/*! 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 3 Evaluate the integrals in Exerci... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 3

Evaluate the integrals in Exercises \(1-28\). $$\int_{-2}^{2} \frac{3}{(x+3)^{4}} d x$$

Short Answer

Expert verified
The integral evaluates to \(\frac{124}{125}\).

Step by step solution

01

Identify the Integral

We have to evaluate the integral \(\int_{-2}^{2} \frac{3}{(x+3)^{4}} dx\). This integral has limits from \(-2\) to \(2\) and the integrand \(\frac{3}{(x+3)^4}\) suggests a substitution can simplify this problem.
02

Choose a Substitution

To simplify the integral, use the substitution \( u = x + 3 \). This implies \( du = dx \) and as \( x \) varies from \(-2\) to \(2\), \( u \) will vary from \(1\) to \(5\).
03

Change the Limits of Integration

After substitution, change the limits of integration: when \( x = -2 \), \( u = 1 \), and when \( x = 2 \), \( u = 5 \). The integral converts to \(\int_{1}^{5} \frac{3}{u^4} du\).
04

Integrate the Simplified Function

The integral \(\int_{1}^{5} \frac{3}{u^4} du\) can be rewritten as \(3\int_{1}^{5} u^{-4} du\). The integral of \(u^{-4}\) is \(-\frac{1}{3} u^{-3} + C\).
05

Evaluate the Definite Integral

Now we evaluate the indefinite integral from \(1\) to \(5\):\[3\left[-\frac{1}{3} u^{-3}\right]_{1}^{5} = 3\left[-\frac{1}{3} \left(\frac{1}{125}\right) - \left(-\frac{1}{3} \right)\right]\]Simplifying gives:\[3\left(-\frac{1}{375} + \frac{1}{3}\right) = 3\left(\frac{125}{375} - \frac{1}{375}\right) = 3\times \frac{124}{375}\]Continuing with the arithmetic:\[= \frac{372}{375}, \text{ which simplifies to } \frac{124}{125}.\]
06

Conclude with Final Result

The evaluated definite integral is \(\frac{124}{125}\).

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.

Substitution in Integration
When solving an integral, substitution can be a powerful technique to simplify the problem. By changing the variable of integration, you can make the integral easier to evaluate.
In the exercise, the substitution chosen was: \( u = x + 3 \).
  • This change simplifies the expressions within the integral.
  • Substitution modifies both the integrand and the limits of integration.
This strategy works because it replaces the complicated expression \((x+3)^{-4}\) with a more manageable \( u^{-4} \). After finding the derivative of \( u \), which is simply \( du = dx \), the integration becomes straightforward. By changing variables, you often reduce the complexity, making it a valuable method in calculus.
Limits of Integration
Adjusting the limits of integration is an essential part of substitution.
  • After substituting \( u = x + 3 \), recalculate the limits based on this new variable.
  • Initial limits of \( x = -2 \) to \( x = 2 \) change accordingly; converting to \( u = 1 \) to \( u = 5 \).
This step is crucial because it ensures that the area calculated correctly represents the desired region under the curve. Keep in mind:
  • The new limits must match the substitution equation.
  • Always transform limits in line with the substitution used.
Using precise limits of integration is key to obtaining accurate results when working with definite integrals after substitution.
Definite Integral Evaluation
Definite integral evaluation finalizes the process, converting the antiderivative back into a number representing area under the curve.In this exercise, you started with \( 3\int_{1}^{5} u^{-4} du \), and then found its antiderivative:
  • The antiderivative of \( u^{-4} \) is \( -\frac{1}{3} u^{-3} + C \).
  • With this, you evaluate it between the new limits \( u = 1 \) and \( u = 5 \).
To find the definite integral:
  • First, plug in the upper limit into the antiderivative.
  • Then, subtract the result of plugging in the lower limit.
This process simplifies the integral to an exact area value. In the example given, this result is \( \frac{124}{125} \). Definite integral evaluation is essential in providing precise solutions in calculus for problems involving change and accumulation.

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

Suppose that \(f\) has a positive derivative for all values of \(x\) and that \(f(1)=0 .\) Which of the following statements must be true of the function $$g(x)=\int_{0}^{x} f(t) d t ?$$ Give reasons for your answers. \(\begin{array}{l}{\text { a. } g \text { is a differentiable function of } x \text { . }} \\ {\text { b. } g \text { is a continuous function of } x \text { . }} \\ {\text { c. The graph of } g \text { has a horizontal tangent at } x=1 \text { . }} \\ {\text { d. } g \text { has a local maximum at } x=1 \text { . }}\\\\{\text { e. } g \text { has a local minimum at } x=1.} \\\ {\text { f. The graph of } g \text { has an inflection point at } x=1.} \\\ {\text { g. The graph of } d g / d x \text { crosses the } x \text { -axis at } x=1}.\end{array}\)

The acceleration of a particle moving back and forth on a line is \(a=d^{2} s / d t^{2}=\pi^{2} \cos \pi t \mathrm{m} / \mathrm{sec}^{2}\) for all \(t .\) If \(s=0\) and \(v=\) 8 \(\mathrm{m} / \mathrm{sec}\) when \(t=0,\) find \(s\) when \(t=1 \mathrm{sec} .\)

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

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

Find the linearization of $$f(x)=2-\int_{2}^{x+1} \frac{9}{1+t} d t$$ at \(x=1.\)
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Statistics

Read Explanation

Decision Maths

Read Explanation

Pure Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Geometry

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.