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Math Help - Today's calculation of integral #2

  1. #1
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    Krizalid's Avatar
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    Today's calculation of integral #2

    Find <br />
\lim_{n \to \infty } \left\{ {\frac{1}{{n^2 }}\sum\limits_{k = 0}^{n - 1} {\left[ {k\int_k^{k + 1} {\sqrt {(x - k)(k + 1 - x)} \,dx} } \right]} } \right\}.<br />
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  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    Quote Originally Posted by Krizalid View Post
    Find <br />
\lim_{n \to \infty } \left\{ {\frac{1}{{n^2 }}\sum\limits_{k = 0}^{n - 1} {\left[ {k\int_k^{k + 1} {\sqrt {(x - k)(k + 1 - x)} \,dx} } \right]} } \right\}.<br />

    expanding the integrand we get

    \int_{k}^{k+1}\sqrt{-x^2+(2k+1)x-(k^2+k)}dx

    Completing the square

    \int_{k}^{k+1}\sqrt{\frac{1}{4}-(x-k-\frac{1}{2})^2}dx

    using the trig sub x-k-\frac{1}{2}=\frac{\sin(\theta)}{2}

    then dx=\frac{\cos(\theta)}{2}d\theta

    \frac{1}{4}\int_{-\pi/2}^{\pi/2}\sqrt{1-\sin^{2}(\theta)}\cos(\theta)d\theta=\frac{1}{8}\i  nt_{-\pi/2}^{\pi/2}1-\cos(2\theta)d\theta=\frac{\pi}{8}

    So now

    <br />
\lim_{n \to \infty } \left\{ {\frac{1}{{n^2 }}\sum\limits_{k = 0}^{n - 1} {\left[ {k\int_k^{k + 1} {\sqrt {(x - k)(k + 1 - x)} \,dx} } \right]} } \right\}.<br />
=\frac{\pi}{8}\lim_{n \to \infty}  \left\{ {\frac{1}{{n^2 }}\sum\limits_{k = 0}^{n - 1} {\left[ {k} \right]} } \right\}

    \frac{\pi}{8}\lim_{n \to \infty}\frac{1}{n^2}\frac{(n-1)(n)}{2}=\frac{\pi}{8} \cdot \frac{1}{2}=\frac{\pi}{16}

    Rock on. I hope there is a better way to do this one. LOL.
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  3. #3
    Math Engineering Student
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    Quote Originally Posted by TheEmptySet View Post
    Rock on. I hope there is a better way to do this one. LOL.
    Yeah, I'll post another solution later.

    Anyway that's the answer.
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  4. #4
    Math Engineering Student
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    Note that the calculation of the integral equals to compute the area of a semicircle whose radius is \frac12. Besides \lim_{n \to \infty } \frac{1}<br />
{{n^2 }}\sum\limits_{k\, = \,0}^{n - 1} k  = \int_0^1 {x\,dx}  = \frac{1}<br />
{2}, which is a Riemann sum.
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