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.
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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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Key Questions Covered:
- Find the area enclosed by the curves y = x² and y = x.
- Find the area of the region bounded by the curve y² = 4x and the line x = 1.
- Find the area of the circle x² + y² = a² using integration.
- Find the area bounded by y = sin x, the x-axis, and the lines x = 0 and x = π.
- Find the area of the region between the parabola y = x² - 2x and the x-axis.
- Find the volume of the solid generated by rotating the region bounded by y = …
- + 2 more questions in the full chapter
Solutions Summary:
| Question | Status |
|---|---|
| Find the area enclosed by the curves y = x² and y = x. | ✓ Solved |
| Find the area of the region bounded by the curve y² = 4x … | ✓ Solved |
| Find the area of the circle x² + y² = a² using integration. | ✓ Solved |
| Find the area bounded by y = sin x, the x-axis, and the l… | ✓ Solved |
| Find the area of the region between the parabola y = x² -… | ✓ Solved |
| Find the volume of the solid generated by rotating the re… | ✓ Solved |
Showing 6 of 8 questions
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/...
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^...
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
Showing 6 of 8 questions. Visit the full page for complete solutions.
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