Definite Integration
Properties of definite integrals
Grade 12
Question:
<p>If \(\int_{0}^{\pi} xf(\sin x) \, dx = A \int_{0}^{\pi/2} f(\sin x) \, dx\), then \(A\) is</p>
<p>\(0\)</p>
<p>\(\pi\)</p>
<p>\(\dfrac{\pi}{4}\)</p>
<p>\(2\pi\)</p>
Step-by-Step Solution
Key Concept: Use the property that ∫₀^π xf(sin x)dx can be transformed by substituting u = π - x, which converts it into a form involving ∫₀^π/2 f(sin x)dx. The key is recognizing that sin(π - x) = sin x, allowing you to relate the two integrals.
<p><strong>Step 1:</strong> Let I = ∫₀^π xf(sin x)dx</p><p><strong>Step 2:</strong> Use substitution x = π - u, so dx = -du. When x = 0, u = π; when x = π, u = 0:<br/>I = ∫_π^0 (π - u)f(sin(π - u))(-du) = ∫₀^π (π - u)f(sin u)du</p><p><strong>Step 3:</strong> Since sin(π - u) = sin u, we have:<br/>I = ∫₀^π πf(sin u)du - ∫₀^π uf(sin u)du = π∫₀^π f(sin u)du - I</p><p><strong>Step 4:</strong> Solving: 2I = π∫₀^π f(sin u)du, so I = (π/2)∫₀^π f(sin u)du</p><p><strong>Step 5:</strong> Since ∫₀^π f(sin u)du = 2∫₀^π/2 f(sin u)du (by symmetry of sin x about x = π/2):<br/>I = (π/2)·2∫₀^π/2 f(sin u)du = π∫₀^π/2 f(sin u)du</p><p>∴ <strong>A = π</strong> (Answer: B)</p>
Correct Answer: B