Integral

Jan 2013
96
0
Prove that \(\displaystyle \int_0^1 \{\sqrt{x} \}=\frac{2}{3}\),where {x}=x-[x] and [x]= the largest integer not exceeding x.
 
Jan 2013
96
0
I solved it! I forgot about \(\displaystyle \{ x \}=x, \forall x \in[0,1)\).