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(B) For a periodic function $f(x)$ with period $T$, the integral over any interval of length $T$ is constant.Let $I(a) = \int_{a}^{a+T} f(x) \, dx$.Di

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$I$ does not depend on $a$

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(B) For a periodic function $f(x)$ with period $T$, the integral over any interval of length $T$ is constant.Let $I(a) = \int_{a}^{a+T} f(x) \, dx$.Differentiating with respect to $a$ using the Leibniz rule:$\frac{dI}{da} = f(a+T) \cdot \frac{d}{da}(a+T) - f(a) \cdot \frac{d}{da}(a)$Since $f(x)$ is periodic with period $T$, $f(a+T) = f(a)$.Thus, $\frac{dI}{da} = f(a) - f(a) = 0$.Since the derivative is $0$, $I$ is independent of $a$ and $I = \int_{0}^{T} f(x) \, dx$.

Let $f(x)$ be a continuous periodic function with period $T$. Let $I = \int_{a}^{a+T} f(x) \, dx$. Then

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