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Math Help - Averege value of a function

  1. #1
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    Averege value of a function

    I'd like to know how to find the averege values of the following functions

    a)f(x)=(x-4)^2 on the interval [2,5]

    and

    b)f(x)=6sin(x)-3sin(2x) on the interval [0,5]
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  2. #2
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    Ave(f)_{[a,b]}  = \frac{{\int\limits_a^b f }}<br />
{{b - a}}
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  3. #3
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    You find the average of a set of numbers by adding them up and dividing by the number of terms. An integral is essentially a sum of all terms between a and b. So, you simply need to find the integral of the function, and divide it by the number of terms. In equation form:

    f(x)_{ave}\vert^b_a = \frac {\int^b_a f(x) dx}{b-a}
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  4. #4
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    more info please

    thanks, i actually know the formula but i was having trouble with these spicific questions, i might just be making some simple mistake, but i'm not sure what i did wrong
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  5. #5
    Senior Member topher0805's Avatar
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    Quote Originally Posted by bam017 View Post
    thanks, i actually know the formula but i was having trouble with these spicific questions, i might just be making some simple mistake, but i'm not sure what i did wrong
    Do you know how to find the integral of a function?
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  6. #6
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    yes

    i'm fimiliar with this type of problem(averege value of a function) and very fimiliar with integration; however, I keep getting the wrong answer on these two problems
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  7. #7
    Senior Member topher0805's Avatar
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    Quote Originally Posted by bam017 View Post
    i'm fimiliar with this type of problem(averege value of a function) and very fimiliar with integration; however, I keep getting the wrong answer on these two problems
    Show us your work. You've probably just made a simple error.
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  8. #8
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    here's what i've got so far

    f(averege)=1/(b-a)$f(x)dx...I'm using $ for the integral sign
    a)1/3$[2,5](x^2-8x+16)dx
    1/3[(x^3/3)-4x^2+16(evaluated from 2 to 5)
    1/3[(5^3/3-4(5)^2-16(5))-(2^3/3-4(2)^2+16(2)]
    1/3[(125/3-100-80)-(8/3-16+32)]
    1/3[(125/3-180)-(8/3 +16)]
    1/3(-415/3-56/3)
    -471/9-ans
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  9. #9
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    Quote Originally Posted by bam017 View Post
    f(averege)=1/(b-a)$f(x)dx...I'm using $ for the integral sign
    a)1/3$[2,5](x^2-8x+16)dx
    1/3[(x^3/3)-4x^2+16(evaluated from 2 to 5)
    1/3[(5^3/3-4(5)^2+16(5))-(2^3/3-4(2)^2+16(2)]
    1/3[(125/3-100+80)-(8/3-16+32)]
    1/3[(125/3-20)-(8/3 +16)]

    1/3(39 - 36) = 1
    see what a sign error can do ?

    btw ... \int (x-4)^2 dx = \frac{(x-4)^3}{3} ... much easier.
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  10. #10
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    awesome

    of course now it seem so obvious
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