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Math Help - average value

  1. #1
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    average value

    Hi everyone,

    Could someone please tell me if this is correct?

    Find the numbers b such that the average value of f(x)=2+6x-3x^2 on the interval [0, b] is equal to 3.

    int. o to b 2+6x-3x^2

    2x+6x^2/2-3x^3/3 from 0 to b=
    2x+3x^2-x^3 from 0 to b=
    2b+3b^2-b^3=3
    b(2+3b-b^2)=3
    2+3b-b^2=3
    -1+3b-b^2=0
    -b^2+3b-1=0
    b=3
    b=.382
    b=2.618

    Therefore, b=2.618

    Thank you very much
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  2. #2
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    Quote Originally Posted by chocolatelover View Post
    Hi everyone,

    Could someone please tell me if this is correct?

    Find the numbers b such that the average value of f(x)=2+6x-3x^2 on the interval [0, b] is equal to 3.

    int. o to b 2+6x-3x^2

    2x+6x^2/2-3x^3/3 from 0 to b=
    2x+3x^2-x^3 from 0 to b=
    2b+3b^2-b^3=3
    b(2+3b-b^2)=3
    2+3b-b^2=3
    I see two things wrong up to this point:
    You haven't divided by b (the size of the interval) and if
    (x - a)(x - b) = c then we do not require that x - a = c or x - b = c. (We only do this when c = 0.)

    Let me write this out to the point you have:
    3 = \frac{1}{b - 0} \int_0^b(2 + 6x - 3x^2)~dx

    3 = \frac{1}{b}(2b + 3b^2 - b^3)

    3b = 2b + 3b^2 - b^3

    Now solve for b.

    -Dan
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  3. #3
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    Thank you very much

    I got 3=1/b(2b+3b^2-b^3)
    3=2b/b+3b^2/b-b^3/b
    0=-1+3b-b^2
    b=.382, 2.618

    Which one would I use? They both can't be right, can they?

    Thank you very much
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by chocolatelover View Post
    I got 3=1/b(2b+3b^2-b^3)
    3=2b/b+3b^2/b-b^3/b
    0=-1+3b-b^2
    b=.382, 2.618

    Which one would I use? They both can't be right, can they?

    Thank you
    Why the doubled post 3 hours apart??

    Why can't there be two solutions? As a simple example look at the average value of sin(x) over the interval 0 to 2\pi. Now look at the average of sin(x) over the interval 0 to 4\pi. Both are 0.

    -Dan
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