Let `T >0` be a fixed real number. Suppose `f` is continuous function such that for all `x in R ,f(x+T)=f(x)dot` If `I=int_0^Tf(x)dx ,` then the value of `int_3^(3+3T)f(2x)dx` is `3/2I` (b) `2I` (c) `3I` (d) `6I`

Options

A. `3/2I`
B. `2I`
C. `3I`
D. `6I`

Explanation

To solve the problem, we need to evaluate the integral \( \int_3^{3+3T} f(2x) \, dx \) given that \( f(x+T) = f(x) \) for all \( x \in \mathbb{R} \) and that \( I = \int_0^T f(x) \, dx \).

### Step-by-Step Solution:

1. **Change of Variables**:
Let \( u = 2x \). Then, the differential \( dx \) can be expressed as:
\[ dx = \frac{du}{2} \]
Now we need to change the limits of integration. When \( x = 3 \), \( u = 2 \cdot 3 = 6 \). When \( x = 3 + 3T \), \( u = 2(3 + 3T) = 6 + 6T \).

Therefore, the integral becomes:
\[ \int_3^{3+3T} f(2x) \, dx = \int_6^{6+6T} f(u) \cdot \frac{du}{2} = \frac{1}{2} \int_6^{6+6T} f(u) \, du \]

2. **Using the Periodicity of \( f \)**:
Since \( f \) is periodic with period \( T \), we can express the integral \( \int_6^{6+6T} f(u) \, du \) using the property of periodic functions:
\[ \int_a^{a+nT} f(x) \, dx = n \int_0^T f(x) \, dx \]
Here, \( a = 6 \) and \( n = 6 \) (since \( 6 + 6T - 6 = 6T \)). Thus:
\[ \int_6^{6+6T} f(u) \, du = 6 \int_0^T f(x) \, dx = 6I \]

3. **Putting it All Together**:
Now substituting back into our integral:
\[ \int_3^{3+3T} f(2x) \, dx = \frac{1}{2} \cdot 6I = 3I \]

Thus, the value of \( \int_3^{3+3T} f(2x) \, dx \) is \( 3I \).

### Final Answer:
The correct option is (c) \( 3I \).

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