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原式=∫[0,1](1-x)dx∫[1,3](x-1)dx+∫[0,3](3-x)dx+∫[3,4](x-3)dx
=(x-x^2/2)[0,1]+(x^2/2-x)[1,3]+(3x-x^2/2)[0,3]+(x^2/2-3x)[3,4]
=1-1/2+(9/2-3)-(1/2-1)+(9-9/2)+(16/2-12)-(9/2-9)
=15/2.
=(x-x^2/2)[0,1]+(x^2/2-x)[1,3]+(3x-x^2/2)[0,3]+(x^2/2-3x)[3,4]
=1-1/2+(9/2-3)-(1/2-1)+(9-9/2)+(16/2-12)-(9/2-9)
=15/2.
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