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(C) Given,$I = \int_0^T f(x) dx$. Since $f(x+T) = f(x)$,$f$ is a periodic function with period $T$. We need to evaluate $J = \int_0^{5T} f(2x) dx$. Le

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$5I$

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(C) Given,$I = \int_0^T f(x) dx$. Since $f(x+T) = f(x)$,$f$ is a periodic function with period $T$. We need to evaluate $J = \int_0^{5T} f(2x) dx$. Let $2x = y$,then $dx = \frac{1}{2} dy$. When $x = 0$,$y = 0$. When $x = 5T$,$y = 10T$. Substituting these into the integral: $J = \int_0^{10T} f(y) \cdot \frac{1}{2} dy = \frac{1}{2} \int_0^{10T} f(y) dy$. Using the property of periodic functions $\int_0^{nT} f(x) dx = n \int_0^T f(x) dx$: $J = \frac{1}{2} 	imes 10 \int_0^T f(y) dy = 5 \int_0^T f(x) dx = 5I$.

Let $T > 0$ be a fixed number. $f: R \rightarrow R$ is a continuous function such that $f(x+T) = f(x)$ for all $x \in R$. If $I = \int_0^T f(x) dx$,then find the value of $\int_0^{5T} f(2x) dx$.

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