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Chapter 5: Problem 61

In Exercises \(55-62,\) graph the function and find its average value over the given interval. $$ g(x)=|x|-1 \quad \text { on } \quad \text { a. }[-1,1], \text { b. }[1,3], \text { and } c .[-1,3] $$

Short Answer

Expert verified
The average values are -0.5, 1, and 0.5 on the intervals [-1,1], [1,3], and [-1,3] respectively.

Step by step solution

01

Understanding the Function

The function given is \( g(x) = |x| - 1 \). This represents an absolute value function that's been shifted down by 1 unit. The graph of \( |x| \) has a V-shape with its vertex at the origin (0,0), and \( g(x) \) lowers this vertex to (0,-1).
02

Evaluate Function on the Interval [-1,1]

Within [-1,1], \( g(x) = |x| - 1 \) translates to \( g(x) = x-1 \) for \( x \geq 0 \) and \( g(x) = -x-1 \) for \( x < 0 \). The graph in this interval is a V-shape, sloping down from (-1,0) to (0,-1) and up to (1,0).
03

Calculate the Average Value for [-1, 1]

The average value of a function \( f(x) \) over an interval \([a, b]\) is given by \[ \frac{1}{b-a} \int_{a}^{b} f(x) \, dx \]. For \([-1,1]\), this becomes \[ \frac{1}{2} \int_{-1}^{1} (|x| - 1) \, dx \]. The integral is split at the vertex: \[ = \frac{1}{2} \left(\int_{-1}^{0} (-x-1) \, dx + \int_{0}^{1} (x-1) \, dx \right) \].Calculating each section:1. \( \int_{-1}^{0} (-x-1) \, dx = \left[ -\frac{x^2}{2} - x \right]_{-1}^{0} = \frac{1}{2} \)2. \( \int_{0}^{1} (x-1) \, dx = \left[ \frac{x^2}{2} - x \right]_{0}^{1} = -\frac{1}{2} \)The total is \( \frac{1}{2} (\frac{1}{2} - \frac{1}{2}) = -\frac{1}{2} \).
04

Evaluate Function on the Interval [1,3]

In the interval [1,3], \( g(x) = x - 1 \) since \( |x| = x \) for \( x \geq 0 \). The function is increasing linearly from (1,0) to (3,2).
05

Calculate the Average Value for [1, 3]

The average value over \([1, 3]\) is given by \[ \frac{1}{2} \int_{1}^{3} (x - 1) \, dx \]. Calculating the integral:\( \int_{1}^{3} (x-1) \, dx = \left[ \frac{x^2}{2} - x \right]_{1}^{3} \)Evaluate:\( = \left( \frac{9}{2} - 3 \right) - \left( \frac{1}{2} - 1 \right)\)\( = \left( \frac{3}{2} \right) - \left( -\frac{1}{2} \right) = 2 \)Thus, the average is \( \frac{2}{2} = 1 \).
06

Evaluate Function on the Interval [-1,3]

In the interval [-1,3], the function switches from \(-x - 1\) to \(x - 1\) at \( x = 0 \). This splits the function into two parts over the interval: decreasing from (-1,0) to (0,-1), then increasing to (3,2).
07

Calculate the Average Value for [-1, 3]

The average over \([-1, 3]\) is \( \frac{1}{4} \int_{-1}^{3} (|x|-1) \, dx \).Split the integral into two sections:1. \( \int_{-1}^{0} (-x-1) \, dx = \frac{1}{2}\)2. \( \int_{0}^{3} (x-1) \, dx = \left[ \frac{x^2}{2} - x \right]_{0}^{3} = \frac{3}{2} \)Total of both parts:\( \frac{1}{4} (\frac{1}{2} + \frac{3}{2}) = \frac{1}{4}(2) = \frac{1}{2} \).

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Key Concepts

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

Absolute Value Function
An absolute value function is a special type of function defined by the formula \( f(x) = |x| \). The absolute value of a number is its distance from zero on the number line, which means it is always non-negative. In mathematical expression, this would be:
  • \( |x| = x \) when \( x \geq 0 \)
  • \( |x| = -x \) when \( x < 0 \)
This absolute value function is characterized by its V-shape on a graph. The point at which the graph changes direction is known as the vertex. For the function given in the exercise, \( g(x) = |x| - 1 \), the whole graph is shifted downward by 1 unit. This means the vertex is now at (0, -1) instead of at the origin. Understanding the graph's formation helps in analyzing the function's behavior across different intervals.
Integration
Integration is a crucial concept in calculus that allows us to find the area under a curve. It helps us in finding the average value of functions over an interval. For finding the average value of a function \( f(x) \) over an interval \([a,b]\), the formula is: \[ \text{Average value} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx \]In our exercise for \( g(x) = |x| - 1 \), integration splits at the vertex for the intervals with both negative and positive x-values. For instance, over the interval \([-1,1]\), the function \( g(x) \) is broken into two parts:
  • \([-1,0]\): Integrate \(-x - 1\)
  • \([0,1]\): Integrate \(x - 1\)
Calculating the integral for each part and summing them gives the total area, aiding in finding the average value over that interval.
Graphing Functions
Graphing functions is an essential skill for visualizing and understanding mathematical relationships and behaviors. It involves plotting points and interpreting the shape and direction of a graph. For the absolute value function \( g(x) = |x| - 1 \), the graph can be understood by breaking it into parts over the specified intervals.
  • For \([-1,1]\): The graph is a V-shape, descending from (-1,0) to (0,-1) and then ascending to (1,0).
  • For \([1,3]\): The graph is a straight line sloping upwards from (1,0) to (3,2).
  • For \([-1,3]\): The graph combines both behaviors, showing a transition from a descending to ascending line at \( x=0 \).
By plotting these, students gain insights into how the function behaves over different domains.
Calculus Problem Solving
Problem-solving in calculus often requires a strategic approach to understanding, breaking down, and tackling questions. Our given problem asks for the average value of a function over several intervals. Here's how to approach similar problems:
  • Understand the function: Gain a clear grasp of the given function's overall behavior and characteristics.
  • Break down the intervals: When dealing with absolute value functions, separate the intervals where inputs change from negative to positive and vice versa.
  • Apply calculus techniques: Use integration for each sub-part of the interval, summing results to compute the overall average value.
  • Graph the function: Visual graphing helps validate algebraic solutions and offers insights into the function's transition over the intervals.
By practicing this approach, students build robust problem-solving frameworks useful in tackling diverse calculus problems effectively.

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Most popular questions from this chapter

Find the areas of the regions enclosed by the lines and curves in Exercises \(51-58 .\) $$ x+y^{2}=0 \quad \text { and } \quad x+3 y^{2}=2 $$

Find the area of the propeller-shaped region enclosed by the curves \(x-y^{1 / 3}=0\) and \(x-y^{1 / 5}=0\)

If you average 30 \(\mathrm{mi} / \mathrm{h}\) on a \(150-\mathrm{mi}\) trip and then return over the same 150 \(\mathrm{mi}\) at the rate of \(50 \mathrm{mi} / \mathrm{h},\) what is your average speed for the trip? Give reasons for your answer.

If \(f\) is a continuous function, find the value of the integral $$I=\int_{0}^{a} \frac{f(x) d x}{f(x)+f(a-x)}$$ by making the substitution \(u=a-x\) and adding the resulting integral to \(I .\)

Upper and lower sums for increasing functions $$ \begin{array}{l}{\text { a. Suppose the graph of a continuous function } f(x) \text { rises steadily }} \\ {\text { as } x \text { moves from left to right across an interval }[a, b] . \text { Let } P} \\ {\text { be a partition of }[a, b] \text { into } n \text { subintervals of equal length }} \\ {\Delta x=(b-a) / n . \text { Show by referring to the accompanying figure }}\\\ {\text { that the difference between the upper and lower sums for }} \\ {f \text { on this partition can be represented graphically as the area }} \\\ {\text { of a rectangle } R \text { whose dimensions are }[f(b)-f(a)] \text { by } \Delta x \text { . }} \\ {\text { (Hint: The difference } U-L \text { is the sum of areas of rectangles }}\\\\{\text { whose diagonals } Q_{0} Q_{1}, Q_{1} Q_{2}, \ldots, Q_{n-1} Q_{n} \text { lie approximately }} \\ {\text { along the curve. There is no overlapping when these rectan- }} \\ {\text { gles are shifted horizontally onto } R . \text { . }}\end{array} $$ $$ \begin{array}{l}{\text { b. Suppose that instead of being equal, the lengths } \Delta x_{k} \text { of the }} \\ {\text { subintervals of the partition of }[a, b] \text { vary in size. Show that }}\end{array} $$ $$ U-L \leq|f(b)-f(a)| \Delta x_{\max } $$ $$ \begin{array}{l}{\text { where } \Delta x_{\max } \text { is the norm of } P, \text { and hence that } \lim _{ \| P | \rightarrow 0}} \\\ {(U-L)=0}\end{array} $$
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