Question

求定积分[上限π,下限0]∫(x^2)sgn(cosx)dx

Answer 1

[0,Pi/2]的时候

sgn(cosx) = 1

[Pi/2,Pi]的时候

sgn(cosx) = -1

所以

∫(x^2) sgn(cosx)dx

=∫[0,Pi/2](x^2) dx-∫[Pi/2,Pi](x^2)dx

=Pi^3/24 - Pi^3/3 + Pi^3/24

=-Pi^3/4