Given the following integral $$ \int_{0}^{1102}\lfloor \sqrt x \rfloor dx $$
What are the ways available to compute this integral? I've tried plugging it into the integral calculator on desmos.com and it gives an answer of 23825
However, when I tried manually expanding this equation, i,e
$$ \int_{0}^{1102}\lfloor \sqrt x \rfloor dx =\\= \int_{1^2}^{2^2} 1 \space\ dx + \int_{2^2}^{3^2} 2 \space\ dx + \int_{3^2}^{4^2} 3 \space\ dx \space +...+ \int_{32^2}^{33^2} 32 \space\ dx + \int_{33^2}^{1102} 33 \space\ dx $$
I get a value of $23837$. Is there something extra I'm accounting for in my expansion and calculation?
Seems like you are correct:
$$ \begin{align} \int_{0}^{1102}\lfloor \sqrt x \rfloor dx &= \sum_{k=1}^{32}\int_{k^2}^{(k+1)^2}kdx + \int_{33^2}^{1102}33dx \\ &= \sum_{k=1}^{32}k(2k+1) + 33(1102-33^2)\\ &=2\sum_{k=1}^{32}k^2 + \sum_{k=1}^{32}k + 429 \\ &=2\frac{32(32+1)(64+1)}{6}+\frac{32\cdot 33}{2} + 429 \\ &= 22880 + 528 + 429 \\ &= 23837 \end{align} $$
One can see that the area (integral) consists of many rectangles where $k^\text{th}$ rectangle ($0$-indexed) has, indeed, area of $$k\big((k+1)^2-k^2)\big)=k(2k+1)$$
Below is the graph of the function: