# Integral

#### yo79

Prove that $$\displaystyle \int_0^1 \{\sqrt{x} \}=\frac{2}{3}$$,where {x}=x-[x] and [x]= the largest integer not exceeding x.

#### yo79

I solved it! I forgot about $$\displaystyle \{ x \}=x, \forall x \in[0,1)$$.