Problem 69 - Calculus Early Transcendentals

Question

Suppose that \(f\) and \(g\) have continuous derivatives on an interval \([a, b] .\) Prove that if \(f(a)=g(a)\) and \(f(b)=g(b),\) then \(\int_{a}^{b} f^{\prime}(x) d x=\int_{a}^{b} g^{\prime}(x) d x\)

Step-by-Step Solution

Answer

Based on the given problem, in order to show that the integral of functions' derivatives f'(x) and g'(x) are equal on the interval [a, b], we first recall the Fundamental Theorem of Calculus and define new function h(x) as the difference between the original functions, with h'(x) = f'(x) - g'(x). We then apply the Mean Value Theorem to h(x) on the interval [a, b] and integrate the derivatives on both sides of the equation. Finally, we arrive at the result that the integral of the derivatives of f(x) and g(x) is equal on the interval [a, b].

Step 1: Fundamental theorem of calculus properties

Recall the Fundamental Theorem of Calculus: if a function \(F\) is the integral of a function \(f\) on the interval \([a, b]\), then the derivative of \(F\) at \(x\) is equal to \(f(x)\): $$F'(x) = f(x)$$ We can see from this theorem that the integral of the derivative of a function on the interval \([a, b]\) is equal to the function itself. With this in mind, let's denote \(F(x) = \int_a^x f'(t) dt\) and \(G(x) = \int_a^x g'(t) dt\). Then according to the Fundamental Theorem of Calculus, we have \(F'(x) = f'(x)\) and \(G'(x) = g'(x)\).

Step 2: Define a new function h(x)

Define a new function \(h(x) = F(x) - G(x)\), which represents the difference between the functions \(F\) and \(G\). By calculating the derivative of \(h\), we can find the relationship between the functions \(f\) and \(g\). According to the properties of the derivative, we have: $$h'(x) = F'(x) - G'(x) = f'(x) - g'(x)$$

Step 3: Conditions of the problem

Now let's apply the conditions given in the problem: \(f(a)=g(a)\) and \(f(b)=g(b)\). Since \(F(x) = \int_a^x f'(t) dt\) and \(G(x) = \int_a^x g'(t) dt\), we have $$F(a) = \int_a^a f'(t) dt = 0 \quad \text{and} \quad G(a) = \int_a^a g'(t) dt = 0$$ Therefore, we have \(h(a) = F(a) - G(a) = 0\). Similarly, we have $$h(b) = F(b) - G(b) = f(b) - g(b) = 0$$

Step 4: Apply Mean Value Theorem on h(x)

Since \(f\) and \(g\) have continuous derivatives on the interval \([a, b]\), \(h'(x) = f'(x) - g'(x)\) is also continuous on this interval. We can apply the Mean Value Theorem for the function \(h(x)\) on the interval \([a, b]\). According to the Mean Value Theorem, there exists at least one point \(c \in (a, b)\) such that \(h'(c) = \frac{h(b) - h(a)}{b - a}\).

Step 5: Integrate h(x) derivative to prove

Now we can use the fact that \(h(a)=h(b)=0\): $$h'(c) = \frac{h(b) - h(a)}{b - a} = \frac{0 - 0}{b - a} = 0$$ Since \(h'(c) = f'(c) - g'(c)\): $$f'(c) - g'(c) = 0$$ Now integrate both sides over the interval \([a, b]\): $$\int_a^b f'(c) dx - \int_a^b g'(c) dx = \int_a^b [f'(c) - g'(c)] dx = 0$$ Thus, we have $$\int_a^b f'(c) dx = \int_a^b g'(c) dx$$ This concludes the proof for the given exercise.

Key Concepts

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a cornerstone of integral calculus and bridges the concept of derivatives with that of integrals. Simply put, it states that if you have a continuous function on an interval, the process of taking an integral followed by a derivative will bring you back to the original function. Moreover, it tells us that the definite integral of a function over an interval can be found using any of its antiderivatives. This theorem is often split into two parts:

• The first part establishes that the integral of a function over an interval can be computed using one of its antiderivatives.
• The second part asserts that the derivative of the integral of a function is the function itself.

This concept is crucial when considering the relationship between the integrals of two functions with the same values at the endpoints of an interval, as examined in the exercise you're working on.

Continuous Derivatives

Continuous derivatives refer to functions that are differentiable everywhere on their domain, and their derivatives are also continuous. This property is significant because it ensures the absence of sudden jumps or breaks in the slope of the tangent line to the function's graph. The continuity of the derivatives plays a pivotal role in the application of the Mean Value Theorem, which states that within a continuous derivative's interval there exists at least one point where the tangent line is parallel to the secant line connecting the endpoints of the interval.

In the exercise, this concept is applied to prove that two functions with the same continuous derivatives and identical values at two points will have the same integral over the interval defined by these points.

Integrals of Functions

Integrals of functions represent the accumulation of quantities, such as area under a curve or other physical quantities like distance and volume. There are two main types of integrals—definite and indefinite. A definite integral has limits of integration and gives a numerical result, representing the total accumulation between those limits. An indefinite integral represents a family of functions and is closely related to antiderivatives. In our exercise, the definite integral quantifies the net change of a function’s derivative over a specific interval. The condition that two functions have the same values at the interval endpoints ensures that their integral over that interval is equal.

Derivative Properties

Derivative properties underpin many techniques and theorems in calculus. These properties describe how derivatives behave with respect to different operations like scaling, addition, subtraction, and the multiplication of functions. Some key properties include:

• Linearity: The derivative of a sum is the sum of the derivatives, and the derivative of a constant times a function is a constant times the derivative of the function.
• Product rule: The derivative of a product of two functions is not simply the product of their derivatives but includes terms involving derivatives of both functions.
• Chain rule: The derivative of a composition of functions is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function.

These properties assist in constructing new functions from known ones (like function h(x) in the exercise) and are central in proving relationships like equality of integrals through differentiation and other operations.