Find $x > 0$ for which $\int_{0}^{x} [t]^2 \ dt = 2 (x-1)$. What are all possible $x > 0$ for which the following equation is satisfied? 
$$\int_{0}^{x} [t]^2 \ dt = 2 (x-1),$$ where $[.]$ denotes the bracket (or floor) function. 
I guess we will have to consider different cases depending on whether $\sqrt{n-1} \leq x \leq \sqrt{n}$ for each positive integer $n$. 
For $0 \leq x \leq 1$, the integral on the left-hand side equals $1$; so the solution in this case is clearly $x = 1$ alone. 
What are all possible solutions for $x > 0$? 
 A: Rewrite your equation as
$$\sum_{n=1}^{[x]}n^2+\int_{[x]}^x[x]^2\,dx$$
This is
$${[x]([x]+1)(2[x]+1)\over 6}+[x]^2\{x\}$$
Multiplying it all out we get
$${1\over 3}[x]^3+\left({1\over 2}+\{x\}\right)[x]^2+{1\over 6}[x]$$
If this has the desired value, we get:
$${1\over 3}[x]^3+\left({1\over 2}+\{x\}\right)[x]^2+{1\over 6}[x]=2([x]+\{x\}-1)$$
$$\iff {1\over 3}[x]^3+\left({1\over 2}+\{x\}\right)[x]^2+1={11\over 6}[x]+2\{x\}$$
Then the polynomial on the Left is $>{1\over 3}[x]^3+{1\over 2}[x]^2+1$ and the one on the left is $< {11\over 6}[x]+2$. Comparing we see $2y^3+3y^2-11y-1>0$ for $y>3$ so for $x>3$ these differences cannot be reconciled, so we need only check $0\le x\le 3$
$$\int_0^x [t]^2\,dt=\begin{cases}0 & 0\le x\le 1 \\x-1 & 1\le x\le 2 \\ 1+4(x-2) & 2\le x\le 3\end{cases}$$
Clearly $x-1\ne 2(x-1)$ unles $x=1$. For the other case, we have $4x-7=2x-2\iff x= {5\over 2}$.
A: Differentiate both sides, we get $[x]^2 = 2$. Le us denote by $\{x\}$ the fractional part of the number $x$, so that $\{x\}=x-[x]$. It is clear that $0\le \{x\} <1$ and $[x]=x-\{x\}$. Thus, the equation becomes $$[x]^2=(x-\{x\})^2=2$$ that is $$x^2-2x\{x\}+\{x\}^2=2.$$ Setting $p=\{x\}$, and write $$x^2-2px+p^2-2=0.$$ Now, using the general law we have
\begin{align}
x_{1,2}  &= \frac{{ - b \pm \sqrt {b^2  - 4ac} }}{{2a}} 
\\
&= \frac{{2p \pm \sqrt {4p^2  - 4\left( 1 \right)\left( {p^2  - 2} \right)} }}{2}
\\
&= \frac{{2p \pm \sqrt 8 }}{2}.
\end{align}
Thus, $x_1 =p+\sqrt{2}$ and $x_2 =p-\sqrt{2}$ however, $a\le p<1$ which means that $x_2<0$ (reject), therefore the required solution is $x=p+\sqrt{2}$, for all $p\in[0,1)$. If you want write as $$\left\{ {x:x = p + \sqrt 2 ,0 \le p < 1} \right\}.$$
A: I assume that $[t]=\lfloor t\rfloor$. For $x\ge0$,
$$
\int_0^x\lfloor t\rfloor^2\,\mathrm{d}t
$$
is convex (its derivative is non-decreasing), the equation
$$
\int_0^x\lfloor t\rfloor^2\,\mathrm{d}t=2(x-1)
$$
can have at most $2$ solutions. One solution is $x=1$ where both sides are $0$.
For $2\le x\le3$, we have
$$
\int_0^x\lfloor t\rfloor^2\,\mathrm{d}t=4x-7
$$
we also have a solution at $x=5/2$ where both sides are $3$.
Thus, the solutions are at $x=1$ and $x=5/2$.
