Problem 50 - A Course in Calculus and Real Analysis

Question

(First Mean Value Theorem for Integrals) Let \(f:[a, b] \rightarrow \mathbb{R}\) be a continuous function and \(g:[a, b] \rightarrow \mathbb{R}\) be a nonnegative integrable function. Use the IVP of \(f\) to show that there is \(c \in[a, b]\) such that $$ \int_{a}^{b} f(x) g(x) d x=f(c) \int_{a}^{b} g(x) d x $$ Give examples to show that neither the continuity of \(f\) nor the nonnegativity of \(g\) can be omitted. [Note: For another version of this result, see Exercise 72 .

Step-by-Step Solution

Answer

Using the Intermediate Value Property (IVP) of the continuous function \(f\) and the nonnegative integrable function \(g\), we can prove that there exists a value \(c \in [a, b]\) such that: $$ \int_{a}^{b} f(x) g(x) d x = f(c) \int_{a}^{b} g(x) d x $$ The continuity of \(f\) and the nonnegativity of \(g\) are necessary conditions for this theorem to hold, as demonstrated by counterexamples where \(f\) is not continuous or \(g\) is not nonnegative on the given interval.

Step 1: Apply the Intermediate Value Property (IVP) of the continuous function \(f\)

As \(f\) is a continuous function on \([a, b]\), it also satisfies the IVP. According to the IVP, there exists \(c \in [a, b]\) such that, for any number \(N\) between \(\min_{x \in [a,b]} f(x)\) and \(\max_{x \in [a,b]} f(x)\), we can find \(f(c) = N\). Thus, the IVP states that any number between the minimum and maximum of \(f\) on the interval \([a, b]\) can be assumed as \(f(c)\).

Step 2: Prove the existence of \(c\) using IVP and \(g\)

Let \(\alpha = \min_{x \in [a,b]} f(x)\) and \(\beta = \max_{x \in [a,b]} f(x)\). Since \(g(x) \geq 0\) for all \(x \in [a, b]\), we have the inequality: $$ \alpha \int_{a}^{b} g(x) d x \leq \int_{a}^{b} f(x) g(x) d x \leq \beta \int_{a}^{b} g(x) d x $$ Now let \(N = \frac{\int_{a}^{b} f(x) g(x) d x}{\int_{a}^{b} g(x) d x}\). It's clear that \(\alpha \leq N \leq \beta\). Due to the IVP, there exists \(c \in [a, b]\) such that \(f(c) = N\). Therefore, we obtain: $$ f(c) = \frac{\int_{a}^{b} f(x) g(x) d x}{\int_{a}^{b} g(x) d x} $$ Multiplying both sides by \(\int_{a}^{b} g(x) d x\) gives us: $$ \int_{a}^{b} f(x) g(x) d x = f(c) \int_{a}^{b} g(x) d x $$ This completes the proof of the First Mean Value Theorem for Integrals.

Step 3: Provide examples to show the necessity of the conditions

Now we will provide examples to show that neither the continuity of \(f\) nor the nonnegativity of \(g\) can be omitted for the theorem to hold. 1. Continuity of \(f\): Let $f(x) = \begin{cases} 1, & x = 0 \\ 0, & x \neq 0 \end{cases}\( and \)g(x) = 1\( for \)x \in [-1, 1]\(. Clearly, \)f\( is not continuous, while \)g$ is nonnegative and integrable. Nevertheless, we have: $$ \int_{-1}^{1} f(x) g(x) d x = \int_{-1}^{0} 0 d x + \int_{0}^{0} 1 d x +\int_{0}^{1} 0 d x = 0 $$ However, there is no \(c \in [-1, 1]\) such that \(f(c) \int_{-1}^{1} g(x) d x = 0\) because \(f(c)\) can never be zero. 2. Nonnegativity of \(g\): Let \(f(x) = x\) and \(g(x) = x\) for \(x \in [-1, 1]\). Here, \(f\) is continuous, but \(g\) is not nonnegative on the entire interval. We compute the integrals as follows: $$ \int_{-1}^{1} f(x) g(x) d x = \int_{-1}^{1} x^2 d x = \frac{2}{3} $$ $$ \int_{-1}^{1} g(x) d x = 0 $$ As a result, the equation \(\int_{a}^{b} f(x) g(x) d x = f(c) \int_{a}^{b} g(x) d x\) can never hold because the left side is nonzero while the right side is always zero. These examples demonstrate the necessity of the continuity of \(f\) and the nonnegativity of \(g\).

Key Concepts

Intermediate Value Property (IVP)

The Intermediate Value Property (IVP) is a fundamental attribute of continuous functions. It states that if you have a function that is continuous over a closed interval, say from a to b, and you pick any number between the function's values at those points, there's guaranteed to be at least one point within that interval where the function will match that chosen number.

Think of it as a guarantee that a continuous line drawn without lifting the pencil from starting point a to end point b will pass through every value in between. For the First Mean Value Theorem for Integrals, we use IVP to find a specific point, let's call it c, where the function's value matches the average value of the function over the interval, when weighted by another function, g(x).

In the context of the exercise provided, we applied IVP to the continuous function f to show that such a c exists. This principle ensures that the theorem is not just some abstract concept, but has a practical application where the average can be pinpointed to an exact location within the interval.

Continuity of functions

Continuity of a function is the idea that its graph can be drawn without lifting the pencil off the paper. More formally, a function f is continuous at a point x if the limit of f as it approaches x equals the function's value at x. Continuity over an interval means the function behaves this way at every point in the interval.

This characteristic is essential for applying the First Mean Value Theorem for Integrals, because if the function were not continuous, there could be jumps or gaps, and we couldn't guarantee the existence of an average value in the same way. For example, if you have a function that jumps from 3 to 1 without passing through 2, then there is no place where the function's value is 2, and hence IVP wouldn't hold. To solidify this idea, the provided exercise illustrated what happens when continuity is not present with a function that's 1 at a single point and 0 everywhere else. Clearly, this discontinuous function breaks the requirement for the theorem.

Nonnegative integrable functions

A nonnegative function is just what it sounds like: it's a function that doesn't dip below zero over the interval we're looking at. Integrability means that you can find the area under the curve of the function over an interval, a fundamental concept in calculus often related to accumulation or total change. In the equation from the First Mean Value Theorem for Integrals, g(x) must be nonnegative and integrable so we can reliably talk about the area it represents, which acts as a sort of weighting function when finding the average value of f(x).

In the step-by-step solution, we've seen that even if g(x) is integrable but not nonnegative, the theorem falls apart. The negative portions of g(x) would essentially 'subtract' from the total area in ways that make finding our average value, or point c, impossible in the way the theorem describes. The provided example with g(x) = x showcases how a function that crosses below zero messes up our equation, resulting in a mismatch between the actual integrated product of f(x)g(x) and the theoretical average value.

Fundamental theorem of calculus

The Fundamental Theorem of Calculus is a cornerstone of integral calculus, linking differentiation and integration in a beautifully symmetrical fashion. It comes in two parts; the first part shows how an antiderivative can be used to evaluate a definite integral. The second part tells us that the derivative of an integral function equals the original function over an interval.

This theorem is crucial for understanding the First Mean Value Theorem for Integrals because it provides the tools to actually do the calculations required. For instance, when we want to evaluate the integral of f(x)g(x) over an interval [a, b], it's the Fundamental Theorem that gives us the means to find that exact area under the curve. Without it, we would be stuck with the concept of integration with no practical way to apply it. Moreover, it reinforces the importance of continuity, as the Fundamental Theorem requires the function to be continuous over the interval to apply these powerful conclusions.