Definite Integration
Periodic Functions
Grade 12
Question:
<p>Let \(f(x)\) be a periodic function with period 3 and \(\int_0^3 f(t) \, dt = 7\). If \(g(x) = \int_0^x f(t+n) \, dt\) where \(n = 3k\), \(k \in \mathbb{N}\), then</p>
<p>(A) \(g'(-2/3) = 7\)</p>
<p>(B) \(g'(-2/3) = -7\)</p>
<p>(C) The integral evaluates to \(7k\)</p>
<p>(D) \(g'(0) = f(n)\)</p>
Step-by-Step Solution
Key Concept: For periodic functions, shifting the argument doesn't change the integral over one period. Use FTC to find the derivative of an integral.
<p><strong>Step 1:</strong> Since $f(x)$ is periodic with period 3, $f(t+3k) = f(t)$.</p><p><strong>Step 2:</strong> By the Fundamental Theorem of Calculus, $g'(x) = f(x+n)$.</p><p><strong>Step 3:</strong> Therefore $g'(0) = f(n) = f(3k) = f(0)$ since $n = 3k$.</p><p><strong>Step 4:</strong> Also, $\int_0^3 f(t+n) \, dt = \int_0^3 f(t) \, dt = 7$ due to periodicity.</p>
Correct Answer: D