calculus

Sketch the region in the first quadrant enclosed by y=6/x, y=2x, and y=.5x. Decide whether to integrate with respect to x or y. Then find the area of the region.

need final area as answer please, and step would be helpful also if you can post.

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asked by Nick
  1. sketch it of course

    where does y = 2x hit y=6/x?
    2 x = 6/x
    x^2 = 3
    x = sqrt 3
    then y = 2 sqrt 3
    so
    from x = 0 to x = sqrt 3 integrate y = (2 x-.5 x) = 1.5 x
    1.5 x^2/2
    .75 ( 3-0) = 2.25

    then where does y = .5 x hit y = 6/x?
    .5 = 6/x^2
    x^2 = 12
    x = 2 sqrt 3
    so integrate y = (6/x - .5 x) from x = sqt 3 to x = 2 sqrt 3
    (6 ln x - x^2/4)
    6 ln (2 sqrt 3 - sqrt 3) - .25(12-3)
    2.62
    so 2.25 + 2.62

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    posted by Damon
  2. Nick, with a little effort on your part to complete the solution I gave you and some small adjustment for this question, you could have saved
    the duplication of this solution by Damon.

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    posted by Reiny
  3. Oh my :(

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    posted by Damon

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