/*! 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 29 $$\text {Evaluate the following ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 29

$$\text {Evaluate the following integrals.}$$ $$\int_{-1}^{2} \frac{5 x}{x^{2}-x-6} d x$$

Short Answer

Expert verified
Question: Evaluate the definite integral of the given function from -1 to 2: $$\int_{-1}^{2} \frac{5x}{x^2 - x - 6}\,dx$$ Answer: The value of the definite integral is $$\int_{-1}^{2} \frac{5x}{x^2 - x - 6}\,dx = 6\ln2$$.

Step by step solution

01

Identify the integral and interval

The given integral is, $$\int_{-1}^{2} \frac{5x}{x^2 - x - 6}\,dx$$ The interval of integration is from -1 to 2.
02

Factor the denominator

First, we will factor the denominator of the rational function (if possible) to simplify it. The denominator is given by, $$x^2 - x - 6$$ which can be factored as, $$(x-3)(x+2)$$
03

Perform partial fraction decomposition

Now that we have factored the denominator, we can perform partial fraction decomposition: $$\frac{5x}{(x-3)(x+2)} = \frac{A}{x-3} + \frac{B}{x+2}$$ Multiplying both sides by the common denominator, we get, $$5x = A(x+2) + B(x-3)$$ Now we will find the values of A and B.
04

Solve for the constants A and B

To find the values of A and B, we will equate the coefficients of the corresponding powers of x and solve the system of equations: For the coefficient of x, we have $$5 = A + B$$ Now we will choose convenient values of x to eliminate one of the coefficients: If x = 3, we get $$15 = 5A$$ Which gives us, $$A = 3$$ Substitute A in the first equation to find B: $$5 = 3 + B$$ Which gives, $$B = 2$$ So the partial fraction decomposition is, $$\frac{5x}{(x-3)(x+2)} = \frac{3}{x-3} + \frac{2}{x+2}$$
05

Integrate the simplified functions

Now we will integrate each term separately over the interval from -1 to 2: $$\int_{-1}^{2} \frac{5x}{x^2 - x - 6}\,dx = \int_{-1}^{2}\left(\frac{3}{x-3}+\frac{2}{x+2}\right)\,dx$$ Here, we can evaluate both integrals separately: $$\int_{-1}^{2} \frac{3}{x - 3}\,dx + \int_{-1}^{2} \frac{2}{x + 2}\,dx$$ The integral of both expressions can be easily found using the ln function: $$3\int_{-1}^{2} \frac{1}{x-3}\,dx + 2\int_{-1}^{2} \frac{1}{x+2}\,dx$$ $$= 3[\ln|x-3|]_{-1}^2+2[\ln|x+2|]_{-1}^{2}$$
06

Evaluate the integrals at the limits

Finally, we will evaluate each expression at the upper and lower limits, then find the difference: $$= 3[\ln|2-3|-\ln|-1-3|]+2[\ln|2+2|-\ln|-1+2|]$$ $$= 3[\ln|-1|-\ln|-4|]+2[\ln4-\ln1]$$ $$= 3(\ln4-\ln1)-2(\ln1-\ln|-1|)$$ $$= 3(\ln4)-2(\ln1)$$ $$= 3(2\ln2)$$ $$= 6\ln2$$ So, the value of the definite integral is, $$\int_{-1}^{2} \frac{5x}{x^2 - x - 6}\,dx = 6\ln2$$

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.

Definite Integrals
A definite integral is a way to calculate the area under a curve within a specific interval. Think of it as summing up tiny rectangles under the curve, from one point to another.
  • The notation \( \int_{a}^{b} f(x) \, dx \) specifies the integral of a function \( f(x) \) from a lower limit \( a \) to an upper limit \( b \).
  • When you evaluate a definite integral, you are finding the net area. This means areas above the x-axis are positive, and areas below are negative.
  • Use the Fundamental Theorem of Calculus to find the definite integral. This involves evaluating an antiderivative at the boundary points \( a \) and \( b \) and finding their difference.
In our example, the definite integral was \( \int_{-1}^{2} \frac{5x}{x^2-x-6} \, dx \). After simplifying it, the goal was to find the result of this calculation.
Partial Fraction Decomposition
Partial fraction decomposition is a technique used to break down complex fractions into simpler ones, making them easier to integrate. Imagine it as splitting a complex puzzle into easier parts.
  • Start by factoring the denominator of the rational function. This will help you express the fraction as a sum of simpler fractions.
  • Set up an equation where the original fraction equals a sum of fractions with unknown coefficients, like \( \frac{A}{x-3} + \frac{B}{x+2} \).
  • Multiply everything by the common denominator to clear the fractions, and then solve for the coefficients by substituting convenient values or comparing coefficients.
In this problem, the equation \( \frac{5x}{(x-3)(x+2)} \) was broken into \( \frac{3}{x-3} + \frac{2}{x+2} \). This simplifies the integration process significantly.
Integration Techniques
Integration techniques are methods used to find the integral of a function. In this example, after decomposing into partial fractions, we move to integration.
  • The integral \( \int \frac{1}{x-a} \, dx \) can be solved with natural logarithms: \( \ln|x-a| + C \), where \( C \) is the constant of integration for indefinite integrals.
  • When dealing with definite integrals, the constant \( C \) cancels out, so you only focus on evaluating the natural logarithm expression at the given limits.
  • Integrating each fraction separately simplifies the process and allows for more straightforward evaluation at the boundaries.
After finding the partial fractions, we used the natural log function to integrate \( \frac{3}{x-3} \) and \( \frac{2}{x+2} \). This makes handling each part of the expression much easier.
Factorization
Factorization is the process of writing a polynomial as a product of its factors. It's like transforming an equation into a more digestible form.
  • Start by identifying if the polynomial is in a form that can be factored, such as quadratic or higher-order polynomials.
  • Look for patterns or use methods such as trial and error, grouping, or the quadratic formula to find factors.
  • Factoring makes other operations, such as integration, much simpler because it often reduces the complexity of the expression.
In this exercise, the expression \( x^2 - x - 6 \) was factored into \( (x-3)(x+2) \). This critical step allowed us to proceed with partial fraction decomposition more easily, simplifying the integration task ahead. By breaking it down into its components, it became manageable and intuitive.

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

Normal distribution of movie lengths A study revealed that the lengths of U.S. movies are normally distributed with a mean of 110 minutes and a standard deviation of 22 minutes. This means that the fraction of movies with lengths between \(a\) and \(b\) minutes (with \(a

The work required to launch an object from the surface of Earth to outer space is given by \(W=\int_{R}^{\infty} F(x) d x,\) where \(R=6370 \mathrm{km}\) is the approximate radius of Earth, \(F(x)=\frac{G M m}{x^{2}}\) is the gravitational force between Earth and the object, \(G\) is the gravitational constant, \(M\) is the mass of Earth, \(m\) is the mass of the object, and \(G M=4 \times 10^{14} \mathrm{m}^{3} / \mathrm{s}^{2}\) a. Find the work required to launch an object in terms of \(m\). b. What escape velocity \(v_{e}\) is required to give the object a kinetic energy \(\frac{1}{2} m v_{e}^{2}\) equal to \(W ?\) c. The French scientist Laplace anticipated the existence of black holes in the 18 th century with the following argument: If a body has an escape velocity that equals or exceeds the speed of light, \(c=300,000 \mathrm{km} / \mathrm{s},\) then light cannot escape the body and it cannot be seen. Show that such a body has a radius \(R \leq 2 G M / c^{2} .\) For Earth to be a black hole, what would its radius need to be?

A powerful tool in solving problems in engineering and physics is the Laplace transform. Given a function \(f(t),\) the Laplace transform is a new function \(F(s)\) defined by $$F(s)=\int_{0}^{\infty} e^{-s t} f(t) d t$$ where we assume s is a positive real number. For example, to find the Laplace transform of \(f(t)=e^{-t},\) the following improper integral is evaluated using integration by parts: $$F(s)=\int_{0}^{\infty} e^{-s t} e^{-t} d t=\int_{0}^{\infty} e^{-(s+1) t} d t=\frac{1}{s+1}$$ Verify the following Laplace transforms, where a is a real number. $$f(t)=\cos a t \rightarrow F(s)=\frac{s}{s^{2}+a^{2}}$$

Estimating error Refer to Theorem 8.1 in the following exercises. Let \(f(x)=\cos x^{2}\) a. Find a Midpoint Rule approximation to \(\int_{-1}^{1} \cos x^{2} d x\) using \(n=30\) subintervals. b. Calculate \(f^{\prime \prime}(x)\) c. Explain why \(\left|f^{*}(x)\right| \leq 6\) on [-1,1] d. Use Theorem 8.1 to find an upper bound on the absolute error in the estimate found in part (a).

Evaluate \(\int_{0}^{\pi / 4} \ln (1+\tan x) d x\) using the following steps. a. If \(f\) is integrable on \([0, b],\) use substitution to show that $$\int_{0}^{b} f(x) d x=\int_{0}^{b / 2}(f(x)+f(b-x)) d x$$ b. Use part (a) and the identity tan \((\alpha+\beta)=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}\) to evaluate \(\int_{0}^{\pi / 4} \ln (1+\tan x) d x\) (Source: The College Mathematics Journal, \(33,4,\) Sep 2004 )
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Decision Maths

Read Explanation

Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Geometry

Read Explanation

Discrete 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.