Calculust 5

$u=x-k \Rightarrow du=dx$

$\int_{k}^{k+1} \sqrt{(x-k)(k+1-x)}dx=\int_{0}^{1} \sqrt{u(1-u)}du=\int_{0}^{1} \sqrt{u-u^2} du=\int_{0}^{1} \sqrt{\dfrac{1}{4}-\left (\dfrac{1}{2}-u \right)^2}du=\dfrac{1}{2} \int_{0}^{1} \sqrt{1-(1-2u)^2} du$

$1-2u=\sin(\theta) \Rightarrow -2du=\cos(\theta) d \theta \Rightarrow du=-\dfrac{1}{2}\cos(\theta)d \theta$

$\cdots=\dfrac{1}{2} \int_{\dfrac{\pi} {2}}^{\dfrac{3 \pi}{2}} \sqrt{1-\sin^2(\theta)} \left (-\dfrac{1}{2}\cos(\theta)\right )d \theta=-\dfrac{1}{4} \int_{\dfrac{\pi} {2}}^{\dfrac{3 \pi}{2}} \cos^2(\theta) d\theta=-\dfrac{1}{4} \left[ \dfrac{\sin(2 \theta)+2x}{4} \right ]_{\dfrac{\pi} {2}}^{\dfrac{3 \pi} {2}}=-\dfrac{\pi}{8}$

As clearly the integral is positive by definition of the square root we can take it to be equal to $\dfrac{\pi} {8}$

$\sum_{k=0}^{n-1} k \times \dfrac{\pi}{8}=\dfrac{\pi}{8}\sum_{k=0}^{n-1} k=\dfrac{\pi}{8} \times \dfrac{n(n-1)}{2}=\dfrac{\pi n (n-1)}{16}$

$\lim_{n \rightarrow \infty} \dfrac{\left ( \dfrac{\pi n (n-1)}{16}\right )}{n^2}=\lim_{n \rightarrow \infty} \dfrac{\pi (n-1)}{16n}=\lim_{n \rightarrow \infty} \dfrac{\pi \left (1-\dfrac{1}{n} \right)}{16}=\dfrac{\pi}{16}$

$\boxed{\boxed{a=16}}$

simply substitute x-k=sin^2(t) and the integral reduces to a k-independant expression. Thus it can be taken out of the summation and the limit. The rest is basic math. Don't think this was as good as other level 5 problems.