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Homework Help: 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


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    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


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    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


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    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.
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