Consider a parabola $y=x^2/4$ and point F(0,1).
Let $A_1(X_1,Y_1),......,A_n(X_n,Y_n)$ be $n$ points on that parabola such that each $X_k >0$ and angle $OFA_k=k\pi/2n$ (where $k=1,2,3,\ldots,n$).
What is the value of $\lim_{n\rightarrow \infty} 1/n \sum_{k=1}^n FA_k$?
I know the answer to this question but I wanted to ask why conceptually we can't just find the answer by integrating the area between the curve and y-axis from 0 to 1?
The answer by integrating the curve comes 4/3 while the original answer is 4/π
There is no equivalence between the area between the curve and y-axis from 0 to 1. I suppose you imagine something like the summation can be transformed into an area calculation by integration in polar coordinates centred at the point F. However, in polar coordinates, the integration is (using the traditional notations) $\tfrac{1}{2}\rho^2\,d\theta$. We can also remark that the summation is ``dimensionnally'' analogous to a length, not to an area.
One can show that \begin{align} F A_k\cos\theta_k&=1-Y_k\\ FA_k\sin\theta_k&=X_k \end{align} with $Y_k=X^2_k/4$ and $\theta_k=k\pi/2n$. From these geometrical expressions we deduce $X_k=2\tan \theta_k/2$ and thus $$FA_k=\frac 1{\cos^2\theta_k/2}$$ Finally the summation can be written as \begin{equation} S_n=\frac{1}{n}\sum_{k=0}^n\frac{1}{\cos^2\frac{k\pi}{2n}} \end{equation} Taking its limit for $n\to\infty$, it is transformed into a Riemann sum: \begin{equation} S=\frac{4}{\pi}\int_0^{\pi/4}\frac{dt}{\cos^2t}=\frac{4}{\pi} \end{equation}