Recent content by c.dube

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    Finding the Mass of a Solid in 3d

    Nevermind, I got an answer. Can anyone confirm 6\pi?
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    Finding the Mass of a Solid in 3d

    Homework Statement Find the mass of the solid bounded by the cylinder x^2+y^2=2x and the cone z^2=x^2+y^2 if the density is \delta = \sqrt{x^2+y^2} [b2. The attempt at a solution[/b] I had some trouble looking at how to set up the limits on this integral. What I came up with was: 2...
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    Convergence of Improper Integrals: Exploring the Power of p

    Ah sweet! Thanks so much for your help it was much appreciated.
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    Convergence of Improper Integrals: Exploring the Power of p

    That it diverges, yes? And so the final answer would be 0\leq p<1?
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    Convergence of Improper Integrals: Exploring the Power of p

    So, on 0\leq p<1, is \frac{e^{-x}}{x^p} integrable and therefore converges? As for p=1, we have the indefinite integral of lnx. But that's undefined at x=0, so am I missing your hint?
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    Convergence of Improper Integrals: Exploring the Power of p

    Ohhhhh ok so we have: 1-p<0 Now, that will go to infinity. The other case: 1-p>0 Will always be zero. So, the function doesn't blow up when p<1. Would that indicate that the answer is 0\leq p<1? And what about the case where p=1? Thanks again for your patience it is very much appreciated.
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    Convergence of Improper Integrals: Exploring the Power of p

    So the indefinite integral is: \frac{x^{-p+1}}{-p+1} + c When x=0, doesn't the integral become 0 for every p except p=1, where you get the indeterminate form 0^0? Gah I really don't know why I'm not getting this.
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    Convergence of Improper Integrals: Exploring the Power of p

    Oh ok so if p=0 then the x^p just becomes one, leaving only e^{-x}, which converges, so the answer would be p=0?
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    Convergence of Improper Integrals: Exploring the Power of p

    From 0 to 1, then, don't we have the exact opposite logic, e.g. \frac{1}{x^p} will always be greater than or equal to one? And therefore it diverges for all p, leaving me with an answer p = null set? Haha thanks for bearing with me I don't know why my brain isn't getting this one.
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    Convergence of Improper Integrals: Exploring the Power of p

    OK that makes sense to me, but where does it leave me? How do I bridge the gap to finding what, if any, p works? Thanks
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    Convergence of Improper Integrals: Exploring the Power of p

    OK, so \int_1^\infty e^{-x} converges because the limit as x goes to infinity is finite. Now, because \frac{1}{x^{p}} is decreasing and the largest case would be one, it is always less than or equal to one (is there a more rigorous way of saying this?). Anyways, moving on to case between 0 and...
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    Convergence of Improper Integrals: Exploring the Power of p

    I'm trying to refresh my knowledge of Calc II, and I'm going through improper integrals right now. The problem I am trying to solve is: For which numbers p\geq0 does \int_0^\infty \frac{e^{-x}}{x^{p}} converge? Justify your answer. So far, I've split up the integral into two halves (0 to 1...
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    Solving integral with substitution

    Look at the part e^{-\ln2} and how you could re-express that. Then run with it. EDIT: Haha as Bohrok said.
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    Solving integral with substitution

    Into two fractions - you have addition on top - and then attack them both separately.
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    Solving integral with substitution

    Hahah I don't know how your book wants you to do it but try splitting it into two integrals.
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