Integral Calculus and Applications — JEE Main Questions

40 JEE Main practice questions on Integral Calculus and Applications, part of Mathematics. Below are 12 of them in full, each with the answer and a written explanation.

Questions & explanations

1. The area of the smaller segment of the circle x² + y² = 4 cut off by the line x = 1 is

  1. (4π - √3)/3
  2. (4π - 3√3)/3
  3. (2π - 3√3)/3
  4. (4π - 3√3)/6

Answer: (4π - 3√3)/3

Area = 2∫_1^2 √(4-x²) dx. Substitute x = 2 sin θ, dx = 2 cos θ dθ, limits θ = π/6 to π/2. Integral becomes 4∫_{π/6}^{π/2} cos²θ dθ = 2∫_{π/6}^{π/2} (1+cos2θ) dθ = 2[θ + sin2θ/2]_{π/6}^{π/2} = 2(π/2 - π/6 - √3/4) = 2π/3 - √3/2 for half, total = 4π/3 - √3 = (4π - 3√3)/3.

2. The area between y = sin x and y = cos x on [0, π/2] is:

  1. √2 - 1
  2. 2(√2 - 1)
  3. 2 - √2
  4. 2√2 - 2

Answer: 2(√2 - 1)

Intersection at x=π/4. On [0,π/4], cos x ≥ sin x; on [π/4,π/2], sin x ≥ cos x. Area = ∫_0^{π/4} (cos x - sin x) dx + ∫_{π/4}^{π/2} (sin x - cos x) dx = [sin x + cos x]_0^{π/4} + [-cos x - sin x]_{π/4}^{π/2} = (√2 - 1) + (1 - √2) = 2(√2 - 1).

3. The area bounded by y = x, y = 2x, and x = 2 is

  1. 1
  2. 4
  3. 2
  4. 3

Answer: 2

The area between two curves is ∫ (upper - lower) dx. Here upper is y=2x, lower is y=x, from x=0 to x=2. Area = ∫₀² (2x - x) dx = ∫₀² x dx = [x²/2]₀² = 2. Alternatively, the region is a triangle with base 2 and height 2, area = ½×2×2 = 2.

4. The area under one arch of the cycloid x = a(t - sin t), y = a(1 - cos t) is

  1. 2πa²
  2. πa²
  3. 3πa²
  4. 4πa²

Answer: 3πa²

Area = ∫ y dx = ∫₀^{2π} a(1-cos t) * a(1-cos t) dt = a² ∫₀^{2π} (1-cos t)² dt. Expand: 1 - 2cos t + cos² t = 1 - 2cos t + (1+cos2t)/2 = 3/2 - 2cos t + (cos2t)/2. Integrate: a²[(3/2)t - 2sin t + (1/4)sin2t]₀^{2π} = a²*(3/2)*2π = 3πa².

5. The area under y = 4 - x² above the x-axis is 32/3. The line y = c divides this area into two equal parts. Then c equals:

  1. 4 - 4^{2/3}
  2. 4 - 4^{1/3}
  3. 4 - 2^{4/3}
  4. 4 - 2^{2/3}

Answer: 4 - 4^{2/3}

Total area = 32/3, half = 16/3. The line y=c intersects parabola at x=±√(4-c). Area above line = ∫_{-√(4-c)}^{√(4-c)} (4-x²-c) dx = (4/3)(4-c)^{3/2}. Set equal to 16/3 → (4-c)^{3/2}=4 → 4-c = 4^{2/3} → c = 4 - 4^{2/3}.

6. The area of the region bounded by y² = 4x, the x-axis, and the line x + y = 3, where x ≥ 0, y ≥ 0, is:

  1. 14/3
  2. 8/3
  3. 10/3
  4. 4

Answer: 10/3

Using horizontal strips, the region is bounded left by the parabola (x = y²/4) and right by the line (x = 3 - y). Intersection gives y = 2. Area = ∫₀² [(3 - y) - y²/4] dy = [3y - y²/2 - y³/12]₀² = 6 - 2 - 2/3 = 10/3.

7. The area of the region in the first quadrant bounded by y = x², y = 2 - x², and y = 0 is:

  1. (4√2 - 7)/3
  2. (4√2 - 5)/3
  3. (4√2 - 1)/3
  4. (4√2 - 4)/3

Answer: (4√2 - 4)/3

The curves intersect at x=1. For 0≤x≤1, region is between y=x² and y=0, area = ∫₀¹ x² dx = 1/3. For 1≤x≤√2, region is between y=2-x² and y=0, area = ∫₁^{√2} (2-x²) dx = (4√2-5)/3. Total = 1/3 + (4√2-5)/3 = (4√2-4)/3.

8. The area of the circle x^2 + y^2 = a^2 is obtained by evaluating which integral?

  1. 2 ∫_0^a √(a^2 - x^2) dx
  2. 4 ∫_0^a √(a^2 - x^2) dx
  3. 4 ∫_0^a (a^2 - x^2) dx
  4. 4 ∫_0^a √(x^2 - a^2) dx

Answer: 4 ∫_0^a √(a^2 - x^2) dx

The area of a circle is given by 4 times the area in the first quadrant. In the first quadrant, y = √(a^2 - x^2). Integrating from x=0 to a gives the first quadrant area, so total area = 4 ∫_0^a √(a^2 - x^2) dx.

9. The area of the region defined by x² ≤ y ≤ x + 2 is

  1. 9/2
  2. 7/2
  3. 11/2
  4. 5/2

Answer: 9/2

The region is between y = x² and y = x + 2. Intersection points are x = -1 and x = 2. Since x + 2 ≥ x² on [-1,2], area = ∫_{-1}^{2} ((x+2) - x²) dx = [x²/2 + 2x - x³/3]_{-1}^{2} = (2+4-8/3) - (1/2-2+1/3) = 9/2.

10. The area enclosed between the parabola y = x² and the chord joining the points (-1, 1) and (2, 4) is:

  1. 4
  2. 11/2
  3. 7/2
  4. 9/2

Answer: 9/2

The chord equation is y = x + 2. Intersection points are x = -1 and 2. On [-1,2], chord is above parabola. Area = ∫₋₁² [(x+2) - x²] dx = [x²/2 + 2x - x³/3]₋₁² = (2+4-8/3) - (1/2-2+1/3) = 10/3 - (-7/6) = 9/2.

11. The area enclosed by y² = 4x and x² = 4y is 16/3. Using horizontal strips, the integral setup is:

  1. ∫₀⁴ (2√x - x²/4) dx
  2. ∫₀⁴ (2√y - y²/4) dy
  3. ∫₀⁴ (y²/4 - 2√y) dy
  4. ∫₀⁴ (x²/4 - 2√x) dx

Answer: ∫₀⁴ (2√y - y²/4) dy

For horizontal strips, express x in terms of y: from y²=4x we get x=2√y (right curve), from x²=4y we get x=y²/4 (left curve). Area = ∫₀⁴ (right - left) dy = ∫₀⁴ (2√y - y²/4) dy, which evaluates to 16/3.

12. The area of the region bounded by y² = x and y = x - 2 is

  1. 8/3
  2. 9/2
  3. 16/3
  4. 7/2

Answer: 9/2

Using horizontal strips: rewrite curves as x = y² and x = y + 2. Intersection: y² = y + 2 → y = -1, 2. Area = ∫_{-1}^{2} ((y+2) - y²) dy = [y²/2 + 2y - y³/3]_{-1}^{2} = (2+4-8/3) - (1/2-2+1/3) = 9/2.

More Mathematics topics

This page shows 12 of 40 questions on this topic. The full set, with progress tracking and five agent perspectives per question, is in the JupiteX app — browse the exam catalogue or browse the Learn library.