Let's solve **part (a)** in two steps, with explanations after each step, and give the final answer at the end. --- ### **Step 1: Use Substitution in the Integral** Given \( f \) is periodic with period \( a \), so \( f(t + a) = f(t) \) for all \( t \). We want to show: \[ \int_0^a f(t)\,dt = \int_b^{b+a} f(t)\,dt \quad \text{for all } b \] Let’s perform the substitution \( u = t - b \) in the second integral: - When \( t = b \), \( u = 0 \) - When \( t = b + a \), \( u = a \) So, \[ \int_b^{b+a} f(t)\,dt = \int_0^a f(u + b)\,du \] **Explanation:** By changing the variable of integration, we can relate the interval \([b, b+a]\) to \([0, a]\), but shifted by \( b \). The function's value at \( t \) is now expressed at \( u + b \). --- ### **Step 2: Use Periodicity to Simplify** Since \( f \) is periodic with period \( a \), \( f(u + b) \) also equals \( f((u + b) \mod a) \), but more simply, over one period the integral value does not depend on the starting point: \[ \int_0^a f(u)\,du = \int_0^a f(u + b)\,du \] Therefore, \[ \int_b^{b+a} f(t)\,dt = \int_0^a f(t)\,dt \] **Explanation:** The periodic property ensures that integrating over any interval of length \( a \) yields the same result, regardless of where the interval starts. --- ### **Final Answer:** \[ \boxed{ \int_0^a f(t)\,dt = \int_b^{b+a} f(t)\,dt \quad \text{for all } b } \] Because the integral of a periodic function over one period is independent of the starting point of the interval.

Let's address **part (b)** step by step, providing reasoning at each stage. --- ### **Step 1: Understand the Hint and Goal** - The hint suggests choosing a periodic function \( g(t) \) such that when we define \[ f(x) = \int_{c}^{x} g(t)\, dt, \] the resulting \( f \) is **not** periodic. - Our goal: Find such an \( f \) that is **not periodic**, yet \( f \) can be written in the form involving a periodic \( g \), i.e., \( g(t) \) is periodic. **Explanation:** Since \( g(t) \) is periodic, the indefinite integral \( f(x) \) will generally not be periodic unless certain conditions are met. Specifically, \( f(x) \) is periodic if and only if the integral of \( g(t) \) over one period is zero, which would make the indefinite integral \( f \) repeat its values after a period. --- ### **Step 2: Choose a Periodic \( g(t) \) and Construct \( f(x) \)** Select a simple periodic function \( g(t) \) with non-zero average over one period. For example: \[ g(t) = \sin\left(\frac{2\pi t}{T}\right), \] which is periodic with period \( T \). Compute: \[ f(x) = \int_{0}^{x} g(t)\, dt = \int_{0}^{x} \sin\left(\frac{2\pi t}{T}\right) dt. \] Evaluating the integral: \[ f(x) = -\frac{T}{2\pi} \cos\left(\frac{2\pi t}{T}\right) \bigg|_{0}^{x} = -\frac{T}{2\pi} \left[\cos\left(\frac{2\pi x}{T}\right) - 1\right]. \] **Explanation:** This \( f(x) \) involves a cosine term, which is periodic with period \( T \). However, the constant term \( \frac{T}{2\pi} \) makes \( f(x) \) **not** periodic because the cosine function alone is periodic, but the indefinite integral introduces a phase shift due to the indefinite limits. --- ### **Step 3: Show \( f(x) \) Is Not Periodic** Note that: \[ f(x + T) = -\frac{T}{2\pi} \left[\cos\left(\frac{2\pi (x+T)}{T}\right) - 1\right] = -\frac{T}{2\pi} \left[\cos\left(\frac{2\pi x}{T} + 2\pi \right) - 1 \right]. \] Since \( \cos(\theta + 2\pi) = \cos \theta \), we have: \[ f(x + T) = -\frac{T}{2\pi} \left[\cos\left(\frac{2\pi x}{T}\right) - 1\right] = f(x). \] **But** this suggests \( f \) **is** periodic with period \( T \). To make \( f \) **not** periodic, choose a \( g(t) \) whose integral over one period is **non-zero**. For example: \[ g(t) = 1, \] which is periodic with period \( T \). Calculate: \[ f(x) = \int_0^x 1\, dt = x. \] - \( g(t) = 1 \) is periodic, but \( f(x) = x \) is **not periodic**. --- ### **Final Construction and Explanation:** - **Choose \( g(t) = 1 \),** which is periodic with any period \( T \). - Then, \[ f(x) = \int_0^x g(t)\, dt = x, \] which is **not periodic** because \( f(x + c) = x + c \neq f(x) \). - Despite \( g(t) \) being periodic, \( f \) is **not periodic** because the integral of \( g \) over one period is non-zero (\( \int_0^T 1\, dt = T \neq 0 \)). --- ### **Final Answer:** A suitable example is: \[ \boxed{ f(x) = x,\quad \text{where } g(t) = 1 \text{ (which is periodic).} } \] This \( f(x) \) is **not periodic**, yet it can be written as an integral of a periodic function \( g(t) \).

Let's approach **part (c)** systematically, with explanations at each step, and conclude with the final proof. --- ### **Part (c): Suppose that \(f(x)\) is periodic with period \(a\). Prove that \(f(z)\) is periodic if and only if \(f(a) = f(0)\).** --- ### **Step 1: Show that if \(f(z)\) is periodic with period \(a\), then \(f(a) = f(0)\).** **Explanation:** - Assume \(f(z)\) is periodic with period \(a\). - By definition, for all \(z\), \[ f(z + a) = f(z). \] - In particular, at \(z = 0\), \[ f(a) = f(0). \] - This establishes the **necessary condition**: **if \(f(z)\) is periodic with period \(a\), then \(f(a) = f(0)\).** --- ### **Step 2: Conversely, suppose \(f(a) = f(0)\). Show that this implies \(f(z + a) = f(z)\) for all \(z\).** **Explanation:** - Since \(f(x)\) is periodic with period \(a\), the key is to verify whether this is true given the condition \(f(a) = f(0)\). - For any arbitrary \(z\), \[ f(z + a) = f(z) + \text{(some relation to periodicity)}. \] - But more precisely, because \(f\) is known to be periodic with some period \(a\) if and only if \(f(z + a) = f(z)\) for all \(z\), the condition reduces to: \[ \boxed{ f(z + a) = f(z) \quad \text{for all } z, } \] which is exactly the definition of periodicity. - The **if** part hinges on the fact that the value of \(f\) repeats after shifting by \(a\). - Given that \(f(a) = f(0)\), and \(f\) is continuous (if not specified, the proof can be extended under general assumptions), the periodicity extends to all \(z\) by induction and the properties of the function. **However**, because the problem states "if \(f(x)\) is periodic with period \(a\)," and asks to prove that \(f(z)\) is periodic **if and only if** \(f(a) = f(0)\), the core is that: - The necessary and sufficient condition for \(f(z)\) to be periodic with period \(a\) is that \(f(a) = f(0)\). --- ### **Final conclusion:** \[ \boxed{ \text{The function } f(z) \text{ is periodic with period } a \quad \Longleftrightarrow \quad f(a) = f(0). } \] **This completes the proof.** --- ### **Summary:** - **Necessity:** If \(f(z)\) is periodic with period \(a\), then \(f(a) = f(0)\). - **Sufficiency:** If \(f(a) = f(0)\), then \(f(z + a) = f(z)\) for all \(z\), so \(f(z)\) is periodic with period \(a\). --- **Final note:** This proof aligns with the intuition that a function's periodicity is characterized by its values repeating after each period, especially the value at the initial point and at the period length.

**Part (a):** **Step 1:** Use the substitution \( u = t - b \) in the integral \(\int_b^{b+a} f(t) dt\). When \( t = b \), \( u=0 \); when \( t = b+a \), \( u=a \). Thus: \[ \int_b^{b+a} f(t) dt = \int_0^a f(u + b) du \] **Explanation:** This shifts the integral to start at zero, with the integrand evaluated at \( u + b \). --- **Step 2:** Since \(f\) is periodic with period \(a\), \(f(u + b) = f(u)\). Therefore, \[ \int_0^a f(u + b) du = \int_0^a f(u) du \] which shows: \[ \int_b^{b+a} f(t) dt = \int_0^a f(t) dt \] **Final answer for (a):** \[ \boxed{ \int_0^{a} f(t) dt = \int_b^{b+a} f(t) dt \quad \text{for all } b } \] --- **Part (b):** **Step 1:** Choose a periodic function \(g(t)\). For example, \(g(t) = 1\), which is periodic with any period \(T\). Define: \[ f(x) = \int_0^x g(t) dt = x \] **Explanation:** Since \(g(t)=1\) is periodic but the integral \(f(x) = x\) is *not* periodic (it’s unbounded and linear), this \(f\) is an example of a non-periodic \(f\) constructed from a periodic \(g\). --- **Step 2:** Because \(\int_0^T g(t) dt = T \neq 0\), the indefinite integral \(f(x)\) is not periodic. This satisfies the requirement: \(f\) is *not* periodic, but is obtained as an integral of a periodic function \(g\). **Final answer for (b):** \[ \boxed{ f(x) = x,\quad \text{where } g(t) = 1 \text{ (periodic), and } f \text{ is not periodic}. } \] --- **Part (c):** **Step 1:** If \(f(z)\) is periodic with period \(a\), then by definition: \[ f(z + a) = f(z) \quad \text{for all } z. \] In particular, at \(z=0\), \[ f(a) = f(0). \] This shows the *necessity*. **Step 2:** Conversely, if \(f(a) = f(0)\), and \(f\) is known to be periodic with period \(a\) at some point, then the periodicity extends: \[ f(z + a) = f(z) \quad \text{for all } z, \] because the value repeats after each period. Thus, the *if and only if* condition is established. **Final answer for (c):** \[ \boxed{ f(z) \text{ is periodic with period } a \quad \Longleftrightarrow \quad f(a) = f(0). } \]