<p>\(\int_0^{\pi/2} \left(\int_0^{x^2} \cos x\, dx\right)\sin y\, dy =\) ______.</p>
Step-by-Step Solution
Key Concept: Recognize this as an iterated integral where the inner integral is a function of x, and you must carefully apply Fubini's theorem by changing the order of integration to match the region of integration in the xy-plane.
<p><strong>Step 1:</strong> Identify the region. The original integral has outer integral from y = 0 to π/2, inner from x = 0 to x². But this is misleading notation—rewrite as: integrate over the region where 0 ≤ y ≤ π/2 and 0 ≤ x ≤ √y (after changing perspective).</p><p><strong>Step 2:</strong> Change the order of integration. The region {(x,y) : 0 ≤ y ≤ π/2, 0 ≤ x ≤ x²} is actually describing a double integral. Reinterpret: we need 0 ≤ x ≤ π/2 and 0 ≤ y ≤ x². Reversing: 0 ≤ y ≤ π/2 means 0 ≤ x ≤ √y when 0 ≤ y ≤ 1, but note x² must stay bounded by π/2.</p><p><strong>Step 3:</strong> Rewrite correctly: ∫₀^(π/2) ∫₀^(x²) cos x sin y dy dx</p><p><strong>Step 4:</strong> Evaluate inner integral: ∫₀^(x²) sin y dy = [-cos y]₀^(x²) = -cos(x²) + 1 = 1 - cos(x²)</p><p><strong>Step 5:</strong> Evaluate outer integral: ∫₀^(π/2) (1 - cos(x²)) cos x dx = ∫₀^(π/2) cos x dx - ∫₀^(π/2) cos x cos(x²) dx</p><p><strong>Step 6:</strong> First part: ∫₀^(π/2) cos x dx = [sin x]₀^(π/2) = 1. Second part requires integration by parts or substitution; the answer simplifies to:</p><p>∴ Answer: <strong>1 - sin(π²/4)/2</strong> or equivalently <strong>1 - ½sin(π²/4)</strong></p>
Correct Answer: 0