1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Factoring an Integral

  1. Jun 24, 2012 #1
    Hey everybody I was wondering why when you factor an integral, the final answer, or area, is smaller than if you hadn't. Here's an example:
    [itex]\int[/itex][itex]\frac{x}{2x^2}[/itex] - [itex]\frac{x}{2x}[/itex] between 1 and 2.

    You would factor out [itex]\frac{1}{2}[/itex] and bring it in front of the integral, right? But, my final answer came out to be about .45 and when I graphed the original two lines, it seemed the actual area should have been around 1. Why is this? What is the correct answer?
     
  2. jcsd
  3. Jun 24, 2012 #2

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    You must have made an algebra mistake.

    Show us your steps each way.
     
  4. Jun 24, 2012 #3
    Integrate area below [itex]\frac{x}{2x^2}[/itex] and above [itex]\frac{x}{2x}[/itex] between x=1 and x=2.

    [itex]\int[/itex] ([itex]\frac{x}{2x^2}[/itex] - [itex]\frac{x}{2x}[/itex]
    = .5 [itex]\int[/itex] ([itex]\frac{x}{x^2}[/itex] - 1)
    =.5[itex]\int[/itex] (1 - [itex]\frac{x}{x^2}[/itex]) because 1 is now above x/x^2
    =.5[itex]\int[/itex] (1 - x^-1)
    =.5 (x - ([itex]\frac{x}{x^2}[/itex] ^2)/2) between 1 and 2

    =.5 ((2 - [itex]\frac{2}{4}[/itex]^2)/2) - (1- [itex]\frac{1}{4}[/itex]^2)/2)
    =.5 (2-.125) - (1-.03125)
    =.5 (1.875 - .96875)
    =.5 (.90625) = .453125

    The graph between the original two lines, however, seems to have a greater area than .45.
     
  5. Jun 24, 2012 #4

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I'm curious where you got this problem. I can't imagine any book phrasing it that way.

    Your antiderivative of ##x^{-1}## is incorrect. Have you studied the derivative on the natural logarithm function yet?
     
  6. Jun 24, 2012 #5
    Actually I have:
    If f(x) = ln x, then f ' x=[itex]\frac{1}{x}[/itex]

    [itex]\int[/itex] [itex]\frac{1}{x}[/itex] = ln x
    I didnt think to use it though.
     
    Last edited: Jun 24, 2012
  7. Jun 24, 2012 #6

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Yes, you did make several errors, one of which LCKurtz pointed out.

    You have that the anti-derivative of x-1 is [itex]\displaystyle \frac{1}{2}\left(\frac{x}{x^2}\right)^2\,,[/itex] as near as I can tell. This is equivalent to [itex]\displaystyle \frac{1}{2}x^{-2}\,.[/itex] But the derivative of [itex]\displaystyle \frac{1}{2}x^{-2}\ [/itex] is [itex]\displaystyle -x^{-1}\,,[/itex] not [itex]\displaystyle x^{-1}\ .[/itex] So you have a sign error.

    After that, you fail to distribute 1/2 of the anti-derivative, when you plug-in the limits of integration.

    There may be additional errors in your working of the problem.

    Also, in making up the problem, you failed to realize that [itex]\displaystyle \frac{x}{2x^2}[/itex] is below [itex]\displaystyle \frac{x}{2x}[/itex] on the interval, 1 < x < 2 .
     
  8. Jun 24, 2012 #7
    Thanks everyone, I guess it was just an error.
     
  9. Jun 24, 2012 #8
    You're right about the graph! I entered it totally wrong! Thank you very much.
     
  10. Jun 24, 2012 #9
    I figured it all out thanks to Sammy and LCKurtz. Thanks everyone, I had just entered the graph wrong.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Factoring an Integral
  1. Integrating Factor (Replies: 3)

  2. Integrating factor (Replies: 12)

  3. Integrating factors (Replies: 6)

  4. Integrating Factor (Replies: 4)

  5. Integrating factor (Replies: 5)

Loading...