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Chapter 5: Problem 17

Suppose that \(f\) is a linear function. Using the graph of \(f,\) explain why the average value of \(f\) on \([a, b]\) is $$ f\left(\frac{a+b}{2}\right) $$

Short Answer

Expert verified
The average value of a linear function on \([a, b]\) is \(f\left(\frac{a+b}{2}\right)\) because it equals the function's midpoint value.

Step by step solution

01

Identify the Characteristics of a Linear Function

A linear function can be expressed as \( f(x) = mx + c \), where \( m \) is the slope and \( c \) is the y-intercept. The graph of a linear function is a straight line.
02

Understand the Average Value of a Function

The average value of a function \( f(x) \) over an interval \([a, b]\) is determined by the formula \( \frac{1}{b-a} \int_a^b f(x) \, dx \).
03

Apply the Formula to a Linear Function

For a linear function, the integral \( \int_a^b f(x) \, dx \) simplifies to \[ \left[ \frac{mx^2}{2} + cx \right]_a^b = \left( \frac{mb^2}{2} + cb \right) - \left( \frac{ma^2}{2} + ca \right) \].
04

Simplify the Expression

Substitute this back into the average value formula: \[ \frac{1}{b-a} \left( \left( \frac{mb^2}{2} + cb \right) - \left( \frac{ma^2}{2} + ca \right) \right) \]. Factor and simplify by computing differences.
05

Relate to \( f(\frac{a+b}{2}) \)

Observe the result simplifies to \( f(\frac{a+b}{2}) \), since the expression equals \( m \frac{a+b}{2} + c \), which is exactly how we compute \( f \) at the midpoint of \( a \) and \( b \).

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Key Concepts

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

Linear Function
A linear function is one of the fundamental building blocks in mathematics, often presented in the form of \( f(x) = mx + c \). Here, \( m \) represents the slope of the line, indicating how steep the line is. The parameter \( c \), known as the y-intercept, is where the line crosses the y-axis.
Linear functions are characterized by their straight-line graphs:
  • Constant Rate of Change: The slope \( m \) ensures that for every unit increase in \( x \), \( f(x) \) increases by \( m \) units.
  • Linearity: This implies that the graph has no curves or turns. The output changes uniformly with the input.
Understanding these characteristics allows us to predict changes in output easily by adjusting \( x \) accordingly, which is particularly useful when analyzing how functions behave over specific intervals.
Definite Integral
The concept of a definite integral is crucial for finding areas under curves, and in this context, determining the average value of a function on a closed interval \([a, b]\).
The definite integral of \( f(x) \) over \([a, b]\) is represented by the expression \( \int_a^b f(x) \, dx \).
  • Calculating Area: When the function is linear, the definite integral helps calculate the area under this linear segment from \( x = a \) to \( x = b \).
  • Fundamental Theorem of Calculus: This states that the definite integral of a function over \([a, b]\) can be found using an antiderivative of the function, providing a powerful link between derivatives and integrals.
  • Practical Application: In practical terms, the integral value represents accumulated quantity, like total distance traveled, total cost, or any other context-specific quantity.
To find the average function value, we normalize this total area by the interval length \( b - a \), offering insights into the function's behavior over that specific span.
Midpoint
The midpoint of an interval in mathematics is a key concept when finding central values such as averages. It is computed simply as \( \frac{a+b}{2} \) for an interval \([a, b]\).
The midpoint here plays a special role in simplifying the computation of a linear function's average value:
  • Symmetry: For linear functions, the value at the midpoint provides a more direct computation of the average value over \([a, b]\). This is because linear functions change uniformly, so evaluating the function at the midpoint gives us the average of all values.
  • Efficient Calculations: By substituting the midpoint back into the function \( f(x) \), you quickly arrive at the equation \( f(\frac{a+b}{2}) \), making calculations straightforward and intuitive.
  • Conceptual Insight: The midpoint essentially breaks the interval into two equal parts, offering a balance point or center that simplifies many analytical processes involving average values.
This understanding of the midpoint not only aids in mathematical calculations but also enhances comprehension of symmetry and balance in linear functions.

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Most popular questions from this chapter

Use a calculating utility to find the midpoint approximation of the integral using \(n=20\) sub-intervals, and then find the exact value of the integral using Part 1 of the Fundamental Theorem of Calculus. $$ \int_{1}^{3} \frac{1}{x} d x $$

Write a short paragraph that describes the various ways in which integration and differentiation may be viewed as inverse processes. (Be sure to discuss both definite and indefinite integrals.)

A traffic engineer monitors the rate at which cars enter the main highway during the afternoon rush hour. From her data she estimates that between 4: 30 P.M. and 5: 30 P. M. the rate \(R(t)\) at which cars enter the highway is given by the formula \(R(t)=100\left(1-0.0001 t^{2}\right)\) cars per minute, where \(t\) is the time (in minutes) since 4: 30 P.M. (a) When does the peak traffic flow into the highway occur? (b) Estimate the number of cars that enter the highway during the rush hour.

Find the area of the region enclosed by the graphs of \(y=\sin ^{-1} x, x=0,\) and \(y=\pi / 2\).

Sketch the region described and find its area. The region below the interval \([-2,-1]\) and above the curve \(y=x^{3} .\)
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