f(x)=∫(0,1)lt(t-x)ldt,x∈(0,1)
=∫(0,x)lt(t-x)ldt+∫(x,1)lt(t-x)ldt
=∫(0,x)t(x-t)dt+∫(x,1)t(t-x)dt
=x∫(0,x)tdt-∫(0,x)t^2dt+∫(x,1)t^2dt-x∫(x,1)tdt
求导f'(x)=∫(0,x)tdt+x^2-x^2-x^2-∫(x,1)tdt+x^2=∫(0,x)tdt-∫(x,1)tdt=x^2-1/2=(x-1/√2)(x+1/√2)
由f'(x)<0,得f(x)单减区间(0,1/√2),由f'(x)>0得f(x)单增区间(1/√2,1),
且f(x)在x=1/√2处取得极小值,代入得f(1/√2)=(1/3)(1-1/√2)
f"(x)=2x>0那么知f(x)凹区间为(0,1)