1到5题!

∫[1→5] (|2-x|+|sinx|) dx
=∫[1→5] |2-x| dx+∫[1→5] |sinx| dx
=∫[1→2] (2-x) dx+∫[2→5] (x-2) dx+∫[1→π] sinx dx-∫[π→5] sinx dx
=(2x-(1/2)x²) |[1→2] + ((1/2)x²-2x) |[2→5] -cosx |[1→π] + cosx |[π→5]
=4-2-2+1/2+25/2-10-2+4-(-1-cos1)+cos5-(-1)
=7+cos1+cos5

1、
f(1+3/2)=f(1-3/2) f(5/2)=f(-1/2)=-f(1/2)
f(1+5/2)=f(1-5/2) f(7/2)=f(-3/2)=-f(3/2)
结果是0
D
2、x=1 y=0
D
3、m-2=0
B
4、
f(0)=0 a=1/2
5、f(0)=0 |a|=1
a^2010=1