设函数f(x)=∫lt(t-x)ldt(0

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)