Let's solve the problem step by step: --- ### **Step 1: Use Substitution for the Integral** Given: \( f \) is periodic with period \( a \), i.e., \( f(x + a) = f(x) \) for all \( x \). We want to show: \[ \int_0^a f(t)dt = \int_b^{b+a} f(t)dt \quad \text{for all } b. \] Let’s use the substitution \( u = t - b \) in the second integral: - When \( t = b \), \( u = 0 \) - When \( t = b + a \), \( u = a \) - \( dt = du \) So, \[ \int_b^{b+a} f(t)dt = \int_0^a f(u + b)du \] **Explanation:** This substitution shifts the interval \([b, b+a]\) back to \([0, a]\), but the argument of \( f \) is shifted by \( b \). --- ### **Step 2: Use Periodicity of \( f \)** Since \( f \) is periodic with period \( a \), \( f(u + b) \) is also periodic in \( u \) with period \( a \): \[ f(u + b + a) = f((u + a) + b) = f(u + b) \] Therefore, \[ \int_0^a f(u + b)du = \int_0^a f(u)du \] **Explanation:** Because \( f \) is periodic with period \( a \), the integral of \( f \) over any interval of length \( a \) is the same, regardless of where the interval starts. --- ## **Final Answer:** \[ \boxed{ \int_0^a f(t)dt = \int_b^{b+a} f(t)dt \quad \text{for all } b } \]