/*! 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 59 In Exercises \(55-62,\) graph th... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 59

In Exercises \(55-62,\) graph the function and find its average value over the given interval. $$ f(t)=(t-1)^{2} \quad \text { on } \quad[0,3] $$

Short Answer

Expert verified
The average value of \(f(t) = (t-1)^2\) over \([0,3]\) is 1.

Step by step solution

01

Understand the Function and the Interval

We start by understanding that the function given is \( f(t) = (t-1)^2 \) and we need to find its average value over the interval \([0,3]\). This is a simple quadratic function, which is a parabola facing upwards.
02

Compute the Integral of the Function

To find the average value of \( f(t) \) over \([0,3]\), we need the integral of the function. Calculate the integral:\[ \int_0^3 (t-1)^2 \, dt. \]To solve, first expand \((t-1)^2\) to \(t^2 - 2t + 1\). Then, integrate each term separately:\[ \int (t^2 - 2t + 1) \, dt = \frac{t^3}{3} - t^2 + t. \]
03

Evaluate the Integral Over the Interval

Evaluate the antiderivative from \(t = 0\) to \(t = 3\):\[ \left[ \frac{t^3}{3} - t^2 + t \right]_0^3 = \left( \frac{3^3}{3} - 3^2 + 3 \right) - \left( \frac{0^3}{3} - 0^2 + 0 \right). \]Calculate the values:- At \(t = 3\), the expression evaluates to \(\frac{27}{3} - 9 + 3 = 9 - 9 + 3 = 3.\)- At \(t = 0\), the expression evaluates to \(0.\)Thus, the definite integral over \([0, 3]\) is \(3 - 0 = 3.\)
04

Calculate the Average Value

The average value of \(f(t)\) over \([0,3]\) is given by the formula:\[ \frac{1}{b-a} \int_a^b f(t) \, dt, \]where \(a = 0\) and \(b = 3\). Substitute these values:\[ \frac{1}{3-0} \cdot 3 = 1. \]Therefore, the average value of the function on this interval is 1.

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 91影视!

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 that focuses on the concept of integration. At its core, integration is about finding the total or whole from parts. When you integrate a function, you essentially add up an infinite number of infinitely small pieces to find the area under a curve. This operation is fundamental in calculus and helps solve various problems related to areas, volumes, and more.
  • The process of integration is often seen as the reverse of differentiation.
  • When you integrate a function, you determine its antiderivative or integral.
  • Integrals are used to calculate areas under curves, which can represent physical quantities like distance, area, and volume.
Integration provides a crucial tool for understanding and describing real-world phenomena that can be expressed in mathematical form. In the context of our problem, we use integration to find the average value of a function over a specific interval.
Definite Integral
A definite integral is a way to calculate the net area under a curve between two specified limits. It provides a specific numerical value, unlike indefinite integrals, which represent a general form of antiderivatives.
  • It is denoted by the integral sign with upper and lower limits, written as \( \int_a^b f(x) \, dx \).
  • The result of a definite integral represents the accumulation of quantities, like the total distance traveled or the total area.
  • In our problem, the definite integral \( \int_0^3 (t-1)^2 \, dt \) calculates the total area under the function \( f(t) \) from \( t = 0 \) to \( t = 3 \).
Evaluating definite integrals involves finding the antiderivative of the function and computing the difference at the upper and lower limits. This operation provides the precise net result, which is crucial in many applications, such as physics, engineering, and economics.
Quadratic Functions
Quadratic functions are second-degree polynomial functions of the form \( f(x) = ax^2 + bx + c \). These functions produce parabolic graphs, which can open upwards or downwards depending on the coefficient \( a \).
  • The general shape of a parabola is symmetrical, and its highest or lowest point is called the vertex.
  • Quadratic functions can be written in different forms, such as standard form and vertex form.
  • In this case, the function \( f(t) = (t-1)^2 \) represents a parabola with its vertex shifted to the right by 1 unit, opening upwards.
Understanding the properties of quadratic functions helps in visualizing their graphs and recognizing their behavior. When solving problems involving these functions, knowing how to manipulate and work with their equations becomes critical.
Evaluation of Integrals
The process of evaluation involves calculating the integral of a function, which requires finding its antiderivative and applying the fundamental theorem of calculus.
  • First, simplify the function if needed, as shown when expanding \((t-1)^2\) to \(t^2 - 2t + 1\).
  • Find the antiderivative of each term separately, which results in \( \frac{t^3}{3} - t^2 + t \).
  • Apply the limits of integration by substituting the upper and lower limits of the interval and evaluating the result.
This step-by-step approach ensures a clear path from the original function through simplification and on to the final computed area or value. In this exercise, the evaluation of the definite integral resulted in a total area of 3. By understanding each part of this process, you gain confidence in handling more complex functions.

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

If your CAS can draw rectangles associated with Riemann sums, use it to draw rectangles associated with Riemann sums that converge to the integrals in Exercises \(89-94 .\) Use \(n=4,10,20,\) and 50 subintervals of equal length in each case. $$ \int_{0}^{1}(1-x) d x=\frac{1}{2} $$

Evaluate the integrals in Exercises \(17-50\) $$ \int \frac{1}{\sqrt{t}} \cos (\sqrt{t}+3) d t $$

Evaluate the integrals in Exercises \(17-50\) $$ \int 3 x^{5} \sqrt{x^{3}+1} d x $$

Find the areas of the regions enclosed by the lines and curves. $$ y=\left|x^{2}-4\right| \quad \text { and } \quad y=\left(x^{2} / 2\right)+4 $$

By using a substitution, prove that for all positive numbers \(x\) and \(y\) $$ \int_{x}^{x y} \frac{1}{t} d t=\int_{1}^{y} \frac{1}{t} d t $$ The Shift Property for Definite Integrals A basic property of definite integrals is their invariance under translation, as expressed by the equation $$ \int_{a}^{b} f(x) d x=\int_{a-c}^{b-c} f(x+c) d x $$ The equation holds whenever \(f\) is integrable and defined for the necessary values of \(x .\) For example in the accompanying figure, show that $$ \int_{-2}^{-1}(x+2)^{3} d x=\int_{0}^{1} x^{3} d x $$ because the areas of the shaded regions are congruent.
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Calculus

Read Explanation

Discrete Mathematics

Read Explanation

Mechanics Maths

Read Explanation

Statistics

Read Explanation

Decision Maths

Read Explanation
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.