<p>The area bounded by the curves \(y = \cos x\) and \(y = \sin x\) between the coordinates \(x = 0\) and \(x = \dfrac{3\pi}{2}\) is</p>
Step-by-Step Solution
Key Concept: The area between two curves requires tracking which function is above the other in different intervals, since sin(x) and cos(x) intersect multiple times. You must split the integral at intersection points where the curves cross.
<p><strong>Step 1: Find intersection points</strong></p><p>Set sin x = cos x ⟹ tan x = 1 ⟹ x = π/4 and x = 5π/4 in [0, 3π/2]</p><p><strong>Step 2: Determine which function is above in each interval</strong></p><p>• [0, π/4]: cos x > sin x</p><p>• [π/4, 5π/4]: sin x > cos x</p><p>• [5π/4, 3π/2]: cos x > sin x</p><p><strong>Step 3: Set up the integral</strong></p><p>Area = ∫₀^(π/4) (cos x - sin x)dx + ∫_(π/4)^(5π/4) (sin x - cos x)dx + ∫_(5π/4)^(3π/2) (cos x - sin x)dx</p><p><strong>Step 4: Evaluate each integral</strong></p><p>∫(cos x - sin x)dx = sin x + cos x + C</p><p>First: [sin x + cos x]₀^(π/4) = (√2/2 + √2/2) - (0 + 1) = √2 - 1</p><p>Second: [sin x + cos x]_(π/4)^(5π/4) = (-√2/2 - √2/2) - (√2/2 + √2/2) = -√2 - √2 = -2√2</p><p>Area of second = |-2√2| = 2√2</p><p>Third: [sin x + cos x]_(5π/4)^(3π/2) = (-1 + 0) - (-√2/2 - √2/2) = -1 + √2</p><p><strong>Step 5: Add all areas</strong></p><p>Total Area = (√2 - 1) + 2√2 + (√2 - 1) = 4√2 - 2</p><p>∴ Answer: D</p>
Correct Answer: D