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Application of Integrals

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Chapter Notes

Overview
This chapter applies the concept of definite integrals to calculate areas bounded by curves, lines, and axes. It extends elementary geometry formulas to find areas enclosed by curves like circles, ellipses, and parabolas, which cannot be found using simple geometric formulas. The one big idea is that the area under a curve y = f(x) from x = a to x = b is given by the definite integral ∫_a^b f(x) dx, and this can be adapted for vertical or horizontal strips. Students must understand how to set up integrals correctly, handle areas below the x-axis, and use symmetry to simplify calculations.
Key Concepts (10)
Elementary area
The area of an arbitrary thin vertical strip of height y and width dx, given by dA = y dx, where y = f(x).
e.g. For the curve y = x² between x = 1 and x = 2, the elementary area of a strip at x = 1.5 is dA = (1.5)² dx = 2.25 dx.
Area under simple curves
The total area A of the region bounded by the curve y = f(x), the x-axis, and the ordinates x = a and x = b is given by A = ∫_a^b y dx = ∫_a^b f(x) dx.
e.g. Area under y = x² from x = 1 to x = 2 is ∫_1^2 x² dx = [x³/3]_1^2 = (8/3 - 1/3) = 7/3 square units.
Area bounded by curve x = g(y)
The area A of the region bounded by the curve x = g(y), the y-axis, and the lines y = c and y = d is given by A = ∫_c^d x dy = ∫_c^d g(y) dy, using horizontal strips.
e.g. For the curve x = y² from y = 0 to y = 2, area = ∫_0^2 y² dy = [y³/3]_0^2 = 8/3 square units.
Area below x-axis
If the curve lies below the x-axis (f(x) < 0) from x = a to x = b, the definite integral gives a negative value, but the numerical area is taken as its absolute value: Area = |∫_a^b f(x) dx|.
e.g. For y = -x² from x = 0 to x = 1, ∫_0^1 (-x²) dx = -1/3, so area = 1/3 square units.
Area with mixed signs
When part of the curve is above the x-axis (A₂ > 0) and part is below (A₁ < 0), the total area is the sum of absolute values: A = |A₁| + A₂.
e.g. For y = sin x from x = 0 to x = 2π, area = |∫_0^π sin x dx| + ∫_π^2π sin x dx = 2 + 2 = 4 square units.
Symmetry in area calculation
If a curve is symmetric about the x-axis or y-axis, the total area can be found by multiplying the area in one quadrant by the appropriate factor (e.g., 2 or 4).
e.g. For the circle x² + y² = a², total area = 4 × area in first quadrant = 4 ∫_0^a √(a² - x²) dx = πa².
Vertical strips
Thin rectangles of width dx and height y = f(x) used to approximate the area under a curve when integrating with respect to x.
e.g. For y = √(a² - x²) from x = 0 to x = a, vertical strips give area = ∫_0^a √(a² - x²) dx.
Horizontal strips
Thin rectangles of width dy and length x = g(y) used to approximate the area when integrating with respect to y.
e.g. For the circle x² + y² = a², horizontal strips give area = ∫_0^a √(a² - y²) dy.
Definite integral as limit of a sum
The definite integral ∫_a^b f(x) dx represents the limit of the sum of elementary areas as the number of strips approaches infinity and their width approaches zero.
e.g. Area under y = x from x = 0 to x = 1 = lim_{n→∞} Σ (i/n)(1/n) = 1/2.
Ordinates
The vertical lines x = a and x = b that bound the region whose area is being calculated.
e.g. In finding area under y = x² from x = 1 to x = 3, the ordinates are x = 1 and x = 3.
Formulas & Equations (6)
A = ∫_a^b y dx = ∫_a^b f(x) dx
Reads as: Area A equals the integral from a to b of y dx, which equals the integral from a to b of f of x dx
Variables: A = area (square units), y = f(x) = height of strip (units), dx = width of strip (units), a = lower limit of x, b = upper limit of x
Use when: Use when the region is bounded by the curve y = f(x), the x-axis, and vertical lines x = a and x = b (vertical strips).
⚠ Forgetting to take absolute value when the curve lies below the x-axis, leading to a negative area.
A = ∫_c^d x dy = ∫_c^d g(y) dy
Reads as: Area A equals the integral from c to d of x dy, which equals the integral from c to d of g of y dy
Variables: A = area (square units), x = g(y) = length of horizontal strip (units), dy = height of strip (units), c = lower limit of y, d = upper limit of y
Use when: Use when the region is bounded by the curve x = g(y), the y-axis, and horizontal lines y = c and y = d (horizontal strips).
⚠ Confusing the limits: using x-limits instead of y-limits when integrating with respect to y.
Area of circle: πa²
Reads as: Area of a circle is pi times a squared
Variables: a = radius of the circle (units)
Use when: Use to verify the result of integrating to find area enclosed by circle x² + y² = a².
⚠ Forgetting that the integral gives πa²/4 for one quadrant, so total area is 4 times that.
Area of ellipse: πab
Reads as: Area of an ellipse is pi times a times b
Variables: a = semi-major axis (units), b = semi-minor axis (units)
Use when: Use to verify the result of integrating to find area enclosed by ellipse x²/a² + y²/b² = 1.
⚠ Writing πa²b or πab² instead of πab.
∫ √(a² - x²) dx = (x/2)√(a² - x²) + (a²/2) sin⁻¹(x/a) + C
Reads as: Integral of square root of a squared minus x squared dx equals x over 2 times square root of a squared minus x squared plus a squared over 2 times inverse sine of x over a plus constant
Variables: a = constant (units), x = variable (units), C = constant of integration
Use when: Use when integrating to find area of a circle or ellipse using vertical strips.
⚠ Forgetting the factor a²/2 in the second term, or misplacing the inverse sine argument.
∫ √(a² - y²) dy = (y/2)√(a² - y²) + (a²/2) sin⁻¹(y/a) + C
Reads as: Integral of square root of a squared minus y squared dy equals y over 2 times square root of a squared minus y squared plus a squared over 2 times inverse sine of y over a plus constant
Variables: a = constant (units), y = variable (units), C = constant of integration
Use when: Use when integrating to find area of a circle or ellipse using horizontal strips.
⚠ Confusing the variable with x when the integration is with respect to y.
How It Works
Finding area under a curve using vertical strips
  1. Identify the curve y = f(x), the x-axis, and the ordinates x = a and x = b.
  2. Consider an elementary vertical strip of height y and width dx at an arbitrary x between a and b.
  3. Write the elementary area as dA = y dx = f(x) dx.
  4. Integrate from x = a to x = b: A = ∫_a^b f(x) dx.
  5. If the curve lies below the x-axis for some interval, take the absolute value of the integral for that part.
💡 Always check if the curve crosses the x-axis within the interval — if it does, split the integral and take absolute values for negative parts.
Finding area bounded by curve x = g(y) using horizontal strips
  1. Identify the curve x = g(y), the y-axis, and the lines y = c and y = d.
  2. Consider an elementary horizontal strip of length x and height dy at an arbitrary y between c and d.
  3. Write the elementary area as dA = x dy = g(y) dy.
  4. Integrate from y = c to y = d: A = ∫_c^d g(y) dy.
  5. If the curve lies to the left of the y-axis for some interval, take the absolute value of the integral for that part.
💡 The limits of integration are y-values, not x-values — a common source of error.
Finding area enclosed by a circle using symmetry
  1. Write the equation of the circle: x² + y² = a².
  2. Note symmetry about both axes, so total area = 4 × area in first quadrant.
  3. In the first quadrant, y = √(a² - x²) from x = 0 to x = a.
  4. Set up the integral: Area = 4 ∫_0^a √(a² - x²) dx.
  5. Evaluate using the standard formula: ∫ √(a² - x²) dx = (x/2)√(a² - x²) + (a²/2) sin⁻¹(x/a).
  6. Apply limits: at x = a, √(a² - a²) = 0, sin⁻¹(1) = π/2; at x = 0, both terms are 0.
  7. Result: Area = 4 × (a²/2)(π/2) = πa².
💡 The formula for the integral of √(a² - x²) must be memorised — it appears frequently in board exams.
Finding area enclosed by an ellipse using symmetry
  1. Write the equation of the ellipse: x²/a² + y²/b² = 1.
  2. Note symmetry about both axes, so total area = 4 × area in first quadrant.
  3. In the first quadrant, y = (b/a)√(a² - x²) from x = 0 to x = a.
  4. Set up the integral: Area = 4 ∫_0^a (b/a)√(a² - x²) dx = (4b/a) ∫_0^a √(a² - x²) dx.
  5. Evaluate using the standard formula: ∫ √(a² - x²) dx = (x/2)√(a² - x²) + (a²/2) sin⁻¹(x/a).
  6. Apply limits: at x = a, √(a² - a²) = 0, sin⁻¹(1) = π/2; at x = 0, both terms are 0.
  7. Result: Area = (4b/a) × (a²/2)(π/2) = πab.
💡 The factor b/a comes from solving y = (b/a)√(a² - x²) — do not forget it.
Finding area when curve crosses x-axis (mixed signs)
  1. Find the x-coordinate(s) where the curve crosses the x-axis by solving f(x) = 0.
  2. Split the interval [a, b] into subintervals where f(x) is entirely positive or entirely negative.
  3. For each subinterval where f(x) is negative, compute the integral and take its absolute value.
  4. For each subinterval where f(x) is positive, compute the integral directly.
  5. Add all absolute values to get the total area.
  6. Example: For y = 3x + 2 from x = -1 to x = 1, the curve crosses x-axis at x = -2/3. Area = |∫_{-1}^{-2/3} (3x+2) dx| + ∫_{-2/3}^1 (3x+2) dx.
💡 Never simply integrate from a to b if the curve crosses the axis — you must split and take absolute values.
Finding area bounded by y = cos x from x = 0 to x = 2π
  1. Identify the intervals where cos x is positive and negative: positive from 0 to π/2 and 3π/2 to 2π; negative from π/2 to 3π/2.
  2. Split the integral into three parts: ∫_0^{π/2} cos x dx, ∫_{π/2}^{3π/2} cos x dx, ∫_{3π/2}^{2π} cos x dx.
  3. Evaluate each: ∫ cos x dx = sin x.
  4. First part: sin(π/2) - sin(0) = 1 - 0 = 1.
  5. Second part: sin(3π/2) - sin(π/2) = (-1) - 1 = -2, take absolute value = 2.
  6. Third part: sin(2π) - sin(3π/2) = 0 - (-1) = 1.
  7. Total area = 1 + 2 + 1 = 4 square units.
💡 The area under one full wave of cos x from 0 to 2π is 4, not 0 — the integral itself is 0, but area is the sum of absolute values.
Exam Traps — Don't Make These Mistakes
⚠ Forgetting absolute value for area below x-axis
✗ Area = ∫_a^b f(x) dx even when f(x) is negative, giving a negative area.
→ Area is always positive; the definite integral gives signed area. If the curve is below the x-axis, the integral is negative, but the actual area is its absolute value.
✓ Area = |∫_a^b f(x) dx| when the curve lies entirely below the x-axis, or split the integral and take absolute values for negative parts.
⚠ Using wrong limits for horizontal strips
✗ For area bounded by x = g(y), y-axis, and y = c, y = d, they write A = ∫_a^b x dy using x-limits instead of y-limits.
→ When integrating with respect to y, the limits must be y-values (c and d), not x-values. The integral is ∫_c^d x dy.
✓ A = ∫_c^d x dy = ∫_c^d g(y) dy, where c and d are the y-coordinates of the bounding lines.
⚠ Not splitting integral when curve crosses x-axis
✗ For y = 3x + 2 from x = -1 to x = 1, they write A = ∫_{-1}^1 (3x+2) dx = [ (3x²/2) + 2x ]_{-1}^1 = (3/2+2) - (3/2-2) = 4, which is wrong.
→ The curve crosses the x-axis at x = -2/3. The integral from -1 to 1 includes both positive and negative parts, and the negative part must be taken as absolute value. The correct area is 13/3, not 4.
✓ Find the x-intercept, split the integral at that point, take absolute value of the negative part: A = |∫_{-1}^{-2/3} (3x+2) dx| + ∫_{-2/3}^1 (3x+2) dx.
⚠ Confusing area of ellipse with area of circle
✗ Area of ellipse x²/a² + y²/b² = 1 is πa² or πb².
→ The area of an ellipse is πab, not πa² or πb². The circle is a special case where a = b = radius.
✓ Area of ellipse = πab, where a is the semi-major axis and b is the semi-minor axis.
⚠ Misapplying symmetry factor
✗ For the circle x² + y² = a², they write total area = 2 × area in first quadrant instead of 4.
→ The circle is symmetric about both axes, so the area in the first quadrant is one-fourth of the total area, not one-half.
✓ Total area = 4 × area in first quadrant = 4 ∫_0^a √(a² - x²) dx.
⚠ Forgetting the constant factor in standard integral
✗ ∫ √(a² - x²) dx = (x/2)√(a² - x²) + sin⁻¹(x/a) + C, omitting the a²/2 factor.
→ The correct formula is (x/2)√(a² - x²) + (a²/2) sin⁻¹(x/a) + C. The a²/2 factor is essential for correct evaluation.
✓ Memorise the exact formula: ∫ √(a² - x²) dx = (x/2)√(a² - x²) + (a²/2) sin⁻¹(x/a) + C.
Compare & Contrast
Method of setting up the integral
Vertical strips (integrating w.r.t. x)
Horizontal strips (integrating w.r.t. y)
Vertical strips use dA = y dx and integrate from x = a to x = b; horizontal strips use dA = x dy and integrate from y = c to y = d. Choose vertical strips when the curve is given as y = f(x) and horizontal strips when the curve is given as x = g(y).
Formula for total area
Area of circle (x² + y² = a²)
Area of ellipse (x²/a² + y²/b² = 1)
Area of circle = πa² (where a is radius); area of ellipse = πab (where a and b are semi-axes). The circle is a special case of ellipse with a = b.
Exam Focus
Marks pattern: 1-2 MCQs (1 mark each), 1 Short Answer (3 marks), 1 Long Answer (5 marks) — total approximately 8-10 marks
Most asked: Finding area bounded by a curve and the x-axis using vertical strips · Finding area bounded by a curve and the y-axis using horizontal strips · Area enclosed by a circle or ellipse using symmetry · Area when curve crosses the x-axis (mixed signs) · Area bounded by trigonometric curves like y = sin x or y = cos x
Don't skip: Long answer (5 marks) on finding area enclosed by an ellipse or circle, often requiring symmetry and standard integral evaluation.; Long answer (5 marks) on area bounded by a line and a curve, or a curve that crosses the x-axis, requiring splitting of integrals.
One-Line Revision (read aloud before the exam)
  1. Area under curve y = f(x) from x = a to x = b is A = ∫_a^b f(x) dx.
  2. Area bounded by curve x = g(y) from y = c to y = d is A = ∫_c^d g(y) dy.
  3. If the curve lies below the x-axis, take the absolute value of the integral.
  4. If the curve crosses the x-axis, split the integral and add absolute values of negative parts.
  5. Area of circle x² + y² = a² is πa².
  6. Area of ellipse x²/a² + y²/b² = 1 is πab.
  7. Use symmetry: for a circle or ellipse, total area = 4 × area in first quadrant.
  8. Vertical strips: dA = y dx, integrate with respect to x.
  9. Horizontal strips: dA = x dy, integrate with respect to y.
  10. Standard integral: ∫ √(a² - x²) dx = (x/2)√(a² - x²) + (a²/2) sin⁻¹(x/a) + C.
  11. For y = sin x from 0 to 2π, total area = 4 (split at π).
  12. For y = cos x from 0 to 2π, total area = 4 (split at π/2 and 3π/2).
  13. The elementary area is the area of a thin strip of height y and width dx.
  14. Always check if the curve is above or below the axis before integrating.
  15. The definite integral is the limit of a sum of elementary areas.

MCQs — 10 Questions with Answers

Q1. The area of the region bounded by the ellipse 4x² + 9y² = 36 is:

  • A. 6π ✓
  • B. 12π
  • C. 18π
  • D. 36π

Answer: A — Rewrite as x²/9 + y²/4 = 1; here a² = 9 (so a = 3) and b² = 4 (so b = 2); Area = πab = π(3)(2) = 6π.

Q2. Find the area bounded by the curve y = x², the x-axis, and the ordinates x = 1 and x = 2.

  • A. 7/3 ✓
  • B. 8/3
  • C. 7/2
  • D. 5/3

Answer: A — Area = ∫₁² x² dx = [x³/3]₁² = (8/3) − (1/3) = 7/3.

Q3. The area under the curve y = cos x from x = 0 to x = π/2 is:

  • A. 0
  • B. 1 ✓
  • C. π/2
  • D. 2

Answer: B — Area = ∫₀^(π/2) cos x dx = [sin x]₀^(π/2) = sin(π/2) − sin(0) = 1 − 0 = 1.

Q4. For the line y = 2x + 1, the x-axis, and ordinates x = 0 and x = 2, the area is:

  • A. 6 ✓
  • B. 8
  • C. 10
  • D. 12

Answer: A — Area = ∫₀² (2x + 1) dx = [x² + x]₀² = (4 + 2) − 0 = 6.

Q5. Which of the following is NOT a correct statement about area calculations?

  • A. Area is always taken as absolute value of the definite integral
  • B. For regions below the x-axis, we integrate normally without absolute value ✓
  • C. Horizontal strips are used when x is expressed as a function of y
  • D. Symmetric regions can be calculated using one quadrant and multiplying by 4

Answer: B — Option B is incorrect because when a region lies below the x-axis, the integral gives a negative value, so we must take its absolute value to get the true area.

Q6. The area of a quarter circle of radius a (in the first quadrant) is:

  • A. πa²/4 ✓
  • B. πa²/2
  • C. πa²
  • D. 2πa²

Answer: A — Full circle area is πa²; one quarter is πa²/4, calculated as ∫₀ᵃ √(a² − x²) dx = (πa²/4).

Q7. If the curve y = |x| is bounded by the x-axis and the lines x = −2 and x = 2, the area is:

  • A. 2
  • B. 4 ✓
  • C. 6
  • D. 8

Answer: B — Area = ∫₋₂⁰ (−x) dx + ∫₀² x dx = [−x²/2]₋₂⁰ + [x²/2]₀² = 2 + 2 = 4 (using symmetry).

Q8. The area bounded by the parabola y² = 4x, the x-axis, and the line x = 1 is:

  • A. 4/3
  • B. 8/3 ✓
  • C. 16/3
  • D. 2

Answer: B — Using horizontal strips: Area = 2∫₀² x dy = 2∫₀² (y²/4) dy = (1/2)[y³/3]₀² = (1/2)(8/3) = 4/3 is incorrect; correct is ∫₀⁴ √(y) dy evaluated gives 8/3.

Q9. If a curve crosses the x-axis at x = c (where a < c < b), and f(c) = 0, then the total area between the curve and x-axis from x = a to x = b is calculated as:

  • A. ∫ₐᵇ f(x) dx only
  • B. |∫ₐᶜ f(x) dx| + |∫ₓᵇ f(x) dx|
  • C. ∫ₐᶜ |f(x)| dx + ∫ₓᵇ |f(x)| dx
  • D. Both B and C are equivalent and correct ✓

Answer: D — When the curve crosses the x-axis, we must split the integral and take absolute values; options B and C are mathematically equivalent approaches to ensure areas are positive.

Q10. **HOTS**: The area between the curves y = x² and y = 2x is:

  • A. 1/3
  • B. 2/3
  • C. 4/3 ✓
  • D. 8/3

Answer: C — Find intersections: x² = 2x gives x = 0 or x = 2; upper curve is y = 2x, lower is y = x²; Area = ∫₀² (2x − x²) dx = [x² − x³/3]₀² = 4 − 8/3 = 4/3.

Flashcards

What is the formula for area under curve y = f(x) between x = a and x = b?

Area = ∫ₐᵇ f(x) dx = ∫ₐᵇ y dx, using vertical strips of height y and width dx.

How do you find area using horizontal strips?

Area = ∫ₓ xdy = ∫ₓ g(y) dy, where x = g(y) and limits are y = c to y = d on the y-axis.

What is the area enclosed by circle x² + y² = a²?

Area = πa², found by integrating 4∫₀ᵃ √(a² − x²) dx using the substitution x = a sin θ.

What is the area enclosed by ellipse x²/a² + y²/b² = 1?

Area = πab, found by integrating 4∫₀ᵃ (b/a)√(a² − x²) dx using symmetry about both axes.

When a curve lies below the x-axis, how is area calculated?

Since f(x) < 0, the integral gives negative value; take absolute value |∫ₐᵇ f(x) dx| for actual area.

If a curve crosses the x-axis between x = a and x = b, what do you do?

Find intersection points, split the integral at these points, and add absolute values: |∫ₐᶜ f(x) dx| + |∫ₐᵈ f(x) dx|.

What does using vertical strips mean in area calculation?

Vertical strip has height y = f(x), width dx; area element dA = y dx; integrate with respect to x.

What does using horizontal strips mean in area calculation?

Horizontal strip has width x = g(y), height dy; area element dA = x dy; integrate with respect to y.

How is area between two curves y₁ = f(x) and y₂ = g(x) found, where f(x) > g(x)?

Area = ∫ₐᵇ [f(x) − g(x)] dx, integrating the difference between upper and lower curves.

Why use absolute value in area calculations?

Area is always positive; definite integral can be negative when region is below x-axis, so |∫ f(x) dx| ensures correct magnitude.

Important Board Questions

Define the area under a curve. Write the formula for area bounded by the curve y = f(x), the x-axis, and the ordinates x = a and x = b. Give one example. [2 marks]

State that area is the definite integral ∫ₐᵇ f(x) dx; use vertical strip method (dA = y dx); example: area under y = x from x = 0 to x = 2 is ∫₀² x dx = 2.

Find the area of the region bounded by the ellipse x²/16 + y²/9 = 1. Show all steps of integration and use the symmetry of the ellipse. [5 marks]

Identify a² = 16 (a = 4) and b² = 9 (b = 3); use formula Area = πab or integrate 4∫₀⁴ (3/4)√(16 − x²) dx; apply substitution x = 4 sin θ or standard integral formula; final answer πab = 12π.

Derive the formula for the area enclosed by the circle x² + y² = a². Explain the role of symmetry and the elementary strip method in the derivation. [6 marks]

Use symmetry to calculate area of first quadrant (AOBA) and multiply by 4; set up ∫₀ᵃ √(a² − x²) dx using vertical strips; apply substitution x = a sin θ to get standard integral ∫(a²/2)(1 + cos 2θ) dθ; evaluate and combine to show final area πa².

Assertion-Reason Questions for Application of Integrals

Q1. Assertion (A):

The area of the region bounded by the curve y = f(x), x-axis and ordinates x = a, x = b is given by ∫ₐᵇ f(x) dx.

Reason (R):

The definite integral ∫ₐᵇ f(x) dx represents the limit of the sum of areas of infinitesimal rectangles under the curve.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both the assertion and reason are correct, and the reason fully explains why the formula works: the definite integral is defined as the limit of Riemann sums, which approximates area by summing rectangular strips.

Q2. Assertion (A):

When a curve lies entirely below the x-axis between x = a and x = b, the area bounded by the curve and x-axis is negative.

Reason (R):

The definite integral ∫ₐᵇ f(x) dx gives a negative value when f(x) < 0, and this negative value represents the area.

Answer: D — A is false but R is true — The assertion is false because area is always positive (a geometric property). While the integral is negative when f(x) < 0, we take its absolute value to obtain the actual area. The reason correctly explains what the integral gives, but incorrectly identifies this as the area itself.

Q3. Assertion (A):

The area of an ellipse with semi-major axis a and semi-minor axis b is πab.

Reason (R):

This area is obtained by evaluating 4∫₀ᵃ (b/a)√(a² - x²) dx using the substitution x = a sin θ.

Answer: B — Both A and R are true, but R is NOT the correct explanation of A — Both the assertion and reason are true. The formula πab is correct, and the integral formula with substitution does yield this result. However, the reason is not the only way to derive this; it could also be derived from the general ellipse equation or scaling arguments from the circle formula.

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Case-Based Questions (CBQ) for Application of Integrals

Case Study:

A city planner is designing a circular public park with radius 50 metres. The park has a circular boundary with equation x² + y² = 2500 (where coordinates are in metres). The planner needs to calculate the total area to determine how much grass seed is required and the cost of maintaining the park. Using integral calculus, the area is computed by taking four times the area in the first quadrant. The park will be divided into four equal sections by two perpendicular pathways along the x and y axes.

Q1. What is the total area of the circular park using the formula for the area of a circle with radius 50?

  • A. 2500π m² ✓
  • B. 5000π m²
  • C. 7500π m²
  • D. 10000π m²

Answer: A

Q2. If the planner calculates the area using 4∫₀⁵⁰ √(2500 - x²) dx, what does the factor '4' represent?

  • A. Four pathways dividing the park
  • B. Multiplication by 4 because the first quadrant area is 1/4 of the total circle ✓
  • C. Four equal sections created by perpendicular pathways
  • D. The radius of the park is 4 times some unit

Answer: B

Q3. Each section of the park (divided by the axes) has an area of:

  • A. 625π m² ✓
  • B. 1250π m²
  • C. 2500π m²
  • D. 5000π m²

Answer: A

Q4. If grass seed costs ₹50 per square metre, the total cost for seeding the entire park is approximately:

  • A. ₹39,27,000 (using π ≈ 3.14)
  • B. ₹78,54,000 (using π ≈ 3.14) ✓
  • C. ₹19,63,500 (using π ≈ 3.14)
  • D. ₹1,25,00,000 (using π ≈ 3.14)

Answer: B

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Previous Year Questions (PYQ) for Application of Integrals

Real CBSE board exam questions from past papers — free to attempt.

CBSE 2026 · Set 65/1/1 · 1 mark

The area bounded by the curve y = x|x|, x-axis and the ordinates x = –1 and x = 1 is given by

  • 0
  • 1/3
  • 2/3
  • 4/3

CBSE 2026 · Set 65/1/1 · 1 mark

Roundabouts are often made on busy roads to ease the traffic and avoid red lights. One such round-about is made such that equation representing its boundary is given by C₁: x² + y² = 64. There is a circular pond with a fountain in the middle of the roundabout whose equation is given by C₂: x² + y² = 4.

Roundabouts are often made on busy roads to ease the traffic and avoid red lights. One such round-about is made such that equation representing its boundary is given by C₁: x² + y² = 64. There is a circular pond with a fountain in the middle of the roundabout whose equation is given by C₂: x² + y² = 4. Represent the given equations C₁ and C₂ with the help of a diagram.

CBSE 2026 · Set 65/1/2 · 1 mark

Roundabouts are often made on busy roads to ease the traffic and avoid red lights. One such roundabout is made such that equation representing its boundary is given by C₁: x² + y² = 64. There is a circular pond with a fountain in the middle of the roundabout whose equation is given by C₂: x² + y² = 4.

Represent the given equations C₁ and C₂ with the help of a diagram.

CBSE 2026 · Set 65/1/3 · 1 mark

Roundabouts are often made on busy roads to ease the traffic and avoid red lights. One such round-about is made such that equation representing its boundary is given by C₁: x² + y² = 64. There is a circular pond with a fountain in the middle of the roundabout whose equation is given by C₂: x² + y² = 4.

Represent the given equations C₁: x² + y² = 64 and C₂: x² + y² = 4 with the help of a diagram.

CBSE 2026 · Set 65/2/1 · 1 mark

Which of the following expressions will give the area of region bounded by the curve y = x² and line y = 16?

  • ∫₀⁴ x² dx
  • 2∫₀⁴ x² dx
  • ∫₀¹⁶ √y dy
  • 2∫₀¹⁶ √y dy

CBSE 2026 · Set 65/2/1 · 5 marks

A rectangle of perimeter 36 cm is revolved around one of its sides to sweep out a cylinder of maximum volume. Find the dimensions of the rectangle.

CBSE 2026 · Set 65/2/1 · 1 mark

A racing track is built around an elliptical ground whose equation is given by 9x² + 16y² = 144. The width of the track is 3 m.

Express y as a function of x from the given equation of ellipse.

CBSE 2026 · Set 65/2/2 · 1 mark

A racing track is build around an elliptical ground whose equation is given by 9x² + 16y² = 144. The width of the track is 3 m.

Express y as a function of x from the given equation of ellipse 9x² + 16y² = 144.

CBSE 2026 · Set 65(B) R Mathematics · 5 marks

Find the area of the region bounded by the line y = 3x + 2, the x-axis and the ordinates x = –1 and x = 1.

CBSE 2026 · Set 65/3/1 · 1 mark

The area of the shaded region of the circle given below is equal to :

  • ∫_1^3 √(9 – y2) dy
  • ∫_1^3 2√(9 – y2) dy
  • ∫_0^3 √(9 – x2) dx
  • ∫_0^3 2√(9 – x2) dx

CBSE 2026 · Set 65/3/1 · 4 marks

There is a triangular park in the society. The park is divided into two sections. In the region OAC, children are allowed to play games like cricket, football, while in the region AOB, activities which involve running are not allowed. The vertices of the triangular park ABC are A(0, 4), B(– 2, 0) and C(3, 0).

There is a triangular park in the society. The park is divided into two sections. In the region OAC, children are allowed to play games like cricket, football, while in the region AOB, activities which involve running are not allowed. The vertices of the triangular park ABC are A(0, 4), B(– 2, 0) and C(3, 0). (i) Write the equation of the boundary line AB of the park. (ii) Write the equation of the boundary line AC of the park. (iii) (a) Using integration, find the area of region OAC, in which children are allowed to play cricket, football.

CBSE 2026 · Set 65/3/2 · 1 mark

The area of the shaded region of the circle given below is equal to:

  • ∫_1^3 √(9 − y²) dy
  • ∫_1^3 2√(9 − y²) dy
  • ∫_0^3 √(9 − x²) dx
  • ∫_0^3 2√(9 − x²) dx

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Class 12 Maths Formula Sheet

Every formula from all 13 chapters on one page, checked against the NCERT textbook and the CBSE 2026-27 syllabus. Free PDF download.

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Class 12 Mathematics Previous Year Questions, Chapter-wise

All 1048 CBSE board questions from 2022–2026, sorted by chapter and question type. Free PDF download.

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