Applications of Integration — Important Questions

8/15

Step 1: For odd n, the reduction formula gives: ∫₀^{π/2} sinⁿx dx = [(n-1)/n] × [(n-3)/(n-2)] × ... × (2/3). Step 2: Here n = 5 (odd): ∫₀^{π/2} sin⁵x dx = (4/5) × (2/3) × 1. Step 3: Calculate: (4/5) × (2/3) = 8/15. Step 4: The final factor is just 1 (we stop here since we've reached the base case). Final Answer: 8/15. Note: The formula ends at 2/3 for n=5. Common mistake: Students confuse the formula for odd n (no π/2 factor) with even n (which includes π/2).

π/16

Step 1: Both m=2 and n=4 are even, so use the formula with a π/2 factor at the end. Step 2: ∫₀^{π/2} sin²x cos⁴x dx = [(4-1)/(2+4)] × [(4-3)/(2+4-2)] × [(2-1)/2] × (π/2). Step 3: = [3/6] × [1/4] × [1/2] × (π/2). Step 4: = (1/2) × (1/4) × (1/2) × (π/2) = π/32... Let me recompute: = (3/6)(1/4)(1/2)(π/2) = (1/2)(1/4)(1/2)(π/2) = π/32. Correction: (3×1×1)/(6×4×2) × π/2 = 3/48 × π/2 = π/32. So the answer is π/32. The correct option is π/32.

πab

Step 1: By symmetry, the total area = 4 × (area in the first quadrant). Step 2: In the first quadrant, y = (b/a)√(a²-x²), so Area = 4∫₀ᵃ (b/a)√(a²-x²) dx. Step 3: Using the standard result ∫₀ᵃ √(a²-x²) dx = πa²/4 (quarter circle of radius a). Step 4: Area = 4 × (b/a) × (πa²/4) = πab. Final Answer: Area of ellipse = πab. Special case: when a = b = r, we get πr² (area of circle). Common mistake: Students write 2πab or forget the factor of 4.

16/3 square units

Step 1: Find intersection points by solving y² = 4x and x² = 4y simultaneously. From x² = 4y, y = x²/4. Substitute: (x²/4)² = 4x → x⁴ = 64x → x = 0 or x = 4. Step 2: Intersection points are (0,0) and (4,4). Step 3: Upper curve: y = 2√x (from y² = 4x). Lower curve: y = x²/4 (from x² = 4y). Step 4: Area = ∫₀⁴ (2√x - x²/4) dx = [2 × (2x^{3/2}/3) - x³/12]₀⁴. Step 5: = [4x^{3/2}/3 - x³/12]₀⁴ = (4×8/3 - 64/12) = 32/3 - 16/3 = 16/3. Final Answer: 16/3 square units.