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Math Help - area between curves

  1. #1
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    area between curves

    Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region.


    From (5/3) to (25/3) 3x^2-5x dx

    I keep getting 407.40737 and it's not right! I dunno what I am doing wrong! BAH!
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  2. #2
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    Quote Originally Posted by MathNeedy18 View Post
    Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region.


    From (5/3) to (25/3) 3x^2-5x dx

    I keep getting 407.40737 and it's not right! I dunno what I am doing wrong! BAH!
    Hello,

    first calculate the intercepts of the two functions: (0, 0), (5/3, 25/3)

    The enclosed area is calculated by the difference of the two functions:

    d(x)=5x-3x^2

    A=\int_0^{\frac53}(5x-3x^2)dx

    I'm sure that you can do the rest.

    (For confirmation only: A=\frac{125}{54}\approx 2.315 )
    Attached Thumbnails Attached Thumbnails area between curves-encl_area.jpg  
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  3. #3
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    So first you need to find the points of intersection: (0,0) and (5/3,25/3).
    Then you can integrate with respect to x. Make sure you take the function on top 5x and subtract the lower function 3x^2.
    \int_0^\frac{5}{3}(5x-3x^2)dx
    I get 125/54.
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  4. #4
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    omg! Thanks! I made the dumbest mistakes! I was finding it from 25/3 to 5/3 instead of 0 to 5/3!!!
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