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Math Help - Curve area

  1. #1
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    Question Curve area

    I got a curve y = n ( 16-x2 )
    Now the question is to find 'n' when the area between the curve and x axis is 33.

    how would i go about it?
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    Quote Originally Posted by taurus View Post
    I got a curve y = n ( 16-x2 )
    Now the question is to find 'n' when the area between the curve and x axis is 33.

    how would i go about it?
    think basic. how do we find the area between the curve and the axis?

    y = n \left( 16 - x^2 \right)

    where does y meet the x-axis? that is, when is y zero?

    \Rightarrow 0 = n \left( 16 - x^2 \right)

    \Rightarrow 0 = 16 - x^2

    \Rightarrow 0 = (4 + x)(4 - x)

    \Rightarrow x = \pm 4

    So we want: A = n \int_{-4}^{4} \left( 16 - x^2 \right)~dx = 2n \int_{0}^{4} \left( 16 - x^2 \right)~dx = 33

    now continue...
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  3. #3
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    just a quick question, wat topic is that rule you are using there?
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    Quote Originally Posted by taurus View Post
    just a quick question, wat topic is that rule you are using there?
    definite integrals
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  5. #5
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    Question

    hmm ok so

    => 2n[(16^2/2)-(4^3/3)]-[(16^2/2)-(0^3/3)] = 33
    => 2n(106.6666667 - 128) = 33
    => 2n(21.33333333) = 33

    But i think im goinf wrong?
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  6. #6
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    Quote Originally Posted by taurus View Post
    hmm ok so

    => 2n[(16^2/2)-(4^3/3)]-[(16^2/2)-(0^3/3)] = 33
    => 2n(106.6666667 - 128) = 33
    => 2n(21.33333333) = 33

    But i think im goinf wrong?
    yes, you are going wrong. find the indefinite integral first, and then plug in the limits using the fundamental theorem of calculus. what is the indefinite integral?
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  7. #7
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    ya the indefinate integral is 2n [(16^2/2)-(x^3/3)]?
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    Quote Originally Posted by taurus View Post
    ya the indefinate integral is 2n [(16^2/2)-(x^3/3)]?
    16 is a constant! what is the integral of any constant c?
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  9. #9
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    oh right 1?. so

    ya the indefinate integral is 2n [1-(x^3/3)]
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  10. #10
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    Quote Originally Posted by taurus View Post
    oh right 1?. so

    ya the indefinate integral is 2n [1-(x^3/3)]
    no. the integral of 16 with repsect to x is 16x.

    remember, \int cx^n~dx = \frac {cx^{n + 1}}{n + 1} + C

    we can think of 16 as 16x^0

    so to integrate a constant, we just attach the variable we are integrating with respect to

    anyway, now continue
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  11. #11
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    2n \int_{0}^{4} \left( 16 - x^2 \right)~dx = 2n \left[ 16x - \frac 13x^3 \right]_{0}^{4} = 2n \cdot \frac {128}3 = 33

    now solve for n
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