Find the exact value of $\int_{0}^{2}x[\frac{1}{x}]dx$. Find the exact value of $\int_{0}^{2}x[\frac{1}{x}]dx$.
Let $[x]$ denote $\lceil{x-\frac{1}{2}}\rceil$.
Using Desmos, I got $2.46736022133$ and WolframAlpha does not give me a solution. My intuition tells me that it might be possible to find an exact value using Trapezoidal Reimann Sums but I am not really sure how to go about doing it. After my attempt, I got stuck but I was at a point where I could plug it into WolframAlpha and it gave me $\frac{\pi^2}{4}$. Why did it come out so nicely?
My attempt:
Where $A_n$ denotes the area of the $nth$ trapezoid from the right:
$$A=\frac{h}{2}(a+b)$$
$$A_n=\frac{\frac{2}{2n-1}-\frac{2}{2n+1}}{2}(\frac{2n}{2n-1}+\frac{2n}{2n+1})$$
$$A_n=\frac{\frac{4n+2}{4n^{2}-1}-\frac{4n-2}{4n^{2}-1}}{2}\left(\frac{4n^{2}+2n}{4n^{2}-1}+\frac{4n^{2}-2n}{4n^{2}-1}\right)$$
$$A_n=\frac{2}{4n^{2}-1}\left(\frac{8n^{2}}{4n^{2}-1}\right)$$
$$A_n=\frac{16n^{2}}{\left(4n^{2}-1\right)^{2}}$$
Then:
$$\int_{0}^{2}x[\frac{1}{x}]dx=\sum_{n=1}^{\infty}A_n=\sum_{n=1}^{\infty}\frac{16n^{2}}{\left(4n^{2}-1\right)^{2}}$$
I do not know how to solve this infinite summation so I plugged it into WolframAlpha and it gave me $\frac{\pi^2}{4}$. How did it get to this conclusion? Is there a more efficient way to solve this?

Continuing where you left off. Note
\begin{align*}
&\sum_{n=1}^\infty\frac{16n^2}{4n^2-1} \\
&= 2\sum_{n=1}^\infty n\left(\frac{1}{(2n-1)^2} - \frac{1}{(2n+1)^2}\right)\\
&= 2\sum_{n=1}^\infty \left(\sum_{k=1}^n 1\right) \left(\frac{1}{(2n-1)^2} - \frac{1}{(2n+1)^2}\right)\\
&= 2\sum_{n=1}^\infty \sum_{k=1}^n \left(\frac{1}{(2n-1)^2} - \frac{1}{(2n+1)^2}\right)\\
&= 2\sum_{k=1}^\infty \sum_{n=k}^\infty \left(\frac{1}{(2n-1)^2} - \frac{1}{(2n+1)^2}\right)\\
&= 2\sum_{k=1}^\infty \sum_{n=k}^\infty \left(\frac{1}{(2n-1)^2}\right) - \sum_{n=k+1}^\infty\left(\frac{1}{(2n-1)^2}\right)\\
&= 2\sum_{k=1}^\infty \frac{1}{(2k-1)^2} \\
&= 2\left(\sum_{k=1}^\infty \frac{1}{k^2} - \sum_{k=1}^\infty \frac{1}{(2k)^2}\right) \\
&= 2\cdot\frac{3}{4}\sum_{k=1}^\infty\frac{1}{k^2} \\
&= 2\cdot\frac{3}{4}\cdot\frac{\pi^2}{6} = \frac{\pi^2}{4}
\end{align*}
