/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 14 Find the average value of each f... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the average value of each function over the given interval. $$ f(x)=2 \text { on }[5,100] $$

Short Answer

Expert verified
The average value of the function is 2.

Step by step solution

01

Understand the Problem

We are tasked with finding the average value of the function \( f(x) = 2 \) over the interval \([5, 100]\). Since \( f(x) \) is a constant function, this simplifies the process.
02

Recall the Formula for Average Value

The average value of a continuous function \( f(x) \) over the interval \([a, b]\) is given by the formula: \[ \text{Average value} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx \] For this problem, \( a = 5 \), \( b = 100 \), and \( f(x) = 2 \).
03

Set Up the Integral

Substitute \( a \), \( b \), and \( f(x) \) into the formula: \[ \text{Average value} = \frac{1}{100-5} \int_{5}^{100} 2 \, dx \]
04

Evaluate the Integral

Calculate the integral: \[ \int_{5}^{100} 2 \, dx = [2x]_{5}^{100} = 2(100) - 2(5) = 200 - 10 = 190 \]
05

Calculate the Average Value

Use the integral result to find the average value: \[ \text{Average value} = \frac{1}{95} \times 190 = 2 \] Since the function is constant, the average value equals the constant.

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Key Concepts

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

Continuous Function
A continuous function is one where small changes in the input result in small changes in the output. This means that the function has no breaks, jumps, or holes in its graph. In other words, you can draw the graph of a continuous function without lifting your pencil off the paper. Continuous functions are important in calculus, especially when dealing with integrals, because they assure us that calculations like finding the area under a curve or average values can be performed without complications.

In the exercise, the function given is a constant, which is naturally continuous. Because it doesn't change at all over the interval, it's trivially continuous. This property simplifies the calculation of the integral quite a bit.
Integral Calculus
Integral calculus is a branch of calculus that deals with the concept of integration. Integration is essentially the opposite of differentiation and is used to find areas, volumes, central points, and many useful things. When you integrate a function, you're essentially calculating the total accumulation of the function's values over a certain interval.

In our given problem, integral calculus simplifies finding the average value of a function over an interval. You integrate the function over the specified bounds and then average it out by dividing by the interval's length. For constant functions, integration is straightforward since the area under a flat line (the constant function) forms a simple rectangle.
Constant Function
A constant function is a simple but important type of function where no matter what input you give, the output remains the same. Graphically, this is represented by a horizontal line. In the function provided in the exercise, \[f(x) = 2\]this is exactly what you have. The function always equals 2, regardless of the input value.
  • Because constant functions are so uniform, calculating their average value over any interval is straightforward.
  • The average value of a constant function over any interval is simply the constant value itself. This is because the total area accumulated is the product of the constant and the length of the interval.
Interval Calculation
Interval calculation in mathematics usually involves finding measurements over a range of values, such as lengths, areas, or averages. An interval refers to the set of all numbers between two given numbers, often represented by notation like \[a, b\].

In calculus, interval calculations are frequently used when dealing with definite integrals. When you calculate a definite integral, you are essentially measuring the area under a curve (or function) over the interval specified. This is exactly what we do in the original exercise, between the bounds 5 and 100.

For a constant function like ours, interval calculations are simplified, as the region under the curve is a perfect rectangle, where the height is the value of the constant function and the base is the length of the interval. This makes the calculation straightforward and gives the same result regardless of the width of the interval; the average of a constant is the constant itself.

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Most popular questions from this chapter

The substitution method can be used to find integrals that do not fit our formulas. For example, observe how we find the following integral using the substitution \(u=x+4\) which implies that \(x=u-4\) and so \(d x=d u\). $$ \begin{aligned} \int(x-2)(x+4)^{8} d x &=\int(u-4-2) u^{8} d u \\ &=\int(u-6) u^{8} d u \\ &=\int\left(u^{9}-6 u^{8}\right) d u \\ &=\frac{1}{10} u^{10}-\frac{2}{3} u^{9}+C \\ &=\frac{1}{10}(x+4)^{10}-\frac{2}{3}(x+4)^{9}+C \end{aligned} $$ It is often best to choose \(u\) to be the quantity that is raised to a power. The following integrals may be found in this way (as well as by the methods of Section 6.1). $$ \int(x-1) \sqrt{x+2} d x $$

Find the average value of each function over the given interval. $$ f(x)=\frac{1}{x} \text { on }[1,10] $$

BUSINESS: Profit from Expansion A company expects profits of \(60 e^{0.02 t}\) thousand dollars per month, but predicts that if it builds a new and larger factory, its profits will be \(80 e^{0.04 t}\) thousand dollars per month, where \(t\) is the number of months from now. Find the extra profits resulting from the new factory during the first two years \((t=0\) to \(t=24\) ). If the new factory will cost \(\$ 1,000,000\), will this cost be paid off during the first two years?

Suppose that a company found its rate of revenue (dollars per day) and its (lower) rate of costs (also in dollars per day). If you integrated "upper minus lower" over a month, describe the meaning of the number that you would find.

$$ \text { These exercises review material that will be helpful in Section } 5.6 \text { . } $$ $$ e^{x^{4}} $$

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