Gentle Sum

This problem can we solve with $Integral Quadrature ( Integral Rieman )$

Let function $y= f(x)$ is continuous on an interval $a \leq x \leq b$ . Divide the interval $a \leq x \leq b$ into $n$ equal subintervals. The lenght $\Delta x$ of each subinterval is $\Delta x= \frac{b-a}{n}$ ; this number is the partition unit. The division points have coordinates: $a=x_0, x_1, x_2, \cdots,x_{k-1},x_k, \cdots, x_{n-1}, x_n=b.$

If the subinterval $n \rightarrow \infty,$ then we will have the area $S,$

$\begin{array}{lll}S&=&\lim_{n \rightarrow \infty} \sum_{k=0}^{n-1} f(x_k){ \Delta x} \\&= &\lim_{n \rightarrow \infty} \sum_{k=1}^{n} f(x_k){ \Delta x} \\&=&\int_{a}^{b} f(x) dx \end{array}$

For question limit above,

$\begin{array}{lll}S&=&\lim_{n\rightarrow \infty} \frac{1}{4n^2} \sum_{k=1}^n \sqrt{4n^2-k^2} \\&=&\lim_{n\rightarrow \infty} \sum_{k=1}^n \frac{1}{4n} \sqrt{4-{(\frac{k}{n})}^2} \end{array}$

Because,

$a=x_0=0,b=x_n=1, \Delta{x}= \frac{a-b}{n}= \frac{1}{n}$ and $x_k= \frac{k}{n} \implies f(x_k)=\sqrt{4-{(x_k)}^2}$

We can have equality,

$\begin{array}{rrr}S&=&\frac{1}{4} \int_{0}^1 \sqrt{4-x^2}dx \end{array}$

Let $x=2 \sin\theta, dx=2 \cos\theta d\theta;$ $x=0, \theta=0 ; x=1, \theta=\frac{\pi}{6}$

$\begin{array}{lrl} S&=&\frac{1}{4} \int_{0}^{\frac{\pi}{6}} 2 \cos\theta . 2 \cos\theta d\theta \\ &=&\int_{0}^{\frac{\pi}{6}} {cos}^2 \theta d\theta \\ &=&\frac{1}{2} \int_{0}^{\frac{\pi}{6}} (1+ \cos 2\theta) d\theta \\ &=&\frac{1}{2} \lgroup \theta +\frac{1}{2 } \sin 2\theta \rgroup_{0}^{\frac {\pi}{6}} \\ &=& \frac{\pi}{12}+\frac{\sqrt{3}}{8} \end{array}$

$P=12,R=8 \implies P+R= \boxed{20}$

$\lim_{n\to\infty}\frac1{4n^2}\sum_{k=1}^{n} \sqrt{4n^2-k^2} = \lim_{n\to\infty}\sum_{k=1}^{n} \frac1{2n}\sqrt{1-\left(\frac{k}{2n}\right)^2},$ which is the definition of the Riemann integral, for $x = k/2n$ . Note that the limits of the integral are given by $1/2n\to 0$ and $n/2n\to 1/2$ : $= \int_0^{1/2} dx \sqrt{1-x}.$ Since $y = \sqrt{1-x}$ describes the top part of the unit circle, this integral is the area of a region enclosed by the unit circle, the coordinate axes, and the line $x = 1/2$ .

This circle fragment can be further viewed as the union of a circle sector (extending $30^\circ$ clockwise from the y-axis) and a triangle. Their common side runs from the origin to the point $(\sin 30^\circ,\cos 30^\circ) = (\tfrac12,\tfrac12\sqrt3)$ . $A_{\text{sector}} = \frac{30^\circ}{360^\circ}\cdot \pi r^2 = \frac{\pi}{12};$ $A_{\text{triangle}} = \tfrac12\cdot \tfrac12\cdot \tfrac12\sqrt3 = \frac{\sqrt 3}{8}.$ Thus the total area = the value of the integral = the value of the limit is equal to $\dots = \frac{\pi}{12} + \frac{\sqrt 3}{8}$ making $P = 12, R = 8$ and the answer is 20.