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Hw help

  1. Jun 23, 2009 #1
    Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Draw a typical approximating rectangle and label its height and width.

    http://www.webassign.net/www28/symImages/2/b/4def6fad2db11c7dd01c3a136b1fba.gif


    Then find the area S of the region.


    1. How do i know whether to integrate with respect to x or y?


    2. is 2/x the same thing as x^-2 ?? (if not how do i rewrite it) ?


    help.

    i graphed it and didn't know what to do with the x = 7.
     
    Last edited by a moderator: Apr 24, 2017
  2. jcsd
  3. Jun 23, 2009 #2

    Office_Shredder

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    x=7 is going to be a vertical line, all points of the form (7,y).

    You can integrate with respect to either x or y, try writing both integrals out and see which one looks easier

    [tex] \frac{2}{x} =/= x^{-2}[/tex] I don't know why you would think that... [tex]\frac{2}{x} = 2* \frac{1}{x}[/tex] If you're unsure of how fractions and exponentials work this is going to be a difficult problem to solve
     
  4. Jun 23, 2009 #3

    HallsofIvy

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    I would start by determining where the two graphs intersect. Where is
    [tex]y= \frac{2}{x}= \frac{2}{x^2}[/tex]?

    Now, if y is a number between those two y values, for what x values is y= 1/x or y= 1/x2?
    If x is number between the two x-values of the intersection points, what are y= 1/x and y= 1/x2?

    Which of those is easier to integrate?
     
  5. Jun 23, 2009 #4
    i set the = and i think i got (0,1) is that right?

    and im not following the second part, sorry.
     
  6. Jun 24, 2009 #5

    HallsofIvy

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    No, that's not right. How did you do it? Did you put x= 0 and x= 1 back into the equation to check?
     
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