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Math Help - Finding the average value of this function

  1. #1
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    Finding the average value of this function

    The function is tsin(t^2) on the interval [0,10]

    So it'd end up being 1/10 the integral from 0 to 10 of tsin(t^2), but that's where I'm stuck. How in the world would I take the antiderivative of that function?
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  2. #2
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    Could somebody confirm or deny that I'm heading in the right direction by saying my next step would be 1/10((-1/2)cos(t^2) evaluated at 10, minus that evaluated at 0?
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  3. #3
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    I'm sorry I wasn't very clear, I'm just wondering if the antiderivative of tsin(t^2) is just -1/2cos(t^2) yes or no?
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  4. #4
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    Are you sure you arent finding the average value of sin(t^2) on the interval [0,10] ?

    Then the average value is:

     \int_{0 }^{10 } tSin( { t}^{2 })

    then substitute u = t^2 and solve.

    You should get around .0688.

    If you are finding the average value of tsin(t^2), the average value is:

     \int_{0 }^{10 }  { t}^{ 2} Sin( { t}^{2 })

    and this integral is now very difficult and involves the Fresnel integral. The value here is about -4.01.
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  5. #5
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    Quote Originally Posted by Mentia View Post
    Are you sure you arent finding the average value of sin(t^2) on the interval [0,10] ?

    Then the average value is:

     \int_{0 }^{10 } tSin( { t}^{2 })

    then substitute u = t^2 and solve.

    You should get around .0688.

    If you are finding the average value of tsin(t^2), the average value is:

     \int_{0 }^{10 }  { t}^{ 2} Sin( { t}^{2 })

    and this integral is now very difficult and involves the Fresnel integral. The value here is about -4.01.

    Umm I thought the formula for fave is 1/b-a times the integral you gave...at least that's what it says in my notes and in my book.
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  6. #6
    Member Mentia's Avatar
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    Wow you are totally right. I was thinking about probability density functions, whoops! Yeah so just do the substitution then divide by 10.
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