由已知等式,令f(x)=3x+k 3x+k=3x-∫[0:1](3x+k)2dx k=-∫[0:1](9x2+6kx+k2)dx =-(3x3+3kx2+k2x)|[0:1] =-[(3·13+3k·12+k2·1)-(3·03+3k·02+k2·0)] =-k2-3k-3 k2+4k+3=0 (k+1)(k+3)=0 k=-1或k=-3 函数f(x)的解析式为f(x)=3x-1或f(x)=3x-3