4.u() zo=r te dt e dt e dt S S J0 S n test n=2 S ]=2 12 所以L]="] SS n=3 n=1 =J。te"dt l]=3p]=3.2=6 S tde SJo 所以=214 4.tnu(t) 0 L t t e d t st 2 0 1 e 1 1 s s s st 0 L t t e d t n n st 0 1 t e d t s n n st 0 de 1 st t s 0 0 e e d 1 t t s st st n 2 2 3 2 2 2 1 2 s s s L t s L t n 3 3 4 3 3 2 3 2 6 s s s L t s L t 1 n n L t s n L t 0 e st n s t 0 1 t e d t s n n st 1 ! n n s n L t n 1 所以 所以