Question:

Since $f(x+T) = f(x)$, the function $f$ is periodic with period $T$. We are given $I = \int_0^T f(x) \, dx$. To evaluate $\int_0^{3T} f(2x) \, dx$, we use a substitution. Let $u = 2x$. Then $du = 2dx$, which implies $dx = \frac{1}{2}du$. We also need to change the limits of integration: When $x = 0$, $u = 2(0) = 0$. When $x = 3T$, $u = 2(3T) = 6T$. Substituting these into the integral: $$ \int_0^{3T} f(2x) \, dx = \int_0^{6T} f(u) \frac{1}{2} \, du = \frac{1}{2}\int_0^{6T} f(u) \, du $$ Since $f$ is periodic with period $T$, for any integer $n$, we have the property $\int_a^{a+nT} f(x) \, dx = n\int_0^T f(x) \, dx$. Applying this property to $\int_0^{6T} f(u) \, du$ with $n=6$: $$ \int_0^{6T} f(u) \, du = 6\int_0^{T} f(u) \, du $$ Given that $I = \int_0^T f(x) \, dx$, we can write: $$ \int_0^{6T} f(u) \, du = 6I $$ Substitute this back into our expression for the original integral: $$ \int_0^{3T} f(2x) \, dx = \frac{1}{2}(6I) = 3I $$ Thus, the value of $\int_0^{3T} f(2x) \, dx$ is $3I$.