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    (Original post by henpen)
    I'm not quite sure what you mean, but

    \displaystyle \int_{-1}^1 \frac{dx}{\sqrt{x}}

    has finite area, despite having an asymptote.

    Edit: Please be clearer, I don't see how the above is not a counterexample to your claim.
    probably because you didnt read his last sentence...
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    Problem 424 */**

    Find the integral of ln(cosx) between x=pi/2 and x=0
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    Problem 424 */**


    The sum of the base-10 logarithms of the divisors of 10^N is 729 792 for some natural number N
    .What is the value of N ?

    Edit: There was a mistake
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    (Original post by IceKidd)
    Problem 424 */**

    Find the integral of ln(cosx) between x=pi/2 and x=0
    This is going to look dreadful, although I have put in some effort putting in arrows and stuff

    Solution 424

    Building on what we've seen previously when the boundaries are pi/2 and 0 and the interchangeability therein of sin(x) and cos(x)..

    I = \displaystyle \int^{\frac{\pi}{2}}_0 \ln \cos (x) dx = \displaystyle \int^{\frac{\pi}{2}}_0 \ln \sin (x) dx \\ \sin (x) \equiv 2 \sin (\frac{x}{2}) \cos(\frac{x}{2}) \\ \Rightarrow I = \frac{\pi}{2} \ln 2\ + \displaystyle \int^{\frac{\pi}{2}}_0 \ln \cos (\frac{x}{2}) + \ln \sin(\frac{x}{2}) dx \\ I = \frac{\pi}{2} \ln 2\ + 2\displaystyle \int^{\frac{\pi}{4}}_0 \ln \cos (x) + \ln \sin(x) dx \\ \displaystyle \int^{\frac{\pi}{4}}_0 \ln \sin (x) dx = \displaystyle \int^{\frac{\pi}{2}}_{\frac{\pi}  {4}} \ln \cos (x) dx \\ I = \frac{\pi}{2} \ln 2\ + 2I \Rightarrow I = -\frac{\pi}{2} \ln 2
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    (Original post by nahomyemane778)
    Problem 424 */**


    The sum of the base-10 logarithms of the divisors of 10^N is 729 for some natural number N
    .What is the value of N ?


    I think you mean 792.
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    Thanks for that felix, I was wondering where I went wrong with N = 1/3 (-2+(19684-81 (59055)^(1/2))^(1/3)+(19684+81 (59055)^(1/2))^(1/3)) :lol:
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    (Original post by Llewellyn)
    Thanks for that felix, I was wondering where I went wrong with N = 1/3 (-2+(19684-81 (59055)^(1/2))^(1/3)+(19684+81 (59055)^(1/2))^(1/3)) :lol:
    Hows durham? Are u top of ur class
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    (Original post by Felix Felicis)
    I think you mean 792.
    How on earth did you possibly know? Have you see this question before?
    Nice to see good old feynman back for christmas
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    (Original post by Llewellyn)
    Thanks for that felix, I was wondering where I went wrong with N = 1/3 (-2+(19684-81 (59055)^(1/2))^(1/3)+(19684+81 (59055)^(1/2))^(1/3)) :lol:
    :rofl:
    (Original post by nahomyemane778)
    How on earth did you possibly know? Have you see this question before?
    Nice to see good old feynman back for christmas
    Nope - I got to the stage where I had an equation with no integer solutions for the value you gave (\frac{1}{2} N(N+1)^2 = 729), checked my working over and realised there's a mistake in the question and 792 just so happened to give a nice value for N. :cool:

    Posted from TSR Mobile
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    (Original post by Felix Felicis)
    :rofl:

    Nope - I got to the stage where I had an equation with no integer solutions for the value you gave (\frac{1}{2} N(N+1)^2 = 729), checked my working over and realised there's a mistake in the question and 792 just so happened to give a nice value for N. :cool:

    Posted from TSR Mobile
    Ahh ok then- well done- I've amended the OP so you can post the solution now.
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    N = 11 using the Bertrand-Chebyshev theorem.
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    (Original post by Zakee)
    N = 11 using the Bertrand-Chebyshev theorem.
    err... what? How does that solve the problem?

    Proof please.
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    (Original post by nahomyemane778)
    err... what? How does that solve the problem?

    Proof please.

    Proof:

    \frac{4^n}{2n }       \le \binom{2n}{n} = \left(\prod_{p \le \sqrt{2n}} p^{R(p,n)}\right) \left(\prod_{\sqrt{2n} < p \le \frac{2n}{3}} p^{R(p,n)}\right)

\lt (2n)^{\sqrt{2n}}  \prod_{1 < p \leq \frac{2n}{3} } p = (2n)^{\sqrt{2n}} \Big( \frac{2n}{3}\Big)\# \le (2n)^{\sqrt{2n}} 4^{2n/3}.\

    By concavity of the right-hand side as a function of n, the last inequality is necessarily verified on an interval. Since it holds true for  n = 11 and not  n = 12 , upper bound is the equating value which is:  n = 11

    The proof is trivial
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    (Original post by Zakee)
    Proof:

    \frac{4^n}{2n }       \le \binom{2n}{n} = \left(\prod_{p \le \sqrt{2n}} p^{R(p,n)}\right) \left(\prod_{\sqrt{2n} < p \le \frac{2n}{3}} p^{R(p,n)}\right)

       < (2n)^{\sqrt{2n}}  \prod_{1 < p \leq \frac{2n}{3} } p = (2n)^{\sqrt{2n}} \Big( \frac{2n}{3}\Big)\# \le (2n)^{\sqrt{2n}} 4^{2n/3}.\

    By concavity of the right-hand side as a function of n, the last inequality is necessarily verified on an interval. Since it holds true for  n = 11 and not  n = 12 , upper bound is the equating value which is:  n = 11

    The proof is trivial
    Well that's some pretty heavy machinery you've used- I dont understand at all to be honest. Is it university level?
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    (Original post by nahomyemane778)
    Well that's some pretty heavy machinery you've used- I dont understand at all to be honest. Is it university level?
    I'm currently in year 13, and I think it may be University level, I'm not sure. I heard from someone that it comes up in FP3 (A-level), but I haven't studied FP3 yet. I'm just extremely interested in Pure Mathematics and have a knack for problem-solving.
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    (Original post by Zakee)
    Proof

    The proof is trivial
    What is that?
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    (Original post by Zakee)
    I'm currently in year 13, and I think it may be University level, I'm not sure. I heard from someone that it comes up in FP3 (A-level), but I haven't studied FP3 yet.
    It's most definitely university level (and quite specialist).
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    Problem 425*/**/***

    Evaluate \displaystyle\int_2^3 7x\,dx
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    I wonder if I am sober enough to do that LOTF...

    Definitely not sober enough to latex it :lol:
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    (Original post by Lord of the Flies)
    Problem 425*/**/***

    Evaluate \displaystyle\int_2^3 7x\,dx

    Solution 425

    Consider a trapezium with base 1, and two parallel sides of lengths 14 and 21. Its area is given by A= \frac{1}{2}(14+21) = 17.5
 
 
 
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