如题所述
解:
∫(0→1)xcosπx dx
=1/π·x·sin(πx) |(0→1)-1/π ∫(0→1)sin(πx)dx
=0+1/π²·cos(πx) |(0→1)
=1/π²·(cosπ-cos0)
=-2/π²
∫(0→1)xcosπx dx
=1/π·x·sin(πx) |(0→1)-1/π ∫(0→1)sin(πx)dx
=0+1/π²·cos(πx) |(0→1)
=1/π²·(cosπ-cos0)
=-2/π²
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第1个回答 2012-12-07
∫[0,1]xcosπxdx
=1/π∫[0,1]xdsinπx
=1/π*xsinπx[0,1]-1/π∫[0,1]sinπxdx
=1/π*xsinπx[0,1]-1/π^2*cosπx[0,1]
=0
=1/π∫[0,1]xdsinπx
=1/π*xsinπx[0,1]-1/π∫[0,1]sinπxdx
=1/π*xsinπx[0,1]-1/π^2*cosπx[0,1]
=0