若f(x)在[a,b]上连续，且f(x)关于x=(ab)/2对称，则有∫xf(x)dx=(ab)/2*∫f(x)dx（积分区间都是[a,b]）证明：由对称性，f(x)=f(ab-x)∫xf(x)dx=∫xf(ab-x)dx(积分区间[a,b])=∫(ab-y)f(y)d(ab-y)(积分区间[b,a])=∫-(ab-y)f(y)dy(积分区间[b,a])=-∫-(ab-y)f(y)dy(积分区间[a,b])=∫(ab-x)f(x)dx(积分区间[a,b])=∫(ab)f(x)dx-∫xf(x)dx(积分区间[a,b])=(ab)∫f(x)dx-∫xf(x)dx(积分区间[a,b])可得：∫xf(x)dx=(ab)/2*∫f(x)dx(积分区间[a,b])
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