Problem 6
Question
Suppose \(F\) and \(G\) are continuously" differentiable functions defined on \([a, b]\) such that \(F^{\prime}(x)=G^{\prime}(x)\) for all \(x \in[a, b] .\) Using the fundamental theorem of calculus, show that \(F\) and \(G\) differ by \(a\) constant. That is, show that there exists a \(C \in \mathbb{R}\) such that \(F(x)-G(x)=C\). The next exercise shows how we can use the integral to "smooth out" a non- differentiable function.
Step-by-Step Solution
Verified Answer
\( F(x) = G(x) + C \), where \( C \) is a constant.
1Step 1: Understand the Problem
We are given two functions, \( F \) and \( G \), that are continuously differentiable on the interval \([a, b]\), and their derivatives are equal: \( F'(x) = G'(x) \). Our task is to show that \( F(x) \) and \( G(x) \) differ by a constant, i.e., \( F(x) - G(x) = C \) for some constant \( C \).
2Step 2: Apply Fundamental Theorem of Calculus
The fundamental theorem of calculus states that if \( H'(x) = 0 \) on an interval \( [a, b] \), then \( H(x) \) is constant on \( [a, b] \). Our goal is to use this theorem to show that the difference \( H(x) = F(x) - G(x) \) is constant.
3Step 3: Differentiate the Difference Function
Let \( H(x) = F(x) - G(x) \). Differentiate \( H(x) \) with respect to \( x \): \( H'(x) = F'(x) - G'(x) \). Since \( F'(x) = G'(x) \) for all \( x \in [a, b] \), this simplifies to \( H'(x) = 0 \) for all \( x \) in the interval.
4Step 4: Conclude with the Constant Difference
Because \( H'(x) = 0 \) for all \( x \in [a, b] \), according to the fundamental theorem of calculus, \( H(x) \) must be constant on \( [a, b] \). Thus, there exists a constant \( C \) such that \( H(x) = C \), or equivalently, \( F(x) - G(x) = C \).
Key Concepts
Continuously Differentiable FunctionsIntegral CalculusDifferentiation
Continuously Differentiable Functions
Continuously differentiable functions are those functions that not only have a derivative everywhere in their domain but also have a derivative that is a continuous function. In simpler terms, if you picture the graph of a function, it will be smooth without any sharp corners or edges. This continuous nature of the derivative helps ensure there are no abrupt changes in the slope anywhere on the interval.
For example, a function like \( f(x) = x^2 \) is continuously differentiable over all real numbers because its derivative \( f'(x) = 2x \) is continuous. This property is crucial when dealing with functions within calculus, as it allows us to apply certain theorems and operations, such as the Fundamental Theorem of Calculus, with assurance that the function behavior is manageable.
For example, a function like \( f(x) = x^2 \) is continuously differentiable over all real numbers because its derivative \( f'(x) = 2x \) is continuous. This property is crucial when dealing with functions within calculus, as it allows us to apply certain theorems and operations, such as the Fundamental Theorem of Calculus, with assurance that the function behavior is manageable.
- Smooth graphs with no jumps or corners
- Derivatives exist and are continuous
- Essential for safely applying calculus operations
Integral Calculus
Integral calculus involves the mathematical analysis of summation and finding things like areas under curves. It essentially reverses the process of differentiation. Given a derivative of a function, integration seeks to find the original function, or at least one form of it, plus a constant of integration.
The Fundamental Theorem of Calculus creates a link between the derivative and the integral. It states that if a function is integrable on an interval and has an antiderivative, the integral of the derivative gives back the original function (up to a constant). This is often written as: \[\int_a^b f'(x) \; dx = f(b) - f(a)\] This property is often used to solve problems like the one in our exercise, where we need to show two functions differ by a constant.
The Fundamental Theorem of Calculus creates a link between the derivative and the integral. It states that if a function is integrable on an interval and has an antiderivative, the integral of the derivative gives back the original function (up to a constant). This is often written as: \[\int_a^b f'(x) \; dx = f(b) - f(a)\] This property is often used to solve problems like the one in our exercise, where we need to show two functions differ by a constant.
- Involves finding areas, volumes, and central values
- Reversing the derivative process
- Critical connection with the Fundamental Theorem
Differentiation
Differentiation is a fundamental concept in calculus that deals with the rate of change. When we differentiate a function, we are finding its derivative, which is another function that describes the instantaneous rate of change of the original function. For example, the derivative of \( f(x) = x^2 \) is \( f'(x) = 2x \), showing how the function changes at any given point.
Differentiation is used to study how functions behave, identify increasing or decreasing trends, and find maximum and minimum points. It is a critical tool in applications ranging from physics to economics because it gives us insights into how systems change.
Differentiation is used to study how functions behave, identify increasing or decreasing trends, and find maximum and minimum points. It is a critical tool in applications ranging from physics to economics because it gives us insights into how systems change.
- Determines rate of change
- Key for analyzing function behavior
- Helps in optimization and finding trends
Other exercises in this chapter
Problem 6
Take \(f:[0, \infty) \rightarrow \mathbb{R}\), Riemann integrable on every interval \([0, b],\) and such that there exist \(M,\) \(a,\) and \(T\), such that \(|
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Let \(c \in(a, b)\) and let \(d \in \mathbb{R}\). Define \(f:[a, b] \rightarrow \mathbb{R}\) as $$ f(x):=\left\\{\begin{array}{ll} d & \text { if } x=c \\ 0 & \
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Show $$ \lim _{x \rightarrow \infty} \frac{\ln (x)}{x}=0 $$
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Let \(f:[a, b] \rightarrow \mathbb{R}\) and \(g:[a, b] \rightarrow \mathbb{R}\) be continuous functions such that \(\int_{a}^{b} f=\int_{a}^{b}\) g. Show that t
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