If f(x) is periodic with period T, then `int_(a+2T)^(b+2T) f(x) dx= int_(a)^(b) f(x) dx`

Answer

To prove that if \( f(x) \) is periodic with period \( T \), then

\[ \int_{a+2T}^{b+2T} f(x) \, dx = \int_{a}^{b} f(x) \, dx, \]

we can follow these steps:

### Step 1: Substitute for the variable
Let \( y = x - T \). Then, \( x = y + T \) and \( dx = dy \).

### Step 2: Change the limits of integration
When \( x = a + 2T \), then \( y = (a + 2T) - T = a + T \).
When \( x = b + 2T \), then \( y = (b + 2T) - T = b + T \).

Thus, we can rewrite the integral:

\[ \int_{a+2T}^{b+2T} f(x) \, dx = \int_{a+T}^{b+T} f(y + T) \, dy. \]

### Step 3: Use the periodicity of the function
Since \( f(x) \) is periodic with period \( T \), we have:

\[ f(y + T) = f(y). \]

### Step 4: Substitute back into the integral
Now we can substitute this back into our integral:

\[ \int_{a+T}^{b+T} f(y + T) \, dy = \int_{a+T}^{b+T} f(y) \, dy. \]

### Step 5: Change the variable of integration
Let’s change the variable of integration from \( y \) to \( z \) where \( z = y \). Thus, we have:

\[ \int_{a+T}^{b+T} f(y) \, dy = \int_{a}^{b} f(z) \, dz. \]

### Step 6: Conclude the proof
Finally, we can conclude that:

\[ \int_{a+2T}^{b+2T} f(x) \, dx = \int_{a}^{b} f(x) \, dx. \]

This completes the proof.