Question

What is the average value of \(f(x)=\frac{1}{x}\) on the interval \([1, p]\) for \(p>1 ?\) What is the average value of \(f\) as \(p \rightarrow \infty ?\)

Short answer

Answer: The average value of the function \(f(x) = \frac{1}{x}\) as the interval approaches infinity is \(0\).

Step-by-step solution

Calculate the average value of f(x) on the interval [1, p]

To find the average value of the function \(f(x) = \frac{1}{x}\) on the interval \([1, p]\), we need to integrate the function over the interval and then divide by the length of the interval. This can be done using the following formula: Average Value \(=\frac{1}{(p - 1)}\int_{1}^{p} f(x)dx\) So, we first calculate the integral of \(f(x)\) from \(1\) to \(p\).

Integrate f(x)

To find the integral of \(f(x)=\frac{1}{x}\), we use the integral: \(\int \frac{1}{x} dx = \ln|x| + C\) Now, we calculate the definite integral from \(1\) to \(p\). \(\int_{1}^{p} \frac{1}{x} dx = \ln|p| - \ln|1| = \ln p\)

Find the average value of f(x) on [1, p]

Now, we divide the integral by the length of the interval \((p-1)\) to find the average value of \(f(x)\) on \([1, p]\). Average Value \(= \frac{1}{(p-1)}\ln p\)

Find the limit of the average value as p approaches infinity

Next, we determine the average value of the function as \(p\) approaches infinity. We do this by finding the limit of the average value: \(\lim_{p\rightarrow \infty} \frac{1}{(p-1)}\ln p\) To solve this limit, we can use L'Hôpital's rule. But first, let's rewrite the limit in a more suitable form: \(\lim_{p\rightarrow \infty} \frac{\ln p}{(p-1)}\) Now, we take the derivative of both the numerator and denominator with respect to \(p\): \(\frac{d}{dp}(\ln p) = \frac{1}{p}\) \(\frac{d}{dp}(p-1) = 1\) So, we apply L'Hôpital's rule: \(\lim_{p\rightarrow \infty} \frac{\frac{1}{p}}{1} = \lim_{p\rightarrow \infty} \frac{1}{p} = 0\) Thus, the average value of \(f(x) = \frac{1}{x}\) as \(p \rightarrow \infty\) is \(0\).

Key concepts

Integration

Integration is a fundamental concept in calculus, and it is used to find the area under a curve or function. It's a technique that helps us calculate the accumulation of quantities. When you integrate a function, you essentially reverse the process of differentiation, going from a derivative back to an original function. This process is crucial in calculating not just areas, but also volumes and other properties like average values.

To integrate a function such as \(f(x) = \frac{1}{x}\), we use the antiderivative. The antiderivative of \(\frac{1}{x}\) is \(\ln|x| + C\), where \(C\) is the constant of integration. However, for definite integrations, like from a point \(a\) to \(b\), we only focus on the specific interval:

• The definite integral of \(\frac{1}{x}\) from 1 to \(p\) is \(\int_{1}^{p} \frac{1}{x} dx = \ln p\)

This helps us understand the behavior of the function over that interval and lays the groundwork for finding the average value across a range.

Definite Integral

A definite integral is an integral where the limits of integration are specified. It helps us find the total accumulation, area, or net change across a specific interval. In our scenario, the definite integral \(\int_{1}^{p} \frac{1}{x} dx = \ln p\) calculates the total area under \(f(x) = \frac{1}{x}\) from \(x = 1\) to \(x = p\).

To comprehend the definite integral's role in finding the average value of a function, consider the formula:

• Average Value \(= \frac{1}{(b-a)} \int_{a}^{b} f(x)dx\)

Here \(a\) and \(b\) are the limits of integration, and the integral delivers the total sum of \(f(x)\) over \([a, b]\). Dividing by \((b-a)\) gives the average height of the function over that interval. For \(f(x) = \frac{1}{x}\) on \([1, p]\), this results in an average value of \(\frac{1}{(p-1)}\ln p\). This shows how definite integrals are integral to calculating meaningful statistics about function behavior.

L'Hôpital's Rule

L'Hôpital's Rule is a powerful tool when dealing with indeterminate forms in limits, such as \(0/0\) or \(\infty/\infty\). When we find these forms, L'Hôpital's Rule allows us to differentiate the numerator and denominator separately before taking the limit.

In our exercise, we utilized L'Hôpital's Rule to find:

• \(\lim_{p\rightarrow \infty} \frac{\ln p}{(p-1)}\)

Initially, as \(p\) approaches infinity, both the numerator \(\ln p\) and denominator \(p-1\) grow unbounded, creating an \(\infty/\infty\) form. By differentiating:

• Numerator: \(\frac{d}{dp}(\ln p) = \frac{1}{p}\)
• Denominator: \(\frac{d}{dp}(p-1) = 1\)

We rewrite and simplify the limit as \(\lim_{p\rightarrow \infty} \frac{1}{p} = 0\). Hence, as \(p\) becomes infinitely large, the average value of the function tends toward zero. L'Hôpital's Rule simplifies these complex limits, making calculations more straightforward and insightful.