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Finding area given definite integral

  1. Jul 14, 2014 #1
    Question :
    View attachment 71328

    My question is how did the 2a and 2b come from??
    Equations:
    Area of trapezoid =(a+b/2)(h)
    Attempt:
    I know that the area of a trapezoid is (a+b/2)(h)
    However why is there now a 2a and 2b in its place? Could it be related to the 2s function??
     
    Last edited by a moderator: Jul 14, 2014
  2. jcsd
  3. Jul 14, 2014 #2
    Hope this helps

    ba 2s ds = [s2]ba
    this then becomes (b2-a2)

    its just integral laws
     
  4. Jul 14, 2014 #3

    SteamKing

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    Beats me.

    We have a homework template here at PF that we ask users to fill out when asking for HW help. It helps save time for the user to provide complete information about the problem and the user's attempt at solution

    Please review the rules in this post about posting HW problems correctly:

    https://www.physicsforums.com/showthread.php?t=617567
     
  5. Jul 14, 2014 #4
    I'm pretty sure the this is what you're looking for:
    $$ \int ^{b}_{a}2xdx=\left[ x^{2}\right] ^{b}_{a}=b^{2}-a^{2} $$


    ~| FilupSmith |~
     
  6. Jul 14, 2014 #5

    HallsofIvy

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    I cannot see the attachment. If your trapezoid has parallel sides of length a and b and height h, its area is NOT (a+ b/2)h. It is (a+ b)h/2 or ((a+ b)/2)h.
     
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