nTr 另解an=(10-x)cos 5 T cOS 5xcos"d=0,(n=1,…) 5 5 5 S(10-x)d=S, 15 n7r 10 (10-x)sin tx=(-1) ,(n=1,2,) 5 5 10 oo 故∫(x)=10-x= ∑ (-1)”:n兀 sIn 兀m=in 5 (5<x<15)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, )