Problem 55 - Calculus Early Transcendentals: Pearson New International Edition

Question

Give a proof of the Mean Value Theorem for Integrals (Theorem A) that does not use the First Fundamental Theorem of Calculus. Hint: Apply the Max-Min Existence Theorem and the Intermediate Value Theorem.

Step-by-Step Solution

Answer

The Mean Value Theorem for Integrals is proven using the Max-Min and Intermediate Value Theorems.

Step 1: Understanding the Mean Value Theorem for Integrals

The Mean Value Theorem for Integrals states that if a function \( f \) is continuous on a closed interval \([a, b]\), then there exists at least one point \( c \in [a, b] \) such that \( f(c) = \frac{1}{b-a} \int_a^b f(x) \, dx \). This means the average value of the function over \([a, b]\) is equal to the value at some point \( c \).

Step 2: Define the Function for Evaluation

Define a function \( F(x) \) as \( F(x) = \int_a^x f(t) \, dt \) for \( x \in [a, b] \). This function represents the area under the curve from \( a \) to any \( x \) within the interval.

Step 3: Apply the Max-Min Existence Theorem

Since \( f \) is continuous on the closed interval \([a, b]\), it attains a maximum and a minimum value. Let \( M = \max_{x \in [a, b]} f(x) \) and \( m = \min_{x \in [a, b]} f(x) \). This implies that \( m \leq f(x) \leq M \) for all \( x \) in the interval.

Step 4: Consider the Average Value of the Function

The average value of \( f \) over \([a, b]\) is given by \( A = \frac{1}{b-a} \int_a^b f(x) \, dx \). By the properties of integrals and continuity, \( m(b-a) \leq \, \int_a^b f(x) \, dx \leq M(b-a) \). Thus, \( m \leq A \leq M \).

Step 5: Apply the Intermediate Value Theorem

The Intermediate Value Theorem states that if \( F \) is continuous on \([a, b]\) and takes on any two values, it also takes on any value between them. Since \( F(x) \) is continuous and \( A = \frac{F(b) - F(a)}{b-a} \), there must exist some \( c \in [a, b] \) such that \( f(c) = A \).

Step 6: Conclude the Proof

By the Intermediate Value Theorem, there exists a \( c \) for which \( f(c) = \frac{1}{b-a} \int_a^b f(x) \, dx \). This completes the proof of the Mean Value Theorem for Integrals without relying on the Fundamental Theorem of Calculus.

Key Concepts

Intermediate Value Theorem

The Intermediate Value Theorem is a fundamental concept in calculus. It applies to continuous functions defined on a closed interval. Here's a friendly rundown of what it means:

• If you have a function that is continuous over an interval \[a, b\],
• And this function takes on two values at different points within that interval,
• Then it must also take on every value between those two values at some point within \[a, b\].

Think of it like this: imagine walking up a gentle hill. If you start at the base (low altitude) and move to the top (higher altitude), you must pass through every elevation between them.
This theorem is crucial because it helps confirm when specific values are hit within the interval, and it underpins the Mean Value Theorem for Integrals by ensuring that average function values are attained.

Max-Min Existence Theorem

The Max-Min Existence Theorem is another key idea linked to continuous functions. This theorem assures us that if a function is continuous on a closed interval \[a, b\], it will definitely achieve both a maximum and a minimum value somewhere within that interval. Here's a breakdown:

• Every continuous function on a closed interval is bounded, meaning there's a highest and lowest point it will hit.
• These maxima and minima are reached at least once within the interval.

Consider the function as a smooth curve extending over a road. As long as the road does not break (which means the function stays continuous), the wheels will go as high and as low as possible.
In proving the Mean Value Theorem for Integrals, this theorem helps us establish the bounds within which the function's average value must fall.

Continuity of Functions

Continuity of a function is an essential prerequisite for many theorems in calculus, including the Mean Value Theorem for Integrals. Let's explore what continuity implies:

• A function is continuous if there are no breaks, jumps, or holes in its graph.
• This means that you can draw the function's graph without lifting your pencil from the paper.

Continuous functions are predictable and well-behaved. Because of continuity, functions will smoothly connect the values they reach without surprise turns or skips.
In the context of the Mean Value Theorem for Integrals, continuity assures that the function behaves nicely over the interval and allows us to find a point where the function's value aligns with its average value over the interval.