Problem 52 - Calculus Early Transcendentals: Pearson New International Edition

Question

Prove or disprove that the integral of the average value equals the integral of the function on the interval: \(\int_{a}^{b} \bar{f} d x=\) \(\int_{a}^{b} f(x) d x\), where \(\bar{f}\) is the average value of the function \(f\) over the interval \([a, b]\).

Step-by-Step Solution

Answer

The statement is true; the integrals are equal.

Step 1: Define the Average Value

The average value of a function \(f(x)\) on an interval \([a, b]\) is given by \(\bar{f} = \frac{1}{b-a} \int_a^b f(x) \, dx\). It represents the mean value of the function over that interval.

Step 2: Consider the Integral of the Average Value

The expression \(\int_a^b \bar{f} \, dx\) can be rewritten using the definition of the average value as \(\int_a^b \frac{1}{b-a} \left( \int_a^b f(x) \, dx \right) \, dx\).

Step 3: Evaluate the Double Integral

The integral \(\int_a^b \bar{f} \, dx = \int_a^b \frac{1}{b-a} C \, dx\), where \(C\) is a constant equal to \(\int_a^b f(x) \, dx\). The result is \(\frac{C}{b-a} \cdot (b-a) = C\).

Step 4: Compare with Initial Integral

Since \(C = \int_a^b f(x) \, dx\), we have shown that \(\int_a^b \bar{f} \, dx\) equals \(\int_a^b f(x) \, dx\).

Step 5: Conclusion: Verify Equality

The integral \(\int_a^b \bar{f} \, dx\) simplifies to \(\int_a^b f(x) \, dx\), thus proving the statement true.

Key Concepts

Average Value of a Function

Understanding the average value of a function is crucial in integral calculus. When you have a function \(f(x)\) defined over an interval \([a, b]\), its average value is calculated to understand how the function behaves overall in that interval. This is important because it gives a single value that represents the entire function's output over that interval. The formula to find the average value \(\bar{f}\) is:

• \( \bar{f} = \frac{1}{b-a} \int_a^b f(x) \, dx \)

This expression takes the integral of the function over \([a, b]\) and scales it by the length of the interval \(b-a\). By using this method, we are effectively distributing the total value of the function across the interval length.
Computing the average value provides insight into function behavior over specified regions, an essential aspect in many applications like physics and economics.

Definite Integrals

Definite integrals play a vital role in computing areas and accumulated quantities. When you encounter a definite integral in calculus, it is essentially finding the signed area under the curve of a function \(f(x)\) from \(x = a\) to \(x = b\). For example, the definite integral \( \int_{a}^{b} f(x) \, dx \) measures such areas or totals over that specific interval.
The process involves calculating the integral of \(f(x)\) while considering the boundaries of \(a\) and \(b\), hence why they are termed 'definite.'

• This technique is not just about finding space under curves; it is broadly used to determine other physical quantities like distance or volume.

Understanding definite integrals is fundamental in integral calculus as they help evaluate real-world problems where variables continuously change over a set range.

Mathematical Proof

In mathematics, a proof is a logical argument demonstrating that a certain statement is universally true. For integral calculus, proving equations or properties, like those involving integrals, helps verify the foundational concepts. For the problem provided, the proof established equality between the two integrals:

• Showing \( \int_a^b \bar{f} \, dx = \int_a^b f(x) \, dx \)

This involves substitution and simplifying expressions to reveal that the statement is true.
Starting with the definition of the average value of a function, we took the equation \( \bar{f} = \frac{1}{b-a} \int_a^b f(x) \, dx \) and substituted it back into the integral form:

• \( \int_a^b \frac{1}{b-a} \left( \int_a^b f(x) \, dx \right) \, dx \)

Simplifying this showed that the average value scaled by the interval provides the same total as the direct integral of the function, thus proving the integrity of the concept.