Application of Integrals (Area Under Curves) — Practice

Question 1

Find the area of the region bounded by y = x², y = 0, x = 0 and x = 3.

Accepted Answer

The curve y = x² is a parabola opening upward. In [0, 3], y = x² ≥ 0. Area = ∫₀³ x² dx = [x³/3]₀³ = 27/3 − 0 = 9 square units.

Question 2

Find the area of the region enclosed by the parabola y = x² and the line y = x + 2.

Accepted Answer

FIND INTERSECTION POINTS: Set x² = x + 2 → x² − x − 2 = 0 → (x−2)(x+1) = 0 → x = −1 and x = 2. So limits are x = −1 to x = 2. CHECK WHICH IS UPPER: At x = 0: parabola y = 0, line y = 2. Line is ABOVE parabola in [−1, 2]. DRAW SKETCH: Parabola y = x² with vertex at origin; line y = x+2 intersecting at (−1, 1) and (2, 4). AREA: A = ∫₋₁² [(x+2) − x²] dx = [x²/2 + 2x − x³/3]₋₁². At x = 2: 4/2 + 4 − 8/3 = 2 + 4 − 8/3 = 6 − 8/3 = 10/3. At x = −1: 1/2 − 2 − (−1/3) = 1/2 − 2 + 1/3 = 3/6 − 12/6 + 2/6 = −7/6. A = 10/3 − (−7/6) = 20/6 + 7/6 = 27/6 = 9/2 square units.

Question 3

Using integration, find the area of the region in the first quadrant enclosed by the circle x² + y² = 4 and the line x = √3·y.

Accepted Answer

FIND INTERSECTION: x = √3·y and x² + y² = 4. Substitute: 3y² + y² = 4 → 4y² = 4 → y = 1 (first quadrant). So x = √3. Intersection point: (√3, 1). The region in the first quadrant is bounded by the line (from origin to (√3,1)) and the circle arc (from (√3,1) to (2,0) on x-axis). SKETCH: Line from O to (√3,1); circle arc from (√3,1) to (2,0). Area = Area under circle from x=0 to x=√3 using y=x/√3 (line) + Area under circle from x=√3 to x=2 using y=√(4−x²) (circle). Wait — reframe: Area = ∫₀^√3 (x/√3) dx + ∫_√3^2 √(4−x²) dx. First integral: (1/√3)[x²/2]₀^√3 = (1/√3)(3/2) = √3/2. Second integral: ∫_√3^2 √(4−x²) dx = [x√(4−x²)/2 + 2 sin⁻¹(x/2)]_√3^2. At x=2: [0 + 2·sin⁻¹(1)] = 2·(π/2) = π. At x=√3: [√3·1/2 + 2·sin⁻¹(√3/2)] = √3/2 + 2·(π/3) = √3/2 + 2π/3. Second integral = π − √3/2 − 2π/3 = π/3 − √3/2. TOTAL AREA = √3/2 + π/3 − √3/2 = π/3 square units.

Practice — Application of Integrals (Area Under Curves)

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