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Find the average value of the function over the given interval. $$ f(x)=1 / x ;[1, e] $$

Short Answer

Expert verified
The average value of the function over the interval is \( \frac{1}{e-1} \).

Step by step solution

01

Understand the Problem

We need to find the average value of the function \( f(x) = \frac{1}{x} \) over the interval \([1, e]\). This involves using the formula for the average value of a function \( f \) on an interval \([a, b]\), which is given by \( \frac{1}{b-a} \int_a^b f(x)\, dx \).
02

Set Up the Integral

According to the formula, we need to calculate the integral \( \int_1^e \frac{1}{x} dx \). This integral will help us find the total area under the curve \( f(x) = \frac{1}{x} \) from \( x = 1 \) to \( x = e \).
03

Evaluate the Integral

The integral of \( \frac{1}{x} \) with respect to \( x \) is \( \ln|x| \). Thus, we need to evaluate \( \int_1^e \frac{1}{x} \, dx = \left[ \ln|x| \right]_1^e \). This simplifies to \( \ln(e) - \ln(1) = 1 - 0 = 1 \) since \( \ln(e) = 1 \) and \( \ln(1) = 0 \).
04

Calculate the Average Value

Finally, use the formula for the average value: \( \frac{1}{b-a} \int_a^b f(x) \, dx \). Here, \( a = 1 \) and \( b = e \), so \( b-a = e-1 \). The average value of the function is \( \frac{1}{e-1} \times 1 = \frac{1}{e-1} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Definite Integral
The concept of a definite integral is fundamental in calculus. It is used to calculate the accumulated quantity or the net value of a function over a specified interval. When you integrate a function from one point to another, you are essentially finding the area under the curve of that function between two points on the x-axis. This is incredibly helpful when you need to understand how a quantity changes over time or across different conditions.

For example, consider the function \( f(x) = \frac{1}{x} \) over the interval \([1, e]\). The definite integral \( \int_1^e \frac{1}{x} \, dx \) calculates the exact "net" area under this curve starting at \( x = 1 \) and ending at \( x = e \).
  • Start with the integral limits which define where your function begins and ends.
  • Apply the specific function inside the integral to calculate this area.
  • The result can give you valuable information such as total change, accumulation, or net consumption over a given period or condition.
Average Value of a Function
Finding the average value of a function over a certain interval is useful in determining what the function's "mean" behavior is over that range. Calculus provides a precise way to find this average using the formula: \( \frac{1}{b-a} \int_a^b f(x)\, dx \). This formula offers a handy shortcut to summarize the behavior of complex functions over a defined segment.

In our example, we sought the average value of the function \( f(x) = \frac{1}{x} \) over the interval \([1, e]\). By determining the definite integral over that interval and dividing by the length of the interval, you can find a concise representation of how the function behaves on average.
  • This process involves calculating the total area under the curve first.
  • Then, find the difference between the upper and lower limits of your interval, \(b - a\).
  • Finally, dividing the area by \(b - a\) gives the average value.
Natural Logarithm
The natural logarithm, often denoted as \( \ln(x) \), is the logarithm to the base \( e \), where \( e \approx 2.71828 \). The properties of natural logarithms make them particularly useful in calculus and solving integrals that involve the function \( \frac{1}{x} \).

In the specific problem we had, knowing that the integral of \( \frac{1}{x} \) results in \( \ln|x| \) was crucial to solving it. This stems from one of the fundamental properties of natural logarithms, which states that the derivative of \( \ln|x| \) is \( \frac{1}{x} \).
  • The integral \( \int \frac{1}{x} \, dx = \ln|x| + C \) (where \(C\) is a constant) showcases how logarithms naturally arise when dealing with hyperbolic and growth-related problems.
  • The property that \( \ln(e) = 1 \) is particularly handy, helping simplify calculations involving this constant.
  • The behavior of logarithms complements exponential functions, making the two concepts symbiotic, often appearing together in applications involving growth and decay.

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