f'(x)=cosx-(cosx-xsinx)=xsinx
当x>0时,由f'(x)=0得：sinx=0,x=kπ为极值点,k>1为整数.
单调增区间(0,π/2)U (2kπ-π/2,2kπ+π/2),k=1,2,.
单调减区间(2kπ+π/2,2kπ+3π/2),k=0,1,2,...
极值点为x=kπ,k=0,1,2,...,2013
f(kπ)=-kπ(-1)^k
因此极值和=-π[0-1+2-3+,+2012-2013]=-π(-1007)=1007π