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(C) Let $J = \int_{3}^{3 + 3T} f(2x) dx$. Substitute $u = 2x$,so $du = 2 dx$ or $dx = \frac{1}{2} du$.When $x = 3, u = 6$. When $x = 3 + 3T, u = 6 + 6

Vedclass · Accepted Answer

$3I$

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(C) Let $J = \int_{3}^{3 + 3T} f(2x) dx$. Substitute $u = 2x$,so $du = 2 dx$ or $dx = \frac{1}{2} du$.When $x = 3, u = 6$. When $x = 3 + 3T, u = 6 + 6T$.Thus,$J = \frac{1}{2} \int_{6}^{6 + 6T} f(u) du$.Since $f$ is periodic with period $T$,$\int_{a}^{a + nT} f(x) dx = n \int_{0}^{T} f(x) dx$ for any $a \in \mathbb{R}$ and $n \in \mathbb{Z}$.Here,the integral is over an interval of length $6T$,which is $6$ times the period $T$.Therefore,$\int_{6}^{6 + 6T} f(u) du = 6 \int_{0}^{T} f(u) du = 6I$.Substituting this back,$J = \frac{1}{2} 	imes 6I = 3I$.

Let $T > 0$ be a fixed number. Suppose $f$ is a continuous function such that for all $x \in \mathbb{R}, f(x + T) = f(x)$. If $I = \int_{0}^{T} f(x) dx$,then the value of $\int_{3}^{3 + 3T} f(2x) dx$ is

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