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.
