Index Numbers — Practice Quiz

Area under y = x from x = 0 to x = 2 is 2, Area under y = x² from x = 0 to x = 1 is 1/3, Area under y = 1 from x = 0 to x = 3 is 3

Option 1: ∫₀² x dx = [x²/2]₀² = 2 ✓. Option 2: ∫₀¹ x² dx = [x³/3]₀¹ = 1/3 ✓. Option 3: ∫₀³ 1 dx = [x]₀³ = 3 ✓. Option 4: ∫₁³ 2x dx = [x²]₁³ = 9-1 = 8 ✗.

32/3 square units

Find x-intercepts: 4 - x² = 0 → x² = 4 → x = ±2. Area = ∫₍₋₂₎² (4 - x²) dx = [4x - x³/3]₍₋₂₎² = (8 - 8/3) - (-8 + 8/3) = 16 - 16/3 = 48/3 - 16/3 = 32/3. Step 1: Find where curve meets x-axis. Step 2: Set up integral from -2 to 2. Step 3: Integrate and evaluate.

4∫₀ᵃ √(a² - x²) dx

For circle x² + y² = a², we get y = √(a² - x²) in first quadrant. Total area = 4 × (area in first quadrant) = 4∫₀ᵃ √(a² - x²) dx = πa². Step 1: Use symmetry of circle. Step 2: Consider first quadrant only. Step 3: Multiply by 4 for total area.

Direct integration of polynomial functions, Substitution method for complex expressions, Using symmetry to simplify calculations, Finding intersection points first

For area calculations: 1) Direct integration works for polynomials, 2) Substitution helps with √(a²-x²) type expressions, 3) Symmetry reduces calculation, 4) Integration by parts is not commonly needed for basic areas, 5) Intersection points define limits.