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Chapter 6: Problem 10

(a) Find the average value of \(f\) on the given interval. (b) Find \(c\) such that \(f_{\text {and }}=f(c)\). (c) Sketch the graph of \(f\) and a rectangle whose area is the same as the area under the graph of \(f\) $$ f(x)=1 / x, \quad[1,3] $$

Short Answer

Expert verified
Average value is \(\frac{1}{2}\ln(3)\). \(c = \frac{2}{\ln(3)}\). Rectangle height is \(\frac{1}{2}\ln(3)\).

Step by step solution

01

Set up the integral for average value

To find the average value of a function on an interval \([a, b]\), use the formula: \[ \text{Average value} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx \]. For \(f(x) = \frac{1}{x}\) on \([1, 3]\), we have: \[ \text{Average value} = \frac{1}{3-1} \int_{1}^{3} \frac{1}{x} \, dx = \frac{1}{2} \int_{1}^{3} \frac{1}{x} \, dx. \]
02

Evaluate the integral

Evaluate the integral \(\int_{1}^{3} \frac{1}{x} \, dx\). This is the natural logarithm of \(x\) evaluated from 1 to 3. Thus, we have: \[ \int_{1}^{3} \frac{1}{x} \, dx = \ln|x| \Big|_{1}^{3} = \ln(3) - \ln(1). \] Since \(\ln(1) = 0\), the result simplifies to \(\ln(3)\).
03

Compute the average value

Substitute the evaluated integral from Step 2 into the average value formula: \[ \text{Average value} = \frac{1}{2} \ln(3). \] This is the average value of the function on the interval \([1, 3]\).
04

Set up and solve for \(c\)

The Mean Value Theorem for Integrals states there exists \(c\) in \([a, b]\) such that \(f(c) = \text{average value}\). So, we have: \[ \frac{1}{c} = \frac{1}{2} \ln(3). \] Solve for \(c\) by rearranging: \[ c = \frac{2}{\ln(3)}. \]
05

Sketch function \(f(x) = \frac{1}{x}\) and rectangle

To sketch: Draw the curve \(y = \frac{1}{x}\) on \([1, 3]\), which is a decreasing curve. To represent the average value as an area, draw a rectangle with width \((3-1)\) and height \(\frac{1}{2}\ln(3)\). This rectangle will have the same area as the area under \(f(x)\) from \(1\) to \(3\).

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Key Concepts

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

Integral Calculus
Integral calculus is a fundamental part of calculus focused on the study of integrals, accumulation of quantities, and areas under and between curves. A key goal in integral calculus is to find the total accumulation of a function over an interval.
  • The definite integral of a function describes the net area under its graph, which can be thought of as the total output or accumulation of said function over a specified interval.
  • In the exercise above, we calculate the definite integral of the function \( f(x) = \frac{1}{x} \) over the interval \([1, 3]\) to determine the area under the curve.
  • This involves evaluating \( \int_{1}^{3} \frac{1}{x} \, dx = \ln(3) \), representing the natural logarithm.
Integral calculus provides practical tools for real-world scenarios like calculating distances, areas, and volumes. It鈥檚 widely used in fields ranging from engineering to economics.
Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus connects differentiation and integration, two primary operations in calculus. It serves as a bridge, allowing us to compute definite integrals using antiderivatives.
  • The theorem has two main parts. The first part states that if a function is continuous over an interval, an antiderivative can represent the integral over that interval.
  • The second part allows us to find the value of a definite integral by evaluating the antiderivative at the bounds of the interval.
  • In our exercise, finding the antiderivative of \( \frac{1}{x} \) as \( \ln |x| \) follows this theorem鈥檚 principles, simplifying the calculation of the integral from \(1\) to \(3\).
Understanding this theorem helps make complex calculus problems more manageable and reveals the deep interrelationships between rates of change and accumulated quantities.
Continuous Functions
A function is continuous if it has no interruptions, jumps, or breaks in its graph. Continuity is crucial for applying the Fundamental Theorem of Calculus and finding definite integrals.
  • Continuous functions allow for smooth integration over an interval, ensuring that all values between the endpoints are covered without gaps.
  • In the given problem, the function \( \frac{1}{x} \) is continuous on the interval \([1, 3]\). This property allows us to safely apply integration techniques and the average value concept.
  • Identifying and working with continuous functions is vital in calculus since they provide predictability and reliability, which are essential for solving calculus problems accurately.
Continuous functions often make it easier to determine specific values, like the average value of a function over an interval, due to their predictable nature.

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Most popular questions from this chapter

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