解：∫{0,x}(t-1)dt=∫{0,x}（0.5t²-t）=0-（0.5x²-x）=-0.5x²+x=-0.5（x²-2x）=-0.5（x²-2x+1）+0.5=-0.5（x-1）²+0.5最大值是0.5
