The question is $\int_{2}^{341} \left(x - \lfloor x \rfloor \right)^2$.

I understand how to solve integrals of floor functions (they get converted to discrete integrals) and even just this part: $(x - \lfloor x \rfloor)$.

I drew the graph and they're just 341 triangles with base 1 and height 1. so the answer is $\frac{341}{2}$.

How does one solve the square part? The answer given is $\frac{341}{3}$.

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    $\begingroup$ Calculate the integrals from $2$ to $3$, $3$ to $4$,...,$340$ to $341$. $\endgroup$ – Kavi Rama Murthy May 1 at 10:08
  • $\begingroup$ Please try to format your posts with MathJax. $\endgroup$ – StubbornAtom May 1 at 10:17

If $x=n+u$ where $n\in\mathbb{Z}$ and $u\in[0,1)$, then $(x-\lfloor x\rfloor)^2=u^2$.

$$\int_n^{n+1}(x-\lfloor x\rfloor)^2dx=\int_0^1u^2du=\frac{1}{3}$$

$$\int_2^{341}(x-\lfloor x\rfloor)^2dx=\sum_{k=2}^{340}\int_n^{n+1}(x-\lfloor x\rfloor)^2dx=\frac{339}{3}=113$$

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    $\begingroup$ When we give complete answers to simple computational questions like this we run the risk of doing OP's homework. It is better to tell OP how to do it and let him complete it. $\endgroup$ – Kavi Rama Murthy May 1 at 10:14

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