# Thread: Today's calculation of integral #2

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

Find $\displaystyle \lim_{n \to \infty } \left\{ {\frac{1}{{n^2 }}\sum\limits_{k = 0}^{n - 1} {\left[ {k\int_k^{k + 1} {\sqrt {(x - k)(k + 1 - x)} \,dx} } \right]} } \right\}.$

2. Originally Posted by Krizalid
Find $\displaystyle \lim_{n \to \infty } \left\{ {\frac{1}{{n^2 }}\sum\limits_{k = 0}^{n - 1} {\left[ {k\int_k^{k + 1} {\sqrt {(x - k)(k + 1 - x)} \,dx} } \right]} } \right\}.$

expanding the integrand we get

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

Completing the square

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

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

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

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

So now

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

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

Rock on. I hope there is a better way to do this one. LOL.

3. 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.

4. Note that the calculation of the integral equals to compute the area of a semicircle whose radius is $\displaystyle \frac12.$ Besides $\displaystyle \lim_{n \to \infty } \frac{1} {{n^2 }}\sum\limits_{k\, = \,0}^{n - 1} k = \int_0^1 {x\,dx} = \frac{1} {2},$ which is a Riemann sum.