求∫（2-x）dx的积分(要求用两种方法)
热心网友
A. ∫（2-x）dx = ∫2dx - ∫xdx = 2x -x^2/2 + C1B. ∫（2-x）dx = -∫（2-x）d(2-x) = -(2-x)^2/2 + C2A = B
热心网友
1)∫（2-x）dx = ∫2dx-∫xdx = 2x -x^2/2 + C2)∫（2-x）dx= -∫（x-2）d(x-2)= -(x-2)^2/2 +C这样算两种吗？
求∫（2-x）dx的积分(要求用两种方法)
A. ∫（2-x）dx = ∫2dx - ∫xdx = 2x -x^2/2 + C1B. ∫（2-x）dx = -∫（2-x）d(2-x) = -(2-x)^2/2 + C2A = B
1)∫（2-x）dx = ∫2dx-∫xdx = 2x -x^2/2 + C2)∫（2-x）dx= -∫（x-2）d(x-2)= -(x-2)^2/2 +C这样算两种吗？