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Math Help - integrals of nonpositive functions

  1. #1
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    integrals of nonpositive functions

    show that if "f" is integrable then

    f(x) < or equal to 0 on [a,b]

    is greater than or = to zero
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  2. #2
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    Quote Originally Posted by 10tomlinsonb View Post
    show that if "f" is integrable then

    f(x) < or equal to 0 on [a,b]

    is greater than or = to zero
    Well that is not true.
    It is true that \int_a^b {f(x)dx}  \leqslant 0 .
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  3. #3
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    sorry that is what i meant to write. can you show me why that is true?
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  4. #4
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    Quote Originally Posted by 10tomlinsonb View Post
    sorry that is what i meant to write. can you show me why that is true?
    That depends upon who much you know.
    Is this a problem in Riemann integrals?
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  5. #5
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    it is a problem in definite integrals
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  6. #6
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    Quote Originally Posted by 10tomlinsonb View Post
    it is a problem in definite integrals
    I am not sure what that implies.

    In most basic calculus textbooks there is a theorem that states:
    If \left( {\forall x \in \left[ {a,b} \right]} \right)\left[ {f(x) \leqslant g(x)} \right] then it is true that \int_a^b {f(x)dx}  \leqslant \int_a^b {g(x)dx} .
    if your text has that theorem then let g(x)=0 and use the theorem.
    Otherwise, you need to use upper sums.
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  7. #7
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    can you show me how you would prove this using upper sums
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  8. #8
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    please help me prove this because i am so confused and i need it for a homework problem
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  9. #9
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    The reason Plato asked if you knew about Riemann sums is that how you prove such a basic property of integrals depends upon how you define integrals in the first place! And "Riemann sums" is a common method of defining integrals. You mention of "upper sums" indicates that's what you have learned, although "upper sums" is only a very small part of that.

    Given that f(x) is integrable on [a, b], then for any partition of [a, b] (dividing it into smaller subintervals), choosing any x_i in the i^{th} interval, with width \Delta x_i, the Riemann sum is \sum f(x_i)\Delta x_i and, as we take "finer and finer" partitions, meaning that the largest \Delta x_i goes to 0 as we take more and more intervals in the partition, and that sum goes to some fixed value, I, whic is the integral, \int_a^b f(x)dx.

    Now, if f(x)\le 0 for all x in [a, b], in particular, f(x_i)\le 0 for every partition and for every x_i in a partition. That, in turn, means that f(x_i)\Delta x_i\le 0 for every partition and every i in a partition, and so \sum f(x_i)\Delta x_i\le 0 for every partition. The integral, the limit of negative numbers can't be positive: \int_a^b f(x)dx\le 0.
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  10. #10
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    my teacher also told that

    (b-a) is max of f(x) therefore,

    ≤ (b-a) ≤ 0

    can you expain why this is
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  11. #11
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    Quote Originally Posted by 10tomlinsonb View Post
    sorry that is what i meant to write. can you show me why that is true?
    A single counter example will show this: let b>a, and f(x)=-1 and you can now show that the integral is negative.

    CB
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