先考察一个子段 [ x k , x k + 1 ] [x_k,x_{k+1}] [xk,xk+1],其中点 x k + 1 2 = 1 2 ( x k + x k + 1 ) x_{k+\frac{1}{2}}=\frac{1}{2}(x_k+x_{k+1}) xk+21=21(xk+xk+1),在该子段上二分前后的两个积分值 T 1 = h 2 [ f ( x k ) + f ( x k + 1 ) ] T 2 = h 4 [ f ( x k ) + 2 f ( x k + 1 2 ) + f ( x k +