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Math Help - Integration, Proof and limit.

  1. #1
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    Integration, Proof and limit.

    \int_{k}^{k+1} \frac{1}{x} dx,    k\geq 0

    Show that \frac{1}{k+1}\leq ln(\frac{k+1}{k}) \leq \frac{1}{k}

    I've found the integral, but how do i prove?

    thank you.
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  2. #2
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    Re: Integration, Proof and limit.

    Of the three terms in the inequality, which one corresponds to the exact value of the integral?
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  3. #3
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    Re: Integration, Proof and limit.

    middle one.
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  4. #4
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    Re: Integration, Proof and limit.

    Right. Is the integrand increasing or decreasing as x increases?
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  5. #5
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    Re: Integration, Proof and limit.

    Draw the graph of y= 1/x between x= k and x= k+1. Look at the area of the rectangle with base from k to k+1 and height 1/k, the area of the rectangle with base from k to k+ 1 and height 1/(k+1), and the area under the graph.
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  6. #6
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    Re: Integration, Proof and limit.

    By the mean value theorem, there exists a D in [K,K+1] such that 1/D=(ln(k+1)-ln(k))/(k+1-k)=ln((k+1)/k)

    k_<D<_k+1 so 1/(k+1)_<1/D<_1/k so the required result follows.
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  7. #7
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    Re: Integration, Proof and limit.

    Quote Originally Posted by BabyMilo View Post
    \int_{k}^{k+1} \frac{1}{x} dx,    k\geq 0
    Show that \frac{1}{k+1}\leq ln(\frac{k+1}{k}) \leq \frac{1}{k}
    The following is found in most calculus textbooks in a list of properties of integrals.
    \text{If }m \leqslant f(x) \leqslant M\text{ for all }x\in[a,b]\text{ then}
    m(b - a) \leqslant \int_a^b {f(x)dx}  \leqslant M(b - a).

    Now notice that K \leqslant x \leqslant K + 1\; \Rightarrow \;\frac{1}{{K + 1}} \leqslant \frac{1}{x} \leqslant \frac{1}{K}.
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