We were given a problem statement saying: Using discrete sum, find the area of the unbounded region limited by the curve $y^2=\frac{x(x-3)^2}{6-x} , x\ge3$ and its asymptote.
I've made the following try: (sorry for the quality, I only have available my Mac's webcam): Picture of the try
But I get completely stuck when I get to the final summation as I have no idea how to come up with the general formula, I was thinking on how to put the summation in such a way it would be possible to solve with the telescoping rules, but I could make no progress at all. I wonder if the whole idea to solve the problem is wrong. Do you have some different ideas to solve this problem? Or can I somehow go on from where I got stuck?
I am adding this in response to the suggestions to put all my try here: The graphic of the idea I am using is here: Computer generated graphic.
Initially we have the equation: $$y^2=\frac{x(x-3)^2}{6-x}$$ As this is symmetric respect the $x$-axis it is enough to find twice the area below: $$y=\frac{\sqrt x(x-3)}{\sqrt{6-x}}$$ between $x=3$ and $x=6$, this second line is the asymptote.
So, to establish the Riemman sum we have: $a=3, b=6-\epsilon,$ where $\epsilon$ is the distance between the $x$-coordinate of the function and the asymptote ($x=6$). And now:$$\Delta x=\frac{b-a}{n}=\frac{6-\epsilon-3}{n}=\frac{3-\epsilon}{n}$$and:$$x_i=a+i\Delta x=3+\frac{i(3-\epsilon)}{n}$$Finally we have the Riemman sum:$$\frac{A_n}{2}=\sum_{i=1}^n f(x_i)\Delta x=\sum_{i=1}^n \frac{\sqrt{3+\frac{i(3-\epsilon)}{n}}\left (3+\frac{i(3-\epsilon)}{n}-3\right)}{\sqrt{6-3-\frac{i(3-\epsilon)}{n}}}\frac{3-\epsilon}{n}=\sum_{i=1}^n \frac{\sqrt{\frac{3n+i(3-\epsilon)}{n}}}{\sqrt{\frac{3n-i(3-\epsilon)}{n}}}\frac{i(3-\epsilon)}{n}\frac{3-\epsilon}{n}$$$$\frac{A_n}{2}=\frac{(3-\epsilon)^2}{n^2}\sum_{i=1}^n i\sqrt{\frac{3n+i(3-\epsilon)}{3n-i(3-\epsilon)}}$$Now, just for simplicity, let: $3-\epsilon=a$, and $3n=b$, then:$$\frac{A_n}{2}=\frac{a^2}{n^2}\sum_{i=1}^n i\sqrt{\frac{a+bi}{a-bi}}$$ And well, (after my marathonic "typing"), here is where I get stuck as I couldn't make any more progress because of the overwhelming (for me) sum whose general formula is necessary to come up with. I hope your help, as it was very tiring for me to write all of this. :)