Average value What is the average value of f(x) = 1/x on the interval [1, p] for p > 1? What is the average value of f as p → ∞?

Verified step by step guidance

1. Recall that the average value of a function \( f(x) \) on the interval \([a, b]\) is given by the formula: \[\text{Average value} = \frac{1}{b - a} \int_a^b f(x) \, dx\] In this problem, \( f(x) = \frac{1}{x} \), \( a = 1 \), and \( b = p \) where \( p > 1 \).
2. Set up the integral for the average value: \[\text{Average value} = \frac{1}{p - 1} \int_1^p \frac{1}{x} \, dx\]
3. Evaluate the integral \( \int_1^p \frac{1}{x} \, dx \). Recall that the integral of \( \frac{1}{x} \) is \( \ln|x| \), so: \[\int_1^p \frac{1}{x} \, dx = \ln(p) - \ln(1) = \ln(p)\]
4. Substitute the integral result back into the average value formula: \[\text{Average value} = \frac{\ln(p)}{p - 1}\]
5. To find the average value as \( p \to \infty \), analyze the limit: \[\lim_{p \to \infty} \frac{\ln(p)}{p - 1}\] Consider the growth rates of the numerator and denominator to determine the behavior of this limit.

Key Concepts

Average Value of a Function

The average value of a function f(x) on an interval [a, b] is given by (1/(b - a)) times the definite integral of f(x) from a to b. It represents the mean height of the function over that interval and is calculated as (1/(b - a)) ∫_a^b f(x) dx.

Definite Integral of 1/x

The integral of 1/x with respect to x is the natural logarithm function, ln|x| + C. For definite integrals, ∫_a^b (1/x) dx = ln(b) - ln(a) = ln(b/a), which is essential for evaluating the average value of f(x) = 1/x over [1, p].

Limit of the Average Value as p Approaches Infinity

To find the behavior of the average value as p → ∞, we analyze the limit of the average value expression. This involves understanding how ln(p)/ (p - 1) behaves as p grows large, which typically requires knowledge of limits and growth rates of logarithmic versus linear functions.