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Chapter 8: Problem 4

Use a table of integrals or a computer algebra system to evaluate the given integral. \(\int_{1}^{2} \frac{1}{x(3 x-2)} d x\)

Short Answer

Expert verified
The integral evaluates to \(-\frac{1}{6} \ln 2\).

Step by step solution

01

Look at the Integral

We want to evaluate the integral \( \int_{1}^{2} \frac{1}{x(3x-2)} dx\). To do this, we can look for a suitable technique like partial fraction decomposition which can simplify the integrand.
02

Decompose into Partial Fractions

Express the fraction \( \frac{1}{x(3x-2)} \) as partial fractions. We set \( \frac{1}{x(3x-2)} = \frac{A}{x} + \frac{B}{3x-2} \). Solve for \( A \) and \( B \) by equating the numerators: \( 1 = A(3x-2) + Bx \), which simplifies to \(1 = (3A + B)x - 2A\). By comparing coefficients, this yields two equations: \(3A + B = 0\) and \(-2A = 1\).
03

Solve for A and B

From \(-2A = 1\), we find \(A = -\frac{1}{2}\). Substitute \(A\) into \(3A + B = 0\) to find \(B\): \(3(-\frac{1}{2}) + B = 0\), so \(B = \frac{3}{2}\). Therefore, \( \frac{1}{x(3x-2)} = \frac{-1/2}{x} + \frac{3/2}{3x-2} \).
04

Integrate Each Term

Now integrate each term separately. The integral becomes \( \int_{1}^{2} \left(-\frac{1}{2x} + \frac{1}{2(3x-2)}\right) dx \). This separates into two integrals: \( -\frac{1}{2} \int_{1}^{2} \frac{1}{x} dx + \frac{1}{2} \int_{1}^{2} \frac{1}{3x-2} dx \).
05

Evaluate the Integrals

The first integral, \( \int \frac{1}{x} dx = \ln |x| \), evaluates from 1 to 2 as \( \ln 2 - \ln 1 = \ln 2 \). For the second integral, use substitution by letting \( u = 3x-2 \), thus \( du = 3dx \), or \( dx = \frac{1}{3}du \). Then the integral becomes \( \frac{1}{2} \left( \frac{1}{3} \int \frac{1}{u} du \right) \), which evaluates to \( \frac{1}{6} \left( \ln |u| \right)\) from \(u=1\) to \(u=4\), which results in \( \frac{1}{6}(\ln 4 - \ln 1) = \frac{1}{6} \ln 4 \).
06

Combine Results

Combining the results of the integrals, we have \( -\frac{1}{2} \ln 2 + \frac{1}{6} \ln 4 \). Simplifying further, \( \ln 4 = 2 \ln 2 \), so \( \frac{1}{6} \ln 4 = \frac{1}{6} \times 2 \ln 2 = \frac{1}{3} \ln 2 \). This results in \(-\frac{1}{2} \ln 2 + \frac{1}{3} \ln 2 = \left(\frac{-3}{6} + \frac{2}{6} \right) \ln 2 = -\frac{1}{6} \ln 2 \).

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Key Concepts

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

partial fraction decomposition
Partial Fraction Decomposition is a technique used in calculus to simplify complex rational expressions into a sum of simpler fractions called partial fractions. This process is particularly useful when integrating rational functions with polynomials in the denominator. To perform partial fraction decomposition, follow these steps:
  • Identify if the fraction can be decomposed. Ensure that the degree of the numerator is less than the degree of the denominator.
  • Express the fraction as a sum of fractions with unknown coefficients over the resolved factors of the denominator.
  • Equate the original fraction to the sum and solve algebraically for the coefficients.
In the given exercise, we applied partial fraction decomposition to \( \frac{1}{x(3x-2)} \) by expressing it as \( \frac{A}{x} + \frac{B}{3x-2} \). Solving for \( A \) and \( B \) gave us the values necessary to split the fraction, making integration more manageable.
definite integrals
A definite integral finds the accumulated value of a function within a specified interval, often representing an area under a curve. The definite integral is expressed as \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) form the bounds of integration.
  • This results in a numerical value rather than a function like in indefinite integrals.
  • The calculation involves evaluating the antiderivative at the upper and lower bounds and subtracting the latter from the former.
  • In our exercise, after applying partial fraction decomposition, we calculated the definite integral over \( x \) from 1 to 2.
By executing proper decomposition and solving within these bounds, we effectively determined the precise numerical area between the curve \( \frac{1}{x(3x-2)} \) and the x-axis over the stated interval.
substitution method
The Substitution Method, or u-substitution, is a common integrative technique that simplifies the process by substituting part of the integrand with a new variable. This technique resembles the chain rule in reverse. Here's how you can apply it:
  • Choose a part of the integrand to substitute with a single variable \( u \), often an inner function in a composite function.
  • Determine \( du \) by differentiating \( u \) with respect to \( x \).
  • Replace the corresponding parts of the integral with \( u \) and \( du \).
  • After integration, replace \( u \) with the original variable.
In our scenario, we used substitution by setting \( u = 3x - 2 \), transforming the related portion of the integrand, which made the integral easier to evaluate. This efficient method often converts a complex integral into a simpler form, enabling more straightforward computation.
logarithmic integration
Logarithmic Integration is a method for integrating functions that lead directly to a logarithmic function. The basic form \( \int \frac{1}{x} \, dx = \ln |x| + C \) underlines this method. It is especially useful when integrating functions that decompose to feature a denominator of the form \( x \) or \( au + b \).
  • Integrals requiring logarithmic integration often stem from partial fraction decompositions.
  • Performing u-substitution can set the stage for this approach, simplifying the integrand into a logarithmic form.
  • The evaluation provides a natural logarithm of the absolute value, which is then assessed over any designated bounds for definite cases.
In the example exercise, logarithmic integration was deployed post-decomposition, using \( \int \frac{1}{x} \, dx \) and substitutions to solve the components, solidifying the final evaluation through inherent properties of logarithmic functions.

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Most popular questions from this chapter

This exercise is designed to support the assertion that $$ \int_{-\infty}^{\infty} e^{-x^{2} / 2} / \sqrt{2 \pi} d x=1 $$ a. Use Simpson's Rule with \(n=100\) to approximate $$ \int_{-5}^{5} e^{-x^{2} / 2} / \sqrt{2 \pi} d x $$ b. Find the smallest positive integer \(b\) such that when you use Simpson's Rule with \(n=100\), the value your computer or calculator gives for \(\int_{-b}^{b} e^{-x^{2} / 2} / \sqrt{2 \pi} d x\) is 1 .

Determine whether the improper integral converges. If it does, determine the value of the integral. \(\int_{1}^{\infty} \frac{1}{\sqrt{x}(\sqrt{x}+1)} d x\)

Determine whether the improper integral converges. If it does, determine the value of the integral. \(\int_{-\infty}^{\infty} \frac{x^{3}}{\left(x^{4}+1\right)^{2}} d x\)

a. Show that $$ \frac{1}{x(x+1)}=\frac{1}{x}-\frac{1}{x+1} $$ b. Does the equation $$ \int_{1}^{\infty} \frac{1}{x(x+1)} d x \stackrel{?}{=} \int_{1}^{\infty} \frac{1}{x} d x-\int_{1}^{\infty} \frac{1}{x+1} d x $$ hold? Explain why or why not.

Approximate \(\int_{0}^{\pi / 2} \sqrt{1+\cos x} d x\) by using Simpson's Rule with \(n=2\). Calculate the exact value of the integral by using the relation $$ \sqrt{1+\cos x}=\sqrt{2} \cos \frac{x}{2} $$ and estimate the error that arises from the approximation.
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