①x∈[0,1]0≤x^n≤1<π/20≤sin(x^n)<1∫sin(x^n)dx≥0②根据x≥0,sinx≤xsin(x^n)≤x^n∫sin(x^n)dx≤∫x^ndx=1/(n+1)*[x^(n+1)]=1/(n+1)∴0≤∫≤1/(n+1) 若f(x)≠0,f(x)^2>0∫f(x)^2dx>0矛盾f(x)=0
