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    A question from Revision exercise 3, question 30.



    First the area of the rectangle: a(1) = a
    The area for A = rectangle - a
    area of rectangle is bounded by y=1 and y=(x-a)^2
    but the result doesn't seem to be correct.
    Please adivce, Thanks!
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    (Original post by bazlawth)
    A question from Revision exercise 3, question 30.



    First the area of the rectangle: a(1) = a
    The area for A = rectangle - a
    area of rectangle is bounded by y=1 and y=(x-a)^2
    but the result doesn't seem to be correct.
    Please adivce, Thanks!
    It looks like you've made a rectangle which sometimes works but then you don't really have a way of finding A or the other part of your rectangle.

    I think they want you to use another method (integration)
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    (Original post by bazlawth)
    A question from Revision exercise 3, question 30.



    First the area of the rectangle: a(1) = a
    The area for A = rectangle - a
    area of rectangle is bounded by y=1 and y=(x-a)^2
    but the result doesn't seem to be correct.
    Please adivce, Thanks!
    As per Sean - in this case you want to find a such that \displaystyle \int_0^{a} (x-a)^2 \, \mathrm{d}x = 9
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    (Original post by SeanFM)
    It looks like you've made a rectangle which sometimes works but then you don't really have a way of finding A or the other part of your rectangle.

    I think they want you to use another method (integration)
    Mixed up sth, now I got the answer. Thanks!
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    nice question
 
 
 
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