设f(x)=分段函数xsinx,(x>0) -1,(x≤0),求∫(0→2π)f(x-π)dx

Question

如题所述

Answer 1

∫(0→2π)f(x-π)dx=∫(0→π)f(x-π)dx+∫(π→2π)f(x-π)dx
=-x| 0→π +∫(π→2π)xsinxdx =-π-∫(π→2π)xdcosx=-π-[(xcosx)--∫(π→2π)cosxdx
=-π-[(xcosx)|π→2π -∫(π→2π)cosxdx]
=-π-(2π+π-0)=-4π追问

答案为0

追答

∫(0→2π)f(x-π)dx=∫(0→π)f(x-π)dx+∫(π→2π)f(x-π)dx
=-x| 0→π +∫(π→2π)xsinxdx =-π+∫(π→2π)xdcosx=-π+[(xcosx)-∫(π→2π)cosxdx
=-π+[(xcosx)|π→2π -∫(π→2π)cosxdx]
=-π+(2π-π-0)=0