\[\Large\displaystyle \int_{0}^{[x]}\left ( t-\left [ t \right ] \right )\mathrm{d}t=\frac{[x]}{2}
\]
\(\Large\mathbf{Proof:}\)
我们来看更一般的形式，令\(m=\left \lfloor a \right \rfloor\)，有
\[\begin{align*}
\int_{a}^{a+k}\left ( t-\left [ t \right ] \right )\mathrm{d}t&=\int_{a}^{a+k}\left \{ t \right \}\mathrm{d}t\\
&=\int_{a}^{m+1}\left \{ t \right \}\mathrm{d}t+\sum_{j=m+1}^{k+m-1}\left ( \int_{j}^{j+1}\left \{ t \right \}\mathrm{d}t \right )+\int_{k+m}^{k+a}\left \{ t \right \}\mathrm{d}t\\
&=\int_{a}^{m+1}\left ( t-m \right ) \mathrm{d}t+\sum_{j=m+1}^{k+m-1}\left ( \int_{j}^{j+1}\left ( t-j \right )\mathrm{d}t\right )+\int_{k+m}^{k+a}\left ( t-k-m \right )\mathrm{d}t\\
&=\frac{k}{2}
\end{align*}\]
所以令 \(a=0~,~k=[x]\)，即得
\[\Large\boxed{\displaystyle \int_{0}^{[x]}\left ( t-\left [ t \right ] \right )\mathrm{d}t=\color{blue}{\frac{[x]}{2}}}
\]