Problem 17
Question
Let \(p\) be a real number and let \(f: \mathbb{R} \rightarrow \mathbb{R}\) be a continuous function such that \(f(x+p)=f(x)\) for all \(x \in \mathbb{R}\). (Such a function is said to be periodic.) Show that the integral \(\int_{a}^{a+p} f(t) d t\) has the same value for every real number \(a\). (Hint: Part (i) of Proposition 6.24.)
Step-by-Step Solution
Verified Answer
The integral \(\int_{a}^{a+p}f(t) dt\) has the same value for every real number \(a\) because the difference in antiderivative values, \(F(a+p) - F(a)\), is constant for any continuous, periodic function \(f(x)\) with period \(p\). We showed this by constructing an antiderivative \(F(x)\) of \(f(x)\), using the Fundamental Theorem of Calculus, and taking advantage of the periodicity of \(f(x)\).
1Step 1: Construct an antiderivative of f
Construct an antiderivative of the function \(f(x)\) which we will call \(F(x)\), such that \(F'(x) = f(x)\) for all \(x \in \mathbb{R}\).
2Step 2: Apply the Fundamental Theorem of Calculus
Using the Fundamental Theorem of Calculus, we have the integral \(\int_{a}^{a+p}f(t) dt = F(a+p) - F(a)\).
3Step 3: Use the periodicity of f
Now, we are going to use the periodicity of \(f(x)\). Since \(f(x+p) = f(x)\) for all \(x \in \mathbb{R}\), then \(F'(x+p) = f(x+p) = f(x) = F'(x)\).
4Step 4: Show that the antiderivative difference is constant
From Step 3, we have \(F'(x+p) = F'(x)\) which implies that the change in antiderivative values is a constant: \[F(a+p) - F(a) = C\] where \(C\) is a constant.
5Step 5: Conclude that the integral has the same value for every real number a
Since the change in antiderivative values \(F(a+p) - F(a)\) is constant (i.e., independent of the choice of \(a\)), it follows that the integral \(\int_{a}^{a+p}f(t) dt\) has the same value for every real number \(a\).
Thus, we have proven that the integral \(\int_{a}^{a+p}f(t) dt\) has the same value for every real number \(a\).
Key Concepts
Fundamental Theorem of CalculusAntiderivativeConstant of Integration
Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus is a key concept connecting differentiation and integration, two primary operations in calculus. It establishes the integral and the antiderivative relationship, effectively creating a bridge between the two processes. Let’s break it down:
* The **first part** of the theorem tells us how to evaluate a definite integral. If you have a continuous function \( f \) on a closed interval \([a, b]\), then the definite integral of \( f \) from \( a \) to \( b \) represents the net area under the curve of \( f \), and can be calculated using any antiderivative \( F \) of \( f \), as \( F(b) - F(a) \).
* The **second part** states if \( F \) is an antiderivative of a function \( f \) over an interval, then the integral of \( f \) from \( a \) to \( x \) equals \( F(x) - F(a) \). This part shows how differentiation and integration are reverse processes.
Why is this important? Well, it allows us to evaluate integrals through antiderivatives without needing to estimate areas under curves manually. By ensuring you know one antiderivative, you can solve a wide range of integral problems efficiently.
* The **first part** of the theorem tells us how to evaluate a definite integral. If you have a continuous function \( f \) on a closed interval \([a, b]\), then the definite integral of \( f \) from \( a \) to \( b \) represents the net area under the curve of \( f \), and can be calculated using any antiderivative \( F \) of \( f \), as \( F(b) - F(a) \).
* The **second part** states if \( F \) is an antiderivative of a function \( f \) over an interval, then the integral of \( f \) from \( a \) to \( x \) equals \( F(x) - F(a) \). This part shows how differentiation and integration are reverse processes.
Why is this important? Well, it allows us to evaluate integrals through antiderivatives without needing to estimate areas under curves manually. By ensuring you know one antiderivative, you can solve a wide range of integral problems efficiently.
Antiderivative
An antiderivative of a function \( f(x) \) is another function \( F(x) \) whose derivative equals \( f(x) \). In other words, \( F'(x) = f(x) \). This makes \( F(x) \) essentially the reverse process of differentiating \( f(x) \).
Finding antiderivatives, or integrating, involves looking for a function whose derivative matches the function you're given. Consider this as a treasure hunt for the original function from its slopes.
Some key points to remember:
Finding antiderivatives, or integrating, involves looking for a function whose derivative matches the function you're given. Consider this as a treasure hunt for the original function from its slopes.
Some key points to remember:
- There is usually not a single antiderivative. Antiderivatives can differ by a constant of integration.
- If \( f(x) \) is a straightforward polynomial or simple function, finding its antiderivative typically involves reversing the process of differentiation rules.
- Complex functions may require more sophisticated methods like substitution or integration by parts.
Constant of Integration
When we discuss antiderivatives, we must also embrace the constant of integration, often represented as \( C \). This concept arises because of the nature of integration:
When you take the derivative of any constant, you get zero. Consequently, if \( F(x) \) is an antiderivative of \( f(x) \), then \( F(x) + C \) is also an antiderivative, where \( C \) is a real number constant. This means there are infinitely many antiderivatives for a given function. This is why when you compute indefinite integrals, you must add \( C \) at the end.
The constant of integration accounts for vertical shifts in the antiderivative’s graph. It highlights the fact that differentiation loses any constant information, while integration tries to recover it. In exercises like the one given, understanding constants of integration is useful to prove statements of periodic functions and ensure thorough grasp of why integrals over cycles yield constant values.
When you take the derivative of any constant, you get zero. Consequently, if \( F(x) \) is an antiderivative of \( f(x) \), then \( F(x) + C \) is also an antiderivative, where \( C \) is a real number constant. This means there are infinitely many antiderivatives for a given function. This is why when you compute indefinite integrals, you must add \( C \) at the end.
The constant of integration accounts for vertical shifts in the antiderivative’s graph. It highlights the fact that differentiation loses any constant information, while integration tries to recover it. In exercises like the one given, understanding constants of integration is useful to prove statements of periodic functions and ensure thorough grasp of why integrals over cycles yield constant values.
Other exercises in this chapter
Problem 15
Let \(f:[a, \infty) \rightarrow \mathbb{R}\) be a bounded function that is integrable on \([a, x]\) for every \(x \geq a\). Let \(F(x):=\int_{a}^{x} f(t) d t\)
View solution Problem 16
Let \(f:[0, \infty) \rightarrow \mathbb{R}\) be continuous and \(f(x) \geq 0\) for all \(x \in[0, \infty)\). If for each \(b>0\), the area bounded by the \(x\)
View solution Problem 18
Let \(f:[a, b] \rightarrow \mathbb{R}\) be continuous. Show that for every \(x \in[a, b]\), $$ \int_{a}^{x}\left(\int_{a}^{u} f(t) d t\right) d u=\int_{a}^{x}(x
View solution Problem 19
Let \(f:[a, b] \rightarrow \mathbb{R}\) be integrable. Define \(F:[a, b] \rightarrow \mathbb{R}\) by $$ F(x):=\int_{x}^{b} f(t) d t \quad \text { for } x \in[a,
View solution