<s>[INST] Area under curve, infinite rectangles I'm trying to calculate the area under the curve of $ y=x^2 $ between $ x=1 $ and $x = 3$ and above $y=0$ using the sum of infinitely many rectangles.
So far I've tackled it by first defining the width of every rectangle to be $\Delta x = \frac{3-1}{n} = \frac{2}{n}$.
After that I proceeded by stating that the area of every rectangle under the curve combined is
$\sum_{i=1}^n (\frac{2i}{n})^2*(\frac{2}{n})$
I'm told that $\sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}$ which I then used in my calculations:
$\sum_{i=1}^n (\frac{2}{n})^2*i^2*(\frac{2}{n}) = \frac{8}{n^3}*(\frac{n(n+1)(2n+1)}{6})$
Simplifying to:
$8 * \frac{(n+1)(2n+1)}{6n^2} = \frac{8n^2+12n+4}{3n^2}$
Proceeding by calculating the limit as $x$ approaches infinity:
$\lim _{x\to \infty }\left(\frac{\left(8n^2+12n+4\right)}{3n^2}\right) = \frac{8}{3}$.
However, the answer should be $\frac{26}{3}$ which is indeed very close to what I have, but I have no idea where the problem lies. Any ideas?
 [/INST] Following up on my comment, you need
$$
   R_n = \sum_{i=1}^n \left(1 + \frac{2i}{n}\right)^2 \cdot \frac{2}{n}
$$
Expand the square and collect the sums of powers:
\begin{align*}
   R_n &= \sum_{i=1}^n \left(1 + \frac{4i}{n} + \frac{4i^2}{n^2}\right)\cdot \frac{2}{n} \\
       &= \frac{2}{n} \sum_{i=1}^n 1
         + \frac{8}{n^2} \sum_{i=1}^n i
         + \frac{8}{n^3} \sum_{i=1}^n i^2 \\
       &= \frac{2}{n} \cdot n + \frac{8}{n^2} \cdot \frac{n(n+1)}{2} 
            + \frac{8}{n^3} \frac{n(n+1)(2n+1)}{6}
\end{align*}
As $n\to \infty$, this tends to $2 + 4 + \frac{8}{3} = \frac{26}{3}$.
</s>