Find solutions for integrals with floor function Find solutions of x satisfying:
$$\int_{0}^{2\lfloor{x+14}\rfloor}\left(\frac{t}{2}-\left\lfloor{\frac{t}{2}}\right\rfloor\right)dt=\int_{0}^{x-\lfloor{x}\rfloor}\lfloor{t+14}\rfloor dt$$
My solution set is below.
How to solve a question like this?
 A: Your work is correct until you get the equation
$$I+14 = 14f$$
Now, since $0 \le f < 1$ you have that
$$0 \le I +14 < 14$$
or equivalently
$$-14 \le I < 0$$
Hence $I \in \{ -14, \dots , -1\}$, while $f=\frac{I}{14}+1$.
Thus you have 14 solutions, namely
$$x=I + f=I+\frac{I}{14}+1 = \frac{15I+14}{14} \qquad \qquad I \in \{ -14, \dots , -1\}$$
Now, calling $k=I+14$ we can rewrite the solutions as
$$x= \frac{15k - 196}{14} \qquad \qquad k \in \{ 0, \dots , 13\}$$
So it seems that you find the correct solutions.
A: My solution to this question:
Take the first integral as $I_1$ and the second as $I_2$.
$$ I_1=\int_{0}^{2\lfloor{x+14}\rfloor}\bigl(\frac{x}{2}-\lfloor{\frac{x}{2}}\rfloor\bigr)dx=\int_{0}^{2\lfloor{x+14}\rfloor}\frac{x}{2}dx -\int_{0}^{2\lfloor{x+14}\rfloor}\lfloor{\frac{x}{2}}\rfloor dx$$
$$I_1=\Bigl(\frac{x^2}{4}\Bigr)_{0}^{2\lfloor{x+14}\rfloor}-\sum_{k=0}^{\lfloor{x+13}\rfloor}\int_{2k}^{2k+2}\lfloor{\frac{x}{2}}\rfloor={\lfloor{x+14}\rfloor}^2-\sum_{k=0}^{\lfloor{x+13}\rfloor}\int_{2k}^{2k+2}(k)dx$$
$$I_1={\lfloor{x+14}\rfloor}^2-\sum_{k=0}^{\lfloor{x+13}\rfloor}(2k)={\lfloor{x+14}\rfloor}^2-\lfloor{x+14}\rfloor \lfloor{x+13}\rfloor=\lfloor{x+14}\rfloor$$
$$I_1=\lfloor{x+14}\rfloor$$
$$I_2=\int_{0}^{x-\lfloor{x}\rfloor}\lfloor{x+14}\rfloor dx=\int_{0}^{x-\lfloor{x}\rfloor}\lfloor{x}\rfloor dx + \int_{0}^{x-\lfloor{x}\rfloor}\lfloor{14}\rfloor dx$$
Since $0\leq x-\lfloor{x}\rfloor<1 \Rightarrow \int_{0}^{x-\lfloor{x}\rfloor}\lfloor{x}\rfloor dx=0$
$$I_2=14x-14\lfloor{x}\rfloor$$
And now: $I_1=I_2$
$$\lfloor{x+14}\rfloor=14x-14\lfloor{x}\rfloor$$
Assume $x=I+f$ where $I \in Z$ and $0 \leq f <1$
S1: $I+14=14I+14f-14I$
$$I+14=14f \Rightarrow 14f \in Z$$
Let $f=\frac{p}{q}$ s.t. $\space p,q \in Z^{+} \cup\{\ 0 \}\,\space  p<q$, p and q are coprime
$$\frac{14p}{q}\in Z \Rightarrow q|14\space or \space p$$
But since p and q are coprime $q|14\Rightarrow q=1,2,7,14$
When $q=1 \Rightarrow p=0$
When $q=2\Rightarrow p=0,1$
When $q=7\Rightarrow p=0,1,...6$
When $q=14\Rightarrow p=0,1...13$
Hence $f\in\{ n| n=\frac{k}{14}; k\in \{0,1,....13 \}\} $
Then from S1 $$x=I+f=15f-14=\frac{15k-196}{14}; k\in \{0,1,....13 \} $$
That completes my solution, please tell me if it is correct or not.
