Download the App!

Get 24/7 study help with the Numerade app for iOS and Android! Enter your email for an invite.

Sent to:
Search glass icon
  • Login
  • Textbooks
  • Ask our Educators
  • Study Tools
    Study Groups Bootcamps Quizzes AI Tutor iOS Student App Android Student App StudyParty
  • For Educators
    Become an educator Educator app for iPad Our educators
  • For Schools

Problem

Evaluate the integral. $ \displaystyle \int_1^…

07:28

Question

Answered step-by-step

Problem 17 Medium Difficulty

Evaluate the integral.

$ \displaystyle \int_1^2 \frac{4y^2 - 7y - 12}{y (y + 2)(y - 3)}\ dy $


Video Answer

Solved by verified expert

preview
Numerade Logo

This problem has been solved!

Try Numerade free for 7 days

JH
J Hardin
Numerade Educator

Like

Report

Textbook Answer

Official textbook answer

Video by J Hardin

Numerade Educator

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Related Courses

Calculus 2 / BC

Calculus: Early Transcendentals

Chapter 7

Techniques of Integration

Section 4

Integration of Rational Functions by Partial Fractions

Related Topics

Integration Techniques

Discussion

You must be signed in to discuss.
Top Calculus 2 / BC Educators
Grace He
Kristen Karbon

University of Michigan - Ann Arbor

Michael Jacobsen

Idaho State University

Joseph Lentino

Boston College

Calculus 2 / BC Courses

Lectures

Video Thumbnail

01:53

Integration Techniques - Intro

In mathematics, integration is one of the two main operations in calculus, with its inverse, differentiation, being the other. Given a function of a real variable, an antiderivative, integral, or integrand is the function's derivative, with respect to the variable of interest. The integrals of a function are the components of its antiderivative. The definite integral of a function from a to b is the area of the region in the xy-plane that lies between the graph of the function and the x-axis, above the x-axis, or below the x-axis. The indefinite integral of a function is an antiderivative of the function, and can be used to find the original function when given the derivative. The definite integral of a function is a single-valued function on a given interval. It can be computed by evaluating the definite integral of a function at every x in the domain of the function, then adding the results together.

Video Thumbnail

27:53

Basic Techniques

In mathematics, a technique is a method or formula for solving a problem. Techniques are often used in mathematics, physics, economics, and computer science.

Join Course
Recommended Videos

00:45

Evaluate the integral.

06:08

Evaluate the integral.
…

02:32

Evaluate the integral.
…

0:00

Evaluate the integral.
…

00:53

Evaluate the indefinite in…

05:16

Evaluate the integral.

…

01:11

Evaluate the integral.
…

01:54

Evaluate the integral: 2 4…

01:41

Evaluate the following int…

02:12

Evaluate the indefinite in…

Watch More Solved Questions in Chapter 7

Problem 1
Problem 2
Problem 3
Problem 4
Problem 5
Problem 6
Problem 7
Problem 8
Problem 9
Problem 10
Problem 11
Problem 12
Problem 13
Problem 14
Problem 15
Problem 16
Problem 17
Problem 18
Problem 19
Problem 20
Problem 21
Problem 22
Problem 23
Problem 24
Problem 25
Problem 26
Problem 27
Problem 28
Problem 29
Problem 30
Problem 31
Problem 32
Problem 33
Problem 34
Problem 35
Problem 36
Problem 37
Problem 38
Problem 39
Problem 40
Problem 41
Problem 42
Problem 43
Problem 44
Problem 45
Problem 46
Problem 47
Problem 48
Problem 49
Problem 50
Problem 51
Problem 52
Problem 53
Problem 54
Problem 55
Problem 56
Problem 57
Problem 58
Problem 59
Problem 60
Problem 61
Problem 62
Problem 63
Problem 64
Problem 65
Problem 66
Problem 67
Problem 68
Problem 69
Problem 70
Problem 71
Problem 72
Problem 73
Problem 74
Problem 75

Video Transcript

Let's evaluate the definite integral From 1 to 2, we have a polynomial divided by a polynomial, and the denominator is already factored. So using what the textbook were called, case one here in the section they refer to this as case one. We have three distinct and linear factor. So the partial fraction decomposition should look something like over y be over y plus two see over y minus three. And then we could go ahead and multiply both sides by this denominator over here. So after multiplying four y squared minus seven Y minus 12 on the left, on the right, we have a no y plus two y minus three. And then for B, we have y and y minus three. And finally for C. Why? But why in a Y plus two. So let's just go ahead and rewrite this left hand or the right hand side. Just multiply things out. Y squared minus Y minus six. This is for a and then for me. Why squared minus three y and for C y squared plus two y. Now, one more thing to do here on this right hand side is to just rewrite it by factoring out first, we'll factor Ry Square. We'll have a plus B plus C one here, one here, one there. Yeah, Then we'll have. For after we found her to lie, we'll have a minus a a minus three B plus to see And then finally, the constant term. On the right hand side, it's just minus six. A. So on the left hand side, the term the constant in front of the White Square is a four on the right hand side, A plus B plus C that gives us an equation. Similarly, for the white term, we have a negative seven on the left, and this is what we have on the right. It gives us our second equation minus a minus three B plus to see equals minus seven. And then on the left hand side, the constant term was minus 12. Whereas on the left hand side it was minus six A. So we can go ahead. And those are three equations. We have a three by three system and A B and C. So, taking this last equation here we see that A s two. So there's already one of our values. And now let's just go ahead and plug this a value into the previous two equations. So we'll replace that A with the two and this over here that will become minus two. And then we can rewrite these equations. So this first one here that just becomes B plus C equals two. And for the other equation, add that negative two to the other side. We have minus three B plus two C equals negative. Seven plus two is minus five. So let's go on to the next page, running out of room here. But we will have this two by two system and B and C to solve, and then we'll have our coefficients. So we had two equations from the previous page. We had B Plus C is, too, and a minus three B plus to see is minus five. So let's just go ahead and solve this for B or C. So taking this first equation C equals two minus B and then plug this into the second equation. We'll have negative three b plus two, and then R C is two minus B, and the right hand side is negative. Five. So let's just go ahead and solve this for B, so we'll have minus three B and then minus two more B. That's minus five B and then we have a plus four. So maybe add the five to the left and then push the five beats on the other side. And then we C B is just 9/5 and then using this equation up here, let's solve proceed. So plugging B equals 9/5. In we have C equals two, which is 10/5 minus B, which is 9/5. That leaves us with 1/5. So now we've found our coefficient A, B and C. On the previous page, we found that a was to sell me. Just rewrite that here from page one. And then Now let's go on to the next page. Mhm. So let's rewrite the original integral 1 to 2. Yeah, why? Why? Plus two y minus three. This is what we intended to evaluate. We just did the partial fraction decomposition. We found a B and C. So now let's go ahead and plug those in. So a over Y becomes too over y be over Y plus two becomes 9/5 over y plus two. And then we found that C was 1/5 over Y minus three d Y Yeah, and this is a much easier integral for us. If this plus two in minus three or throwing you off, you go ahead. And do you use up here? U equals Y plus two or for the second one, U equals my mind. History. Yeah, let's go ahead and evaluate those integral the first one to natural log Absolute value y The second one That's a five in the denominator. 9/5 and that natural log y plus two plus 1/5. Natural log. Absolute value. Why my mystery and then our endpoints 1 to 2. Let's go ahead and plug in those endpoints. So plug in the two first we have natural log of four plus 1/5, and then we have two minus three, which is minus one. But then absolute value gives us a one so natural log of one. Now we plug in one for why we have 1/5 natural log of negative, too. But then we take absolute value. That's a two there. And now we cancel and simplify as much as we can. We know that natural log of one is zero. So these go away zero and zero. That's the whole the whole term there. And now we combine as much as we can, so we have to natural log two minus 1/5. Natural log, too. We can write. That is nine number five l N two. And then we have 9/5. Ln four minus 9/5 l N three. Mm. And here we can. That can be our final answer. It's a number, but we can go ahead and combine. So here, let's write this for is two squared. And then using the properties of the algorithm, we can pull out this two in front of the nine and then multiply. So we have 9/5 Ln two and then two times nine. So we have 18 there, over five, minus 9/5. That should have been Ellen. Mm. Ln of two minus 9/5 l n three. And then just combine those first two fractions and we have 27/5 natural log of two minus nine, Number five Ln of three. And there's a final answer

Get More Help with this Textbook
James Stewart

Calculus: Early Transcendentals

View More Answers From This Book

Find Another Textbook

Study Groups
Study with other students and unlock Numerade solutions for free.
Math (Geometry, Algebra I and II) with Nancy
Arrow icon
Participants icon
83
Hosted by: Ay?Enur Çal???R
Math (Algebra 2 & AP Calculus AB) with Yovanny
Arrow icon
Participants icon
52
Hosted by: Alonso M
See More

Related Topics

Integration Techniques

Top Calculus 2 / BC Educators
Grace He

Numerade Educator

Kristen Karbon

University of Michigan - Ann Arbor

Michael Jacobsen

Idaho State University

Joseph Lentino

Boston College

Calculus 2 / BC Courses

Lectures

Video Thumbnail

01:53

Integration Techniques - Intro

In mathematics, integration is one of the two main operations in calculus, with its inverse, differentiation, being the other. Given a function of a real variable, an antiderivative, integral, or integrand is the function's derivative, with respect to the variable of interest. The integrals of a function are the components of its antiderivative. The definite integral of a function from a to b is the area of the region in the xy-plane that lies between the graph of the function and the x-axis, above the x-axis, or below the x-axis. The indefinite integral of a function is an antiderivative of the function, and can be used to find the original function when given the derivative. The definite integral of a function is a single-valued function on a given interval. It can be computed by evaluating the definite integral of a function at every x in the domain of the function, then adding the results together.

Video Thumbnail

27:53

Basic Techniques

In mathematics, a technique is a method or formula for solving a problem. Techniques are often used in mathematics, physics, economics, and computer science.

Join Course
Recommended Videos

00:45

Evaluate the integral. $ \displaystyle \int^3_1 \frac{y^3 - 2y^2 - y}{y^2} …

06:08

Evaluate the integral. $\int_{1}^{2} \frac{4 y^{2}-7 y-12}{y(y+2)(y-3)} d y$

02:32

Evaluate the integral. 2 4y2 − 2y − 12 y(y + 2)(y − 3) dy 1

0:00

Evaluate the integral. $$ \int_{1}^{3} \frac{y^{3}-2 y^{2}-y}{y^{2}} d y $$

00:53

Evaluate the indefinite integral. $ \displaystyle \int y^2 (4 - y^3)^{2/3} \…

05:16

Evaluate the integral. $ \displaystyle \int \frac{y}{(y + 4)(2y - 1)}\ dy $

01:11

Evaluate the integral. $\int_{1}^{2}(1+2 y)^{2} d y$

01:54

Evaluate the integral: 2 4y2 6y 12 dy Yly + 2)(y - 3)

01:41

Evaluate the following integrals. $$\int \frac{1}{\left(y^{2}+1\right)\left(y^{…

02:12

Evaluate the indefinite integral. $$ \int y^{2}\left(4-y^{3}\right)^{2 / 3} d y…

Add To Playlist

Hmmm, doesn't seem like you have any playlists. Please add your first playlist.

Create a New Playlist

`

Share Question

Copy Link

OR

Enter Friends' Emails

Report Question

Get 24/7 study help with our app

 

Available on iOS and Android

About
  • Our Story
  • Careers
  • Our Educators
  • Numerade Blog
Browse
  • Bootcamps
  • Books
  • Notes & Exams NEW
  • Topics
  • Test Prep
  • Ask Directory
  • Online Tutors
  • Tutors Near Me
Support
  • Help
  • Privacy Policy
  • Terms of Service
Get started