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?

*

*If $[x]:=n$ iff $n-\tfrac12< x\leq n+\tfrac12$ (i.e. $[x]=\lceil x-\tfrac12\rceil$, which seems to be what the OP had in mind), then
\begin{align}
\int^2_0  x[1/x]\,dx & =\sum^\infty_{n=1}\int_{[\tfrac{2}{2n+1},\tfrac2{2n-1})}nx\,dx\\
&=\sum^\infty_{n=1}\Big(\frac{2n-1+1}{(2n-1)^2}-\frac{2n+1-1}{(2n+1)^2}\Big)\\
&=\sum^\infty_{n=1}\Big(\frac{1}{2n-1}-\frac{1}{2n+1}+\frac{1}{(2n-1)^2}+\frac{1}{(2n+1)^2}\Big)\\
&=1+2\sum^\infty_{n=1}\frac{1}{(2n-1)^2}-1=2\Big(\sum^\infty_{n=1}\frac{1}{n^2}-\frac14\sum^\infty_{n=1}\frac1{n^2}\Big)\\
&=\frac{\pi^2}{4}
\end{align}


*The same estimate holds when $[x]:=n$ iff $n-\tfrac12 \leq x<n+\tfrac12$ (i.e., $[x]=\lfloor x+\tfrac12\rfloor$) since $\lceil x-\tfrac12\rceil=\lfloor x+\tfrac12\rfloor$ a.s.


*

*If by $[\cdot]$ the OP means the ceiling function $\lceil x\rceil=n$ where $n-1<x\leq n$, then
$$\int^2_0 x\lceil 1/x \rceil\,dx=\int^1_0  x\lceil 1/x \rceil\,dx +\int^2_1 x\,dx=\int^1_0 x\lceil 1/x\rceil \,dx + \frac32$$
From
\begin{align}
\int^1_0  x\lceil 1/x \rceil\,dx & =\sum^\infty_{n=2}\int_{[\tfrac{1}{n},\tfrac1{n-1})}nx\,dx=\sum^\infty_{n=2}\frac{n}{2}\Big(\frac{1}{(n-1)^2}-\frac{1}{n^2}\Big)\\
&=\sum^\infty_{n=2}\frac{2n-1}{2n(n-1)^2}=\sum^\infty_{n=2}\frac{1}{(n-1)^2}+\frac{n-1-n}{2n(n-1)^2}\\
&=\frac12\sum^\infty_{n=1}\frac{1}{n^2}+\frac12\sum^\infty_{n=2}\frac{1}{n(n-1)} =\frac{\pi^2}{12}+\frac12
\end{align}
The value of the integral in this case is  $\frac{\pi^2}{12}+2\approx 2.822467$


*If the OP means $[\cdot]$ to be the integer part function (floor function) $\lfloor x\rfloor =n$ iff $n\leq x<n+1$, then
$$\int^2_0 x\lfloor 1/x \lfloor\,dx=\int^1_0  x\lfloor 1/x \rfloor\,dx +\int^2_1 0\cdot x\,dx=\int^1_0 x\lfloor 1/x\rfloor \,dx $$
The value of the integral is then given by
\begin{align}
\int^1_0  x\lfloor 1/x \rfloor\,dx & =\sum^\infty_{n=1}\int_{(\tfrac{1}{n+1},\tfrac1n]}nx\,dx=\sum^\infty_{n=1}\frac{n}{2}\Big(\frac{1}{n^2}-\frac{1}{(n+1)^2}\Big)\\
&=\sum^\infty_{n=1}\frac{2n+1}{2n(n+1)^2}=\sum^\infty_{n=1}\frac{1}{(n+1)^2}+\frac{n+1-n}{2n(n+1)^2}\\
&=\frac12\sum^\infty_{n=2}\frac{1}{n^2}-\frac12\sum^\infty_{n=1}\frac{1}{n(n+1)} =\frac{\pi^2}{12}
\end{align}
