求定积分[上限π,下限0]∫(x^2)sgn(cosx)dx

网友回答

[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
======以下答案可供参考======
供参考答案1:
∫(0,π)(x^2)sgn(cosx)dx=∫(0,π/2)(x^2)dx-∫(π/2,π)(x^2)dx=-(1/4)π^3