<p>If \(f(t)\) is an odd function, then prove that \(\phi(x) = \int_0^x f(t)\,dt\) will be an even function.</p>
Step-by-Step Solution
Key Concept: Use the property that odd functions satisfy f(-t) = -f(t), then substitute u = -t in the integral to transform ϕ(-x) into a form that equals ϕ(x).
<p><strong>Step 1: Express ϕ(-x)</strong></p><p>ϕ(-x) = ∫₀⁻ˣ f(t)dt</p><p><strong>Step 2: Substitute u = -t (so dt = -du)</strong></p><p>When t = 0, u = 0; when t = -x, u = x</p><p>ϕ(-x) = ∫₀ˣ f(-u)·(-du) = -∫₀ˣ f(-u)du</p><p><strong>Step 3: Apply the odd function property</strong></p><p>Since f is odd: f(-u) = -f(u)</p><p>ϕ(-x) = -∫₀ˣ [-f(u)]du = ∫₀ˣ f(u)du</p><p><strong>Step 4: Conclude</strong></p><p>ϕ(-x) = ∫₀ˣ f(u)du = ϕ(x)</p><p>∴ <strong>ϕ(x) is an even function</strong> (since ϕ(-x) = ϕ(x) for all x in the domain)</p>
Correct Answer: Proof-based