求定积分∫x^2cosxdx,上限是2π,下限是0

Answer 1

x²，cosx
2x，sinx，+
2，-cosx，-
0，-sinx，+
∫x²cosxdx
=(x²)(sinx)-(2x)(-cosx)+(2)(-sinx)+C
=x²sinx+2xcosx-2sinx+C
∫(0,2π)x²cosxdx
={0+2*2π-0}-0
=4π