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Math Help - average value

  1. #1
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    average value

    The average value of the function f(x) = (x-1)^2 on the interval from x=1 to x=5 is:


    a) -16/3
    b) 16/3
    c) 64/3
    d) 66/3
    e) 256/3
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  2. #2
    GAMMA Mathematics
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    Quote Originally Posted by DINOCALC09 View Post
    The average value of the function f(x) = (x-1)^2 on the interval from x=1 to x=5 is:


    a) -16/3
    b) 16/3
    c) 64/3
    d) 66/3
    e) 256/3
    Integrate the function and then divide by four.

    \int_1^5 (x-1)^2 dx = \frac{(x-1)^3}{3}|_1^5

    \int_1^5 (x-1)^2 dx = \frac{4^3}{3}

    \int_1^5 (x-1)^2 dx=64/3

    Divide that total area by four to get the average value of 16/3.
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  3. #3
    Lord of certain Rings
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    Quote Originally Posted by DINOCALC09 View Post
    The average value of the function f(x) = (x-1)^2 on the interval from x=1 to x=5 is:


    a) -16/3
    b) 16/3
    c) 64/3
    d) 66/3
    e) 256/3
    I think
    (b) \frac{16}{3}

    \frac{\int^4_0{t^2}\ dt }{4} =\frac14 \frac{4^3}{3} = \frac{16}{3}
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    why did they divide by 4? and how do they know what to integrate?


    Definition: Let f(x) be an integrable function on the closed interval [a,b]. Then the average value of f(x) on [a,b] is given by:

    \mbox{Average Value} = \frac 1{b - a} \int_a^b f(x)~dx
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