<p>If \(f(x)\) is continuous and \(n \in \mathbb{N}\) then the value of \(\int_{-1}^{1} [f(x) - f(-x)]^{2n+1}\,dx\) is</p>
Step-by-Step Solution
Key Concept: Recognize that f(x) - f(-x) is an odd function, and any odd function raised to an odd power remains odd. The integral of an odd function over a symmetric interval [-1,1] always equals zero.
<p><strong>Step 1:</strong> Define g(x) = f(x) - f(-x). We need to verify if g(x) is odd.</p><p><strong>Step 2:</strong> Check: g(-x) = f(-x) - f(-(-x)) = f(-x) - f(x) = -[f(x) - f(-x)] = -g(x). So g(x) is an odd function.</p><p><strong>Step 3:</strong> Since g(x) is odd, [g(x)]^(2n+1) = [f(x) - f(-x)]^(2n+1) is also odd (odd function raised to odd power = odd function).</p><p><strong>Step 4:</strong> By the property of odd functions: ∫_{-a}^{a} h(x)dx = 0 when h(x) is odd.</p><p><strong>Step 5:</strong> Therefore, ∫_{-1}^{1} [f(x) - f(-x)]^(2n+1)dx = 0</p><p>∴ Answer: C (which is 0)</p>
Correct Answer: C