Definite Integration
Periodic Functions — Multiple Correct
Grade 12
Question:
<p>Suppose \(f\) is continuous with period \(T>0\). Which statements are correct?</p>
<li>\(\displaystyle\int_0^{nT}f(x)\,dx = \frac{1}{n}\int_0^T f(x)\,dx\)</li>
<li>\(\displaystyle\int_0^{nT}f(x)\,dx = n\int_0^T f(x)\,dx\)</li>
<li>\(\displaystyle\int_a^{a+T}f(x)\,dx = \int_0^T f(x)\,dx\) for all \(a\)</li>
<li>\(\displaystyle\int_a^{a+nT}f(x)\,dx = n\int_0^T f(x)\,dx\) for all \(a,\,n\in\mathbb{N}\)</li>
Step-by-Step Solution
Key Concept: Standard periodic integral properties: integral over any full period is the same; n periods give n times one period.
<div class='solution'>
<p><strong>A:</strong> ✗ Wrong — it should be \(n\times\), not \(1/n\times\).</p>
<p><strong>B:</strong> ✓ By periodicity, each of the \(n\) sub-intervals \([kT,(k+1)T]\) contributes \(\int_0^T f\). Sum = \(n\int_0^T f\).</p>
<p><strong>C:</strong> ✓ Classic result: the integral of a periodic function over any complete period is constant. Sub \(u=x-a\): \(\int_a^{a+T}f(x)dx=\int_0^T f(u+a)du\). Since \(f(u+a)\) has the same integral as \(f(u)\) over one period (both cover one full period), result follows.</p>
<p><strong>D:</strong> ✓ Combine B and C: \(\int_a^{a+nT}f=\int_a^{a+T}f + \int_{a+T}^{a+2T}f+\cdots = n\int_0^T f\).</p>
</div>
Correct Answer: ['B', 'C', 'D']