高数课程妥媒血课件 镭理工大理>> [另解] an=55 (10-x)cos T 5 5 2/5 nTr 15 nTr cOS r cos dx=0,(n=1,2,) 5 5 5 (10-x)d=0 5 nTcr 10 bn =5(10-x)sin tx=(-1) ,(n=1,2, 5 故∫(x)=10-x= 10÷(-1)":n兀 SIn H=1 n 5 (5<x<15) Http://www.heut.edu.cn[另解]   = − 15 5 5 (10 )cos 5 1 dx n x an x   = − 15 5 5 (10 )sin 5 1 dx n x b x n    −  = 15 5 15 5 5 cos 5 1 5 2 cos dx n x dx x n x = 0, =  − 15 0 5 (10 ) 5 1 a x dx = 0, , 10 ( 1) n n = − (n = 1,2, )   = −   = − = 1 5 sin 10 ( 1) ( ) 10 n n x n n 故 f x x (5  x  15) (n = 1,2, )