/*! 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 36 Find the value of \(k\) that mak... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 36

Find the value of \(k\) that makes the antidifferentiation formula true. [Note: You can check your answer without looking in the answer section. How?] $$\int \frac{5}{2-3 x} d x=k \ln |2-3 x|+C$$

Short Answer

Expert verified
\k = -\frac{5}{3}\.

Step by step solution

01

Rewrite the Integral in a Suitable Form

Consider the integral \(\frac{5}{2-3x}dx\). Recognize that this is of the form \( \frac{f'(x)}{f(x)}dx\), which can be integrated using logarithm rules.
02

Perform the u-Substitution

Let \(u = 2 - 3x\). Then the derivative is \(du = -3dx\) or \(dx = -\frac{1}{3}du\).
03

Substitute and Integrate

Substitute \(u\) into the integral: \(\frac{5}{2-3x}dx = \frac{5}{u} \times -\frac{1}{3}du\). This simplifies to \(-\frac{5}{3} \times \frac{1}{u} du\). Now integrate: \(\frac{5}{3} \times \frac{-1}{u} du = -\frac{5}{3}\text{ln}|u| + C\).
04

Substitute Back u

Replace u with \(2 - 3x\) to get \(-\frac{5}{3}\text{ln}|2-3x|+C\).
05

Identify the Coefficient k

Compare with \(k \text{ln}|2-3x|+C\). It follows that \k = -\frac{5}{3}\.

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.

u-substitution
The concept of u-substitution is crucial for antidifferentiation. It simplifies complex integrals by converting them into a more recognizable form.
In our problem, we have the integral \[ \frac{5}{2-3x} dx \]. By recognizing a pattern, we can substitute variables to make integration easier.
Let鈥檚 set \[ u = 2 - 3x \] which turns the integral鈥檚 denominator into a simpler variable. The derivative of \[ u \] is \[ du = -3dx \] or \[ dx = -\frac{1}{3} du \]. By substituting, we get a new integral with \[ u \] replacing \[ 2 - 3x \]. This approach is a key step in breaking down more difficult integrals.
integration by parts
Another method for antidifferentiation is integration by parts. While not used in this specific problem, it鈥檚 handy for different functions.
The formula arises from the product rule for differentiation and states \[ \text{鈭珆 u dv = uv - \text{鈭珆 v du \].
This technique is especially useful for integrating products of functions or for functions that simplify upon differentiation. Knowing when and how to apply integration by parts opens up a greater range of problems you can solve.
logarithmic integration
Logarithmic integration is employed when facing integrals in the form \[ \frac{f'(x)}{f(x)} dx \].
This method was crucial in our exercise. The function \[ \frac{5}{2-3x} \] fits the logarithmic form by recognizing \[ 2-3x \] is \[ f(x) \] and its derivative \[ -3 \].
After substituting \[ u = 2 - 3x \], our integral changes to \[ \frac{5}{u} \times -\frac{1}{3} du \], which highlights the natural logarithm rule \[ \text{鈭珆 \frac{du}{u} = \text{ln}|u| + C \]. This simplifies our problem dramatically.
definite integrals
Definite integrals differ from indefinite integrals as they calculate the accumulated value under a curve within specific limits \[ [a, b] \].
No reconsidering constants \[ C \] since they cancel in the calculation. For example, if charged with evaluating \[ \text{鈭珆_{1}^{2} \frac{5}{2-3x} dx \], the approach remains similar to our original problem but includes evaluating endpoints. Substituting from our indefinite example will involve altering limits to match substitution steps. Practicing definite integrals reinforces our understanding by tying theoretical methods to practical boundaries.
integration techniques
Various integration techniques exist to tackle different integrals. Here are a few:
  • u-substitution 鈥� Simplifies by replacing variables.
  • Integration by parts 鈥� Handy for products of functions.
  • Partial fractions 鈥� Breaks down rational functions.
  • Trigonometric integrals 鈥� Uses identities for trigonometric functions.
  • Improper integrals 鈥� Handles infinite limits or discontinuities.
Exploring these techniques equips students with a robust toolkit for tackling various integrals. Familiarity with each ensures adaptability in solving myriad mathematical problems.

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

Determine the average value of \(f(x)\) over the interval from \(x=a\) to \(x=b,\) where. $$f(x)=\frac{1}{\sqrt{x}} ; a=1, b=9$$

You took a \(\$ 200,000\) home mortgage at an annual interest rate of \(3 \% .\) Suppose that the loan is amortized over a period of 30 years, and let \(P(t)\) denote the amount of money (in thousands of dollars) that you owe on the loan after \(t\) years. A reasonable estimate of the rate of change of \(P\) is given by \(P^{\prime}(t)=-4.1107 e^{0.03 t}\) (a) Approximate the net change in \(P\) after 20 years. (b) What is the amount of money owed on the loan after 20 years? (c) Verify that the loan is paid off in 30 years by computing the net change in \(P\) after 30 years.

The rate of change of a population with emigration is given by \(P^{\prime}(t)=\frac{7}{300} e^{t / 25}-\frac{1}{80} e^{t / 16}\) where \(P(t)\) is the population in millions, \(t\) years after the year 2000 . (a) Estimate the change in population as \(t\) varies from 2000 to 2010 (b) Estimate the change in population as \(t\) varies from 2010 to 2040. Compare and explain your answers in (a) and (b).

Find the value of \(k\) that makes the antidifferentiation formula true. [Note: You can check your answer without looking in the answer section. How?] $$\int \frac{3}{2+x} d x=k \ln |2+x|+C$$

An investment grew at an exponential rate \(R(t)=700 e^{0.07 t}+1000,\) where \(t\) is in years and \(R(t)\) is in dollars per year. Approximate the net increase in value of the investment after the first 10 years (as \(t\) varies from 0 to 10 ).
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Decision Maths

Read Explanation

Logic and Functions

Read Explanation

Statistics

Read Explanation

Calculus

Read Explanation

Applied Mathematics

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.