Chapter 8: Application of Integrals
Q1. Find the area of the region bounded by y² = 9x, x = 2, x = 4 and the x-axis in the first quadrant.
Solution
y² = 9x ⇒ y = 3√x (first quadrant)
Area = ∫24 3√x dx = 3 ∫24 x1/2 dx
= 3 [2x3/2/3]24 = 2[x3/2]24
= 2[8 − 2√2] = (16 − 4√2) sq. units
Q2. Find the area enclosed between the parabola y² = 4ax and the line y = mx.
Solution
Intersection: (mx)² = 4ax ⇒ m²x² = 4ax ⇒ x = 0 or x = 4a/m²
Area = ∫04a/m² (mx − 2√(ax)) dx ... correcting: integrate line minus parabola:
Area = ∫04a/m² [mx − 2√a · √x] dx
= [mx²/2 − 2√a · 2x3/2/3]04a/m²
= m/2 · 16a²/m⁴ − 4√a/3 · 8a²/m³
= 8a²/m³ − 32a²/(3m³)
= 8a²/(3m³) sq. units
Q3. Find the area of the ellipse x²/16 + y²/9 = 1.
Solution
By symmetry: Area = 4 ∫04 y dx where y = 3√(1 − x²/16)
Area = 4 × πab/4 = π · 4 · 3 = 12π sq. units
Q4. Find the area bounded by y = |x + 1| + 1, x = −3, x = 3 and the x-axis.
Solution
For x ≥ −1: y = x + 1 + 1 = x + 2
For x < −1: y = −(x + 1) + 1 = −x
Area = ∫−3−1 (−x) dx + ∫−13 (x + 2) dx
= [−x²/2]−3−1 + [x²/2 + 2x]−13
= (−1/2 − (−9/2)) + ((9/2 + 6) − (1/2 − 2))
= 4 + 8 = 12 sq. units
Q5. Find the area of the region bounded by x² + y² = 4 and y = x.
Solution
Intersection: x² + x² = 4 ⇒ 2x² = 4 ⇒ x = ±√2
Area = 2 ∫0√2 (√(4−x²) − x) dx
= 2 [x√(4−x²)/2 + 2 sin−1(x/2) − x²/2]0√2
= 2[(√2 · √2)/2 + 2 sin−1(√2/2) − 1]
= 2[1 + 2(π/4) − 1] = π sq. units