求积分∫(1~4)(4-x)sin(x-1)dx
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求积分∫(1~4)(4-x)sin(x-1)dx
∫[1→4] (4-x)sin(x-1) dx
=4∫[1→4] sin(x-1) dx-∫[1→4] xsin(x-1) dx
=-4cos(x-1) + ∫[1→4] x d(cos(x-1))
=-4cos(x-1) + xcos(x-1) - ∫[1→4] cos(x-1)dx
=-4cos(x-1) + xcos(x-1) - sin(x-1) |[1→4]
=-4cos3 + 4cos3 - sin3 + 4cos0 - cos0 + sin0
=3 - sin3
=4∫[1→4] sin(x-1) dx-∫[1→4] xsin(x-1) dx
=-4cos(x-1) + ∫[1→4] x d(cos(x-1))
=-4cos(x-1) + xcos(x-1) - ∫[1→4] cos(x-1)dx
=-4cos(x-1) + xcos(x-1) - sin(x-1) |[1→4]
=-4cos3 + 4cos3 - sin3 + 4cos0 - cos0 + sin0
=3 - sin3
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