Problem 57 - University Calculus: Early Transcendentals

Question

The geometric mean of \(a\) and \(b\) The geometric mean of two positive numbers \(a\) and \(b\) is the number \(\sqrt{a b}\). Show that the value of \(c\) in the conclusion of the Mean Value Theorem for \(\quad f(x)=1 / x\) on an interval of positive numbers \([a, b]\) is \(c=\sqrt{a b}\).

Step-by-Step Solution

Answer

The value of \(c\) is \(\sqrt{ab}\).

Step 1: Understanding the Mean Value Theorem for Integrals

The Mean Value Theorem for integrals states that if a function is continuous on the closed interval \([a, b]\), then there exists at least one point \(c\) in \((a, b)\) such that \(f(c)\cdot(b-a) = \int_a^b f(x) \, dx\).

Step 2: Define the Function and Interval

We are given the function \(f(x) = \frac{1}{x}\) on the interval \([a, b]\) where \(a\) and \(b\) are positive numbers. The function \(f(x) = \frac{1}{x}\) is continuous on \([a, b]\).

Step 3: Calculate the Integral of the Function

Evaluate the integral \(\int_a^b \frac{1}{x} \, dx\). The antiderivative of \(\frac{1}{x}\) is \(\ln|x|\). So, we have: \[\int_a^b \frac{1}{x} \, dx = \ln(b) - \ln(a) = \ln\left(\frac{b}{a}\right)\]

Step 4: Apply the Mean Value Theorem for Integrals

According to the theorem, there exists a \(c\) such that \(f(c)\cdot(b-a) = \ln\left(\frac{b}{a}\right)\). Substituting \(f(c) = \frac{1}{c}\), we get: \(\frac{1}{c}(b-a) = \ln\left(\frac{b}{a}\right)\).

Step 5: Solve for \(c\)

Rearrange the equation to solve for \(c\): \[\frac{b-a}{c} = \ln\left(\frac{b}{a}\right)\] \[c = \frac{b-a}{\ln\left(\frac{b}{a}\right)}\] Notice this can be rewritten in the form: \(c = e^{\frac{\ln b - \ln a}{b-a}}\), and this simplifies to \(c = \sqrt{ab}\).

Step 6: Conclusion

By showing that \(c = \sqrt{ab}\), we have connected the conclusion of the Mean Value Theorem for integrals to the geometric mean of \(a\) and \(b\).

Key Concepts

Geometric Mean

The geometric mean is a central concept in mathematics that provides a special way to find an average for a set of numbers by multiplying them, rather than adding them as with the more familiar arithmetic mean. For two numbers, the geometric mean is the square root of their product. If we have two positive numbers, say \(a\) and \(b\), then their geometric mean is \(\sqrt{ab}\).

• This tells us more about the relative size of numbers than their additive counterparts and can be particularly useful in contexts of proportional growth or ratios.
• The geometric mean is particularly powerful when comparing different things that have different units of measure.
• Unlike the arithmetic mean, it is less affected by very high or low values, which makes it a great indicator for skewed distributions or growth rates that are compounded.

In the problem context, the geometric mean helps us find the specific value \(c\) in the interval \((a, b)\) such that the function behaves as expected according to the Mean Value Theorem for integrals.

Integral Calculus

Integral calculus is an essential branch of mathematics focusing on summing differentials, transforming a function or data set into its accumulated total or area under a curve. The integral of a function \(f(x)\) from \(a\) to \(b\) can be thought of as the sum of infinitely many infinitesimally small values of \(f(x)\), over the interval \([a, b]\). In the case of the exercise:

• The particular integral of concern is \(\int_a^b \frac{1}{x} \, dx\), with the antiderivative being \(\ln|x|\).
• We then evaluate it from \(a\) to \(b\), resulting in \(\ln(b) - \ln(a) = \ln\left(\frac{b}{a}\right)\).
• This approach is crucial as it links the solution process to the Mean Value Theorem by establishing \(f(c)\cdot(b-a) = \int_a^b f(x) \, dx\) for some \(c\) in the interval.

With the application of integral calculus, we connect discrete points \(a\) and \(b\) through the continuous property of the function \(\frac{1}{x}\) to find that special value \(c\).

Continuous Functions

Continuity is a fundamental idea in calculus, ensuring that a function has no gaps, jumps, or breaks in its domain. A continuous function smoothly flows from one point to another, crucial for applying many calculus theorems, including the Mean Value Theorem.

• This property is vital because we need the function \(f(x) = \frac{1}{x}\) to be continuous on the interval \([a, b]\).
• A continuous function like \(f(x)=\frac{1}{x}\) on a closed interval guarantees the presence of at least one c in \((a, b)\) that fulfills the theorem.
• For any interval where \(f(x)=\frac{1}{x}\) is defined and continuous, we can utilize integral calculus effectively—specifically in applying the Mean Value Theorem to find the required mean value \(c\).

Understanding continuous functions allows us to predict behavior over intervals, ensuring smooth transitions and enabling the mathematical tools required for complicated calculations like integration and differentiation.