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