Problem 13 - Calculus: Early Transcendentals

Question

If \( f \) is continuous and \( \displaystyle \int_{1}^3 f(x) dx = 8 \), show that \( f \) takes on the value 4 at least once on the interval \( [1, 3] \).

Step-by-Step Solution

Answer

By the Mean Value Theorem for Integrals, \( f \) equals 4 at least once on \([1, 3]\).

Step 1: Understand 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 \( c \) in \([a, b]\) such that \( f(c) = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx \). This means the function reaches its average value at some point in the interval.

Step 2: Apply the Mean Value Theorem to the Integral

Given the integral \( \int_{1}^{3} f(x) \, dx = 8 \) and the interval \([1, 3]\), we can calculate the average value of the function on this interval. The length of the interval is \( 3 - 1 = 2 \). Thus, the average value is \( \frac{1}{2} \times 8 = 4 \).

Step 3: Conclude Using the Mean Value Theorem

The average value of \( f \) on the interval \([1, 3]\) is 4. According to the Mean Value Theorem for Integrals, since \( f \) is continuous, there exists some \( c \) in \([1, 3]\) where \( f(c) = 4 \). This demonstrates that \( f \) takes on the value 4 at least once in the interval.

Key Concepts

Understanding Continuous Functions

Continuous functions are the building blocks of calculus that allow us to work smoothly over an interval without any breaks or gaps. Imagine drawing a curve on a graph without lifting your pencil. That's a continuous function.
A function is continuous over a closed interval, \([a, b]\), if:

• It is defined at every point within the interval, including the endpoints.
• For every point \(x\) within the interval, small changes in \(x\) result in small changes in \(f(x)\). There are no sudden jumps or sharp corners in the graph.

Understanding continuity is vital because it guarantees that the properties and theorems, like the Mean Value Theorem for Integrals, can be applied, ensuring we can find a specific point within the interval where the function behaves predictably.

Finding the Average Value of a Function

The average value of a function over an interval gives us a sense of its overall behavior across that range. Specifically, in mathematics, the average value of a continuous function \(f\) over the interval \([a, b]\) is calculated by dividing the definite integral of the function by the width of the interval.
Consider the formula:\[\text{Average value} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx\]In this expression:

• \(\int_{a}^{b} f(x) \, dx\) represents the accumulation of function values, or the total "area" under the curve from \(a\) to \(b\).
• \(b-a\) is simply the length of the interval.

So, the average value effectively "flattens" the function’s effect across the interval, representing a singular height where the function would equalize. The Mean Value Theorem for Integrals guarantees that on a continuous interval, the function will indeed touch this average value at least once.

Understanding Definite Integrals

Definite integrals are a powerful tool for measuring accumulation, whether it's area, distance, or any other quantity.
They're represented as:\[\int_{a}^{b} f(x) \, dx\]In this notation:

• \(a\) and \(b\) are the limits of integration. They indicate the starting and ending points of the interval over which we are summing the function values.
• \(f(x)\) is the continuous function we are integrating. Its values across the interval are added up.

For example, integrating the function \(f(x)\) from \(x=1\) to \(x=3\) accumulates the total effect or area of \(f(x)\) over this range. This area can be interpreted as the "accumulated change" of the function within the interval. Integrals are key in finding the function's average value, which, according to the Mean Value Theorem for Integrals, is reached somewhere in the interval if the function is continuous.