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    Q10) Where did I go wrong in my working out? The answer at the back of my textbook says it is 1/2Pi - 1.

    :confused:

    Edit: Right, I took Penguin's, Keshim's and ghostwalker's advice and did v2 - v1. And this is what I achieved:



    Could the textbook possibly be wrong?


    EDIT: CASE CLOSED! =D
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    (Original post by APonderingMind)
    Q10) Where did I go wrong in my working out? The answer at the back of my textbook says it is 1/2Pi - 1.

    :confused:
    tan^2(x) is the the lower curve, so your meant to do the integral of 1/2sec^2(x)-tan^2(x)
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    Well the way I'd do it is different to you..
    I'd intergrate the equation y=1/2sec^2x - tan^2x between the limits of 0 and Pi/4 and then times this answer by two to get the area in between the two curves..
    I would do it now but I'm off out so if you haven't worked it out later then quote me on here and I'll get round to helping out a bit better
    In the meantime try my way? But good luck :teehee:
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    (Original post by kashim91)
    tan^2(x) is the the lower curve, so your meant to do the integral of 1/2sec^2(x)-tan^2(x)
    I would still get -1 - 1/2Pi if I did it that way, not the correct answer
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    i can't remember how to do C3 maths...it was so long ago :doctor:
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    The sec^2 curve is on the top in the region you are considering, so you want V_2-V_1
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    (Original post by Penguinsaysquack)
    Well the way I'd do it is different to you..
    I'd intergrate the equation y=1/2sec^2x - tan^2x between the limits of 0 and Pi/4 and then times this answer by two to get the area in between the two curves..
    I would do it now but I'm off out so if you haven't worked it out later then quote me on here and I'll get round to helping out a bit better
    In the meantime try my way? But good luck :teehee:
    Hmm... still didn't get the same answer as the book.
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    (Original post by APonderingMind)
    Hmm... still didn't get the same answer as the book.
    Your mistake is on the second line - the -1 should be a +1.
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    (Original post by APonderingMind)
    Hmm... still didn't get the same answer as the book.
    You made an error on your first line of working -tan^2(x) becomes 1-sec^2(x)
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    (Original post by ghostwalker)
    You made an error on your first line of working -tan^2(x) becomes 1-sec^2(x)
    (Original post by Mark13)
    Your mistake is on the second line - the -1 should be a +1.

    tan(x^2) = Sec(x^2) -1

    so -(tanx^2) = -(secx^2 -1) = 1 - sec(x^2)

    Oopsy... sign error. Thankies
 
 
 
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