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Originally Posted by Krizalid Find expanding the integrand we get Completing the square using the trig sub then So now Rock on. I hope there is a better way to do this one. LOL.
Originally Posted by TheEmptySet 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.
Note that the calculation of the integral equals to compute the area of a semicircle whose radius is Besides which is a Riemann sum.
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\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 />](http://latex.codecogs.com/png.latex?<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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Originally Posted by Krizalid 
