/*! 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 27 Find the average value of each o... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find the average value of each of the given functions on the given interval. \(f(x)=x\) on [0,10]

Short Answer

Expert verified
The average value of the function is 5.

Step by step solution

01

Understand the Average Value Formula

The formula for the average value of a function \( f(x) \) on the interval \([a, b]\) is given by: \[\text{Average Value} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx\]Here, \( a \) is 0 and \( b \) is 10. Our function is \( f(x) = x \). This means the formula in this case is:\[\text{Average Value} = \frac{1}{10-0} \int_{0}^{10} x \, dx\].
02

Set Up the Integral

Set up the integral of \( f(x) = x \) from \( 0 \) to \( 10 \): \[\int_{0}^{10} x \, dx\].This integral will help us find the total area under the curve from 0 to 10.
03

Evaluate the Integral

To evaluate the integral \( \int_{0}^{10} x \, dx \), use the power rule for integration, which says \[\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\].In our case, \( n \) is 1, so \[\int x \, dx = \frac{x^2}{2} + C\].Apply the limits from 0 to 10:\[\left. \frac{x^2}{2} \right|_{0}^{10} = \frac{10^2}{2} - \frac{0^2}{2} = \frac{100}{2} = 50\].
04

Calculate the Average Value

Now substitute the value of the integral back into the average value formula:\[\text{Average Value} = \frac{1}{10-0} \times 50 = \frac{1}{10} \times 50 = 5\].Hence, the average value of the function on \([0, 10]\) is 5.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Integral Calculus
Integral calculus is a branch of mathematics focused on the concept of integration, which is used to determine the total accumulation of quantities. It's the counterpart of differential calculus, and together they form the foundation of calculus. In this context, integration deals with finding areas under curves, among many other applications.

In our exercise, we used integration to find the total area under the function curve of \( f(x) = x \) over the interval [0, 10].
This area represents a cumulative value that is crucial in understanding things like total distance traveled when thinking about velocity functions. The indefinite integral provides a general form of the antiderivative, while the definite integral provides the actual area within a specific range, such as we did from 0 to 10 in this problem.
Power Rule for Integration
The power rule for integration is a fundamental principle you'll frequently come across in integral calculus. It's especially handy when dealing with polynomial functions.
The power rule states that to integrate a function of the form \( x^n \), you increase the power by one, and then divide by the new power, specifically:
  • \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \) for \( n eq -1 \)

The 'C' is a constant of integration, which appears when dealing with indefinite integrals. But in this case, because we're calculating a definite integral, it cancels out.

For the linear function \( f(x) = x \), we apply the power rule with \( n = 1 \). This becomes \( \int x \, dx = \frac{x^2}{2} + C \). Integrating this over our specified interval, we calculate the value of the definite integral, vital for finding that average value.
Definite Integrals
Definite integrals are a specific type of integrals used when calculating the total accumulation of a quantity over a certain interval.
This is achieved by setting definite limits, \( a \) and \( b \), to the integral, thus focusing on a section of the function's curve.

In our exercise, we computed the definite integral of \( f(x) = x \) over the interval [0, 10].
By evaluating the integral expression \( \left. \frac{x^2}{2} \right|_{0}^{10} \), we determined the area under the curve, which equals 50.
Definite integrals are crucial as they provide us a way to calculate exact values of accumulation over specified ranges, which is essential in physics, engineering, and other disciplines that require precise computation of quantities.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Consumer Surplus Shafer and colleagues \(^{17}\) created a mathematical model of a demand function for recreational boating in the Lake Erie/Presque Isle Bay Area of Pennsylvania that is given by the equation \(p(q)=417 e^{-0.1062}\), where \(q\) is the number of visitor trips, that is, the number of individuals who participated in any one recreational power boating trip, and \(p\) is the cost (or price) per person per trip. The average number of visitor trips was \(25 .\) Find, as the researchers did, the consumer surplus.

Suppose copper is being extracted from a certain mine at a rate given by \(\frac{d}{d t} P(t)=100 e^{-0.2 t},\) where \(P(t)\) is measured in tons of copper and \(t\) is measured in years from opening the mine. At the beginning of the sixth year a new mining innovation is introduced to boost production to a rate given by \(Q^{\prime}(t)=500 / t .\) Find the increase in production of copper due to this innovation during the second 5 years of its use over what copper production would have been without its use.

Acceleration If a ball is thrown upward, then from physics it can be shown that the acceleration is given by \(a(t)=-32\) \(\mathrm{ft} / \mathrm{sec} .\) If a ball is thrown upward with an initial velocity of \(10 \mathrm{ft} / \mathrm{sec},\) find \(v(t)\)

Show that the curves are Lorentz curves. (You will need to find \(f^{\prime \prime}(x) .\) ) Then find the coefficient of inequality. $$ f(x)=0.01 x^{2}+0.99 x $$

Demand Find the demand function for a table cloth manufacturer if marginal demand, in dollars, is given by \(p^{\prime}(x)=\) \(-x^{-3 / 2},\) where \(x\) is the number of thousands of table cloths sold. Assume that \(p(1)=100\).

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.