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Math Help - help on prove a integral is bigger than 0

  1. #1
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    help on prove a integral is bigger than 0

    hi, how can i prove the attached function is bigger than 0, if p(x) not equal to 0.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by yht0251 View Post
    hi, how can i prove the attached function is bigger than 0, if p(x) not equal to 0.
    Note that your integrand is always positive on that interval. I mean, depending on the context of the question I would think that'd be enough. If your integral diverged it would approach....what? Can you tell?
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    Quote Originally Posted by Drexel28 View Post
    Note that your integrand is always positive on that interval. I mean, depending on the context of the question I would think that'd be enough. If your integral diverged it would approach....what? Can you tell?
    why positive integrand gives positive integral?
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by yht0251 View Post
    why positive integrand gives positive integral?
    Non-rigorously because if a curve is above the x-axis then the area below it will be positive.
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    Quote Originally Posted by Drexel28 View Post
    Non-rigorously because if a curve is above the x-axis then the area below it will be positive.
    can it be proved in a mathematical way?
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    Is p continuous?

    Since p(x)^2 doesn't distinguish between the sign of p(x), we may assume that p(x) is nonnegative everywhere. Now, let's suppose that c is a value at which p(c)=L\ne 0. By the assumed continuity of p, we have, from the epsilon-delta definition of limit,

    For all \epsilon, there exists a \delta such that 0<|x-c|<\delta implies |f(x)-L|<\epsilon.

    This just states that \lim_{\small x\rightarrow c}f(x)=f(c).

    What we need to do is find a way to show that there is some positive area under the curve y=p(x). What happens when we let \epsilon=\frac{L}{2}?
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Scott H View Post
    Is p continuous?

    Since p(x)^2 doesn't distinguish between the sign of p(x), we may assume that p(x) is nonnegative everywhere. Now, let's suppose that c is a value at which p(c)=L\ne 0. By the assumed continuity of p, we have, from the epsilon-delta definition of limit,

    For all \epsilon, there exists a \delta such that 0<|x-c|<\delta implies |f(x)-L|<\epsilon.

    This just states that \lim_{\small x\rightarrow c}f(x)=f(c).

    What we need to do is find a way to show that there is some positive area under the curve y=p(x). What happens when we let \epsilon=\frac{L}{2}?
    what's your point? Can't the function theoretically be integrable and not continuous?
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  8. #8
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Scott H View Post
    Is p continuous?

    Since p(x)^2 doesn't distinguish between the sign of p(x), we may assume that p(x) is nonnegative everywhere. Now, let's suppose that c is a value at which p(c)=L\ne 0. By the assumed continuity of p, we have, from the epsilon-delta definition of limit,

    For all \epsilon, there exists a \delta such that 0<|x-c|<\delta implies |f(x)-L|<\epsilon.

    This just states that \lim_{\small x\rightarrow c}f(x)=f(c).

    What we need to do is find a way to show that there is some positive area under the curve y=p(x). What happens when we let \epsilon=\frac{L}{2}?
    what's your point? Can't the function theoretically be integrable and not continuous?

    @OP-what class are you in?
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  9. #9
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    Quote Originally Posted by Drexel28 View Post
    Non-rigorously because if a curve is above the x-axis then the area below it will be positive.
    thank you , i think i got the point. integral is some kind of adding , when everything added is positive, the sum is positive.
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