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Chapter 5: Problem 5

Find the average value of each function over the given interval. \(f(x)=\frac{1}{x^{2}}\) on [1,5]

Short Answer

Expert verified
The average value is \( \frac{1}{5} \).

Step by step solution

01

Understand the formula for average value

The average value of a continuous function \( f(x) \) over the interval \([a, b]\) is given by the formula: \[ \text{Average value} = \frac{1}{b-a} \int_a^b f(x) \, dx \] Here, \(a = 1\) and \(b = 5\).
02

Set up the integral

Substitute \( f(x) = \frac{1}{x^2} \) and the limits of the interval into the formula: \[ \frac{1}{5-1} \int_1^5 \frac{1}{x^2} \, dx \]
03

Calculate the integral

Evaluate the integral \( \int \frac{1}{x^2} \, dx \). The antiderivative of \( \frac{1}{x^2} \) is \( -\frac{1}{x} \), so:\[ \int_1^5 \frac{1}{x^2} \, dx = \left[ -\frac{1}{x} \right]_1^5 = -\frac{1}{5} + 1 = \frac{4}{5} \]
04

Calculate the average value

Use the result from Step 3 to find the average value:\[ \frac{1}{4} \times \frac{4}{5} = \frac{1}{5} \]
05

Conclude the solution

The average value of the function \( f(x) = \frac{1}{x^2} \) over the interval \([1, 5]\) is \( \frac{1}{5} \).

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Key Concepts

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

Definite Integral
In calculus, the definite integral is a fundamental concept that allows us to compute the accumulated area under a curve over a certain interval on the x-axis. For example, to find the area under a function \( f(x) \) from \( x = a \) to \( x = b \), you would use the definite integral.
  • The notation \( \int_a^b f(x) \, dx \) represents this area calculation.
  • The limits \( a \) and \( b \) are where you start and stop measuring the area under the curve.
  • Definite integrals are used to find solutions to problems involving areas, volumes, and other quantities that accumulate continuously.
To find the definite integral of a function, we often calculate the antiderivative first and then apply the limits of integration. This is a crucial step in many calculus problems, including finding the average value of a function.
Antiderivative
An antiderivative, or an indefinite integral, is a function whose derivative is the original function you started with. In other words, if \( F(x) \) is an antiderivative of \( f(x) \), then the derivative of \( F(x) \) is \( f(x) \).
  • The process of finding an antiderivative involves reversing differentiation.
  • For functions like \( \frac{1}{x^2} \), their antiderivative is \( -\frac{1}{x} \), since differentiating \( -\frac{1}{x} \) gives you \( \frac{1}{x^2} \).
  • This step is key in evaluating definite integrals, as it allows for simpler calculations by substituting back into the original integral limits.
Antiderivatives provide a basis for solving integration problems and are essential in finding the average value of functions over given intervals in calculus.
Continuous Function
Continuous functions are those that have no breaks, holes, or jumps in their graphs. They are fundamental to calculus because most of the theorems and operations, like finding integrals, depend on the function being continuous over the interval.
  • A function \( f(x) \) is continuous over an interval \([a, b]\) if you can draw it without lifting your pen off the paper.
  • Being continuous ensures that the definite integral calculations are valid and that you can apply the fundamental theorem of calculus.
  • Average value calculations assume that the function is continuous over the interval of interest.
Understanding the concept of continuous functions helps ensure that the operations we perform in calculus, like evaluating integrals or finding average values, are always applicable and produce accurate results.

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Most popular questions from this chapter

Prove the integration formula $$ \int u^{n} d u=\frac{1}{n+1} u^{n+1}+C \quad(n \neq-1) $$ as follows. a. Differentiate the right-hand side of the formula with respect to \(x\) (remembering that \(u\) is a function of \(x\) ). b. Verify that the result of part (a) agrees with the integrand in the formula (after replacing \(d u\) in the formula by \(u^{\prime} d x\) ).

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.

SOCIOLOGY: Marriages The marriage rate (marriages per year) in the United States has been declining recently, with about \(1.97 e^{-0.0102 t}\) million marriages per year, where \(t\) is the number of years since 2014 . Assuming that this rate continues, find the total number of marriages in the United States from 2014 to 2024

If two curves cross twice, you can find the area contained by them by evaluating one definite integral (integrating "upper minus lower"). What if the curves cross three times - how many integrations of "upper minus lower" would you need? What if the curves cross ten times?

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 as explained on the left (as well as by the methods of Section 6.1). $$ \int \frac{x}{\sqrt[3]{x+1}} d x $$
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