Application of Integrals — Questions & Answers

Sample questions from Application of Integrals

1. Q1. The total geometric area enclosed between the curve y = x^4 and the curve y = x^2 is:

   • A.2/15 sq units
   • B.4/15 sq units✓
   • C.8/15 sq units
   • D.1/15 sq units
   Solution

   x^4 = x^2 gives x = 0, ±1, and x^2 ≥ x^4 on (−1, 1). Area = 2∫ from 0 to 1 of (x^2 − x^4) dx = 2(1/3 − 1/5) = 4/15 sq units.

2. Q2. The area of the smaller region of the circle x^2 + y^2 = 16 cut off by the line y = 2 sqrt(2) is:

   • A.4π sq units
   • B.4π − 8 sq units✓
   • C.8π − 8 sq units
   • D.2π − 8 sq units
   Solution

   The line meets the circle where x^2 = 16 − 8 = 8, i.e. x = ±2 sqrt(2); it lies above the centre, so a true minor segment results. Area = 2∫ from 2 sqrt 2 to 4 of sqrt(16 − y^2) dy = 2[y sqrt(16 − y^2)/2 + 8 sin^(−1)(y/4)] from 2 sqrt 2 to 4 = 2[(0 + 4π) − (4 + 2π)] = 4π − 8 sq units.

3. Q3. The area bounded by x = y², y = 1, y = 3, x = 0 is:

   • A.26/3✓
   • B.13/3
   • C.4
   • D.8
   Solution

   A is correct. Area = ∫₁^3 y² dy = [y³/3]₁^3 = 27/3 − 1/3 = 26/3 sq units.

4. Q4. The area enclosed between y = x² and y = x³ from x = 0 to x = 1 is:

   • A.1/6
   • B.1/4
   • C.1/3
   • D.1/12✓
   Solution

   D is correct. On [0,1], x² ≥ x³. Area = ∫₀^1 (x²−x³) dx = [x³/3−x⁴/4]₀^1 = 1/3−1/4 = 4/12−3/12 = 1/12 sq units.

5. Q5. The area between x-axis and y = sin x over one full period [0, 2π] is:

   • A.0
   • B.2
   • C.4✓
   • D.2π
   Solution

   C is correct. ∫₀^(2π) |sin x| dx = ∫₀^π sin x dx + ∫_π^(2π) (−sin x) dx = 2 + 2 = 4 sq units. Note: ∫₀^(2π) sin x dx = 0 (net signed area), but the geometric area is 4.

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