如题所述
原式=∫[1,2]dx∫[0,π/2]xsinydy
=∫[1,2]xdx
=[1/2x^2][1,2]
=3/2追问
=∫[1,2]xdx
=[1/2x^2][1,2]
=3/2追问
求解第一步到第二步
追答∫[1,2]dx∫[0,π/2]xsinydy=∫[1,2]dx [x(-cosy)][0,π/2]
=∫[1,2]dx(-x*(cos π/2-cos0))
=∫[1,2]dx (-x * -1)
=∫[1,2]xdx
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