0.5:
y = e^0.5 ≈ 1.65
y = e^(-0.5) ≈ 0.61
So e^x is above e^(-x)
Step 2: Set up the integral.
Area = ∫₀¹ (e^x - e^(-x)) dx
Step 3: Evaluate.
= [e^x - (-e^(-x))]₀¹
= [e^x + e^(-x)]₀¹
= (e¹ + e^(-1)) - (e⁰ + e⁰)
= (e + 1/e) - 2
= e + 1/e - 2 square units
Numerically:
= 2.718 + 0.368 - 2
= 1.086 square units
Final Answer: (e + 1/e - 2) square units ≈ 1.09 square units
Q8. Find the area of the ellipse x²/a² + y²/b² = 1.
To find the area of the ellipse x²/a² + y²/b² = 1:
Step 1: Use symmetry.
The ellipse is symmetric about both axes.
Area = 4 × (area in first quadrant)
Step 2: Express y in terms of x.
From x²/a² + y²/b² = 1:
y = (b/a)√(a² - x²) (first quadrant)
Step 3: Set up integral for first quadrant.
Area (first quadrant) = ∫₀ᵃ (b/a)√(a² - x²) dx
Step 4: Evaluate.
Let x = a sin θ, dx = a cos θ dθ
When x = 0, θ = 0; when x = a, θ = π/2
= ∫₀^(π/2) (b/a) × a cos θ × a cos θ dθ
= ab ∫₀^(π/2) cos²θ dθ
= ab ∫₀^(π/2) (1 + cos 2θ)/2 dθ
= (ab/2) × [θ + sin 2θ/2]₀^(π/2)
= (ab/2) × π/2
= πab/4
Step 5: Multiply by 4 for total area.
Total Area = 4 × πab/4 = πab
Final Answer: πab square units
Application of 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. The area bounded by the curve y = 2x, the x-axis, and the ordinates x = 1 and x = 3 is:
- A. 2 sq units
- B. 4 sq units
- C. 6 sq units
- D. 8 sq units
Answer: D. 8 sq units — The area is given by the definite integral ∫(2x)dx from 1 to 3. This evaluates to [x²] from 1 to 3, which is 3² - 1² = 9 - 1 = 8.
MCQ 2. Find the area bounded by the parabola y = x², the x-axis, and the lines x = 0 and x = 2.
- A. 8/3 sq units
- B. 4/3 sq units
- C. 2 sq units
- D. 8 sq units
Answer: A. 8/3 sq units — The area is ∫(x²)dx from 0 to 2. This evaluates to [x³/3] from 0 to 2 = (2³/3) - (0³/3) = 8/3.
MCQ 3. The area of the region bounded by x² + y² = 4, the x-axis, and the lines x = 0 and x = 2 in the first quadrant is:
- A. π sq units
- B. 2π sq units
- C. 4π sq units
- D. π/2 sq units
Answer: A. π sq units — The equation x² + y² = 4 represents a circle of radius 2. The area in the first quadrant from x=0 to x=2 is one-fourth of the total circle's area, which is (1/4)πr² = (1/4)π(2²) = π.
MCQ 4. The area of the region bounded by the ellipse x²/16 + y²/9 = 1 is:
- A. 12π sq units
- B. 6π sq units
- C. 24π sq units
- D. 36π sq units
Answer: A. 12π sq units — The area of an ellipse x²/a² + y²/b² = 1 is given by the formula πab. Here, a=4 and b=3, so the area is π(4)(3) = 12π.
MCQ 5. Find the area bounded by the parabola y = x² and the line y = 4.
- A. 32/3 sq units
- B. 16/3 sq units
- C. 8/3 sq units
- D. 4 sq units
Answer: A. 32/3 sq units — The intersection points are x² = 4, so x = ±2. The area is ∫(4 - x²)dx from -2 to 2, which by symmetry is 2∫(4 - x²)dx from 0 to 2 = 32/3.
MCQ 6. The area bounded by the curve y = |x|, the x-axis, and the lines x = -1 and x = 1 is:
- A. 1 sq unit
- B. 2 sq units
- C. 1/2 sq unit
- D. 0 sq units
Answer: A. 1 sq unit — The area is ∫|x|dx from -1 to 1. Due to symmetry, this is 2∫x dx from 0 to 1 = 2[x²/2] from 0 to 1 = 1.
MCQ 7. The area of the region bounded by the parabolas y = x² and x = y² is:
- A. 1/3 sq unit
- B. 2/3 sq units
- C. 1 sq unit
- D. 4/3 sq units
Answer: A. 1/3 sq unit — The intersection points are (0,0) and (1,1). The area is ∫(√x - x²)dx from 0 to 1, which evaluates to 1/3.
MCQ 8. The area bounded by the curve y = |x - 1|, the x-axis, and the lines x = 0 and x = 2 is:
- A. 1 sq unit
- B. 2 sq units
- C. 1/2 sq unit
- D. 3/2 sq units
Answer: A. 1 sq unit — The area is ∫|x-1|dx from 0 to 2. This integral can be split into ∫-(x-1)dx from 0 to 1 and ∫(x-1)dx from 1 to 2. Each integral evaluates to 1/2, so the total area is 1.
MCQ 9. The area of the region bounded by the curve y = cos x, the x-axis, and the lines x = 0 and x = π/2 is:
- A. 0 sq units
- B. 1 sq unit
- C. π/2 sq units
- D. 2 sq units
Answer: B. 1 sq unit — The area is ∫cos(x)dx from 0 to π/2. This evaluates to [sin(x)] from 0 to π/2 = sin(π/2) - sin(0) = 1 - 0 = 1.
MCQ 10. Find the area bounded by the curve x = y², the y-axis, and the lines y = 1 and y = 3.
- A. 8/3 sq units
- B. 26/3 sq units
- C. 28/3 sq units
- D. 30/3 sq units
Answer: B. 26/3 sq units — The area is ∫(y²)dy from 1 to 3. This evaluates to [y³/3] from 1 to 3 = (3³/3) - (1³/3) = 27/3 - 1/3 = 26/3.
MCQ 11. The area of the triangle with vertices (0,0), (1,0), and (1,2) using integration is:
- A. 1 sq unit
- B. 2 sq units
- C. 1/2 sq unit
- D. 3/2 sq units
Answer: A. 1 sq unit — The line joining (0,0) and (1,2) is y = 2x. The area is ∫(2x)dx from 0 to 1 = [x²] from 0 to 1 = 1.
MCQ 12. The area bounded by the parabola y² = x and the line x = 4 is:
- A. 8/3 sq units
- B. 16/3 sq units
- C. 32/3 sq units
- D. 64/3 sq units
Answer: C. 32/3 sq units — The parabola is symmetric about the x-axis. The area is 2∫√x dx from 0 to 4 = 2[(2/3)x^(3/2)] from 0 to 4 = (4/3)[4^(3/2)] = (4/3)(8) = 32/3.
MCQ 13. The area bounded by the curve y = x² and the line y = x is:
- A. 1/6 sq unit
- B. 1/3 sq unit
- C. 1/2 sq unit
- D. 1 sq unit
Answer: A. 1/6 sq unit — The intersection points are x² = x, which gives x=0 and x=1. The area is ∫(x - x²)dx from 0 to 1 = [x²/2 - x³/3] from 0 to 1 = (1/2 - 1/3) = 1/6.
MCQ 14. The area of the region bounded by the curve y² = 4x and the line y = 2x is:
- A. 1/6 sq unit
- B. 1/3 sq unit
- C. 1/2 sq unit
- D. 2/3 sq units
Answer: B. 1/3 sq unit — Intersection points: (2x)² = 4x => 4x² = 4x => x=0 or x=1. The area is ∫(√4x - 2x)dx from 0 to 1 = ∫(2√x - 2x)dx from 0 to 1 = 1/3.
MCQ 15. The area of the region bounded by y = 0, x = 0, x = 1, and y = e^x is:
- A. e sq units
- B. 1 sq unit
- C. e-1 sq units
- D. e+1 sq units
Answer: C. e-1 sq units — The area is given by ∫(e^x)dx from 0 to 1. This evaluates to [e^x] from 0 to 1 = e¹ - e⁰ = e - 1.
MCQ 16. The area bounded by the curve x = 4 - y², the y-axis, and the lines y = 0 and y = 2 is:
- A. 8/3 sq units
- B. 16/3 sq units
- C. 32/3 sq units
- D. 64/3 sq units
Answer: B. 16/3 sq units — The area is given by ∫(4 - y²)dy from 0 to 2. This evaluates to [4y - y³/3] from 0 to 2 = (4(2) - 2³/3) - 0 = 8 - 8/3 = 16/3.
MCQ 17. The area enclosed by the curve y = √x, the x-axis, and the line x = 9 is:
- A. 9 sq units
- B. 18 sq units
- C. 27 sq units
- D. 18/3 sq units
Answer: B. 18 sq units — The area is ∫√x dx from 0 to 9. This evaluates to [(2/3)x^(3/2)] from 0 to 9 = (2/3)(9^(3/2)) = (2/3)(27) = 18.
MCQ 18. The area bounded by the curve y = x³ and the x-axis from x = -1 to x = 1 is:
- A. 0 sq units
- B. 1/2 sq unit
- C. 1 sq unit
- D. 2 sq units
Answer: B. 1/2 sq unit — The function y=x³ crosses the x-axis at x=0. The area is ∫|x³|dx from -1 to 1, which is ∫(-x³)dx from -1 to 0 + ∫(x³)dx from 0 to 1. Each integral evaluates to 1/4, so the total area is 1/2.
MCQ 19. The area bounded by the curve x = 2y and the y-axis from y = 0 to y = 2 is:
- A. 2 sq units
- B. 4 sq units
- C. 6 sq units
- D. 8 sq units
Answer: B. 4 sq units — The area is given by ∫(2y)dy from 0 to 2. This evaluates to [y²] from 0 to 2 = 2² - 0² = 4.
MCQ 20. The area of the region bounded by y = x + 2, the lines x = -1 and x = 1, and the x-axis is:
- A. 2 sq units
- B. 3 sq units
- C. 4 sq units
- D. 5 sq units
Answer: C. 4 sq units — The line y=x+2 is above the x-axis in the interval [-1, 1]. The area is ∫(x+2)dx from -1 to 1. This evaluates to [x²/2 + 2x] from -1 to 1 = (1/2 + 2) - (1/2 - 2) = 4.
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.