搜搜做题作业网已知f“(x)在闭区间a到b上连续且f(0)=2,f(派)=1,则∫(0到派)【f(x)+f"(x)】sinxdx=?
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已知f“(x)在闭区间a到b上连续且f(0)=2,f(派)=1,则∫(0到派)【f(x)+f"(x)】sinxdx=?
高数大神~拜托了
高数大神~拜托了
∫[0,π][f(x)+f"(x)]sinxdx
=∫[0,π]f(x)sinxdx+∫[0,π]f"(x)sinxdx
=∫[0,π]f(x)sinxdx+∫[0,π] sinxdf'(x)
=∫[0,π]f(x)sinxdx+sinxf'(x)[0,π] -∫[0,π] f'(x)dsinx
=∫[0,π]f(x)sinxdx-∫[0,π] f'(x)cosxdx
=∫[0,π]f(x)sinxdx-∫[0,π] cosxdf(x)
=∫[0,π]f(x)sinxdx-cosxf(x)[0,π] +∫[0,π] f(x)dcosx
=∫[0,π]f(x)sinxdx-cosxf(x)[0,π] -∫[0,π] f(x)sinxdx
=-cosxf(x)[0,π]
=f(π)+f(0)
=3
=∫[0,π]f(x)sinxdx+∫[0,π]f"(x)sinxdx
=∫[0,π]f(x)sinxdx+∫[0,π] sinxdf'(x)
=∫[0,π]f(x)sinxdx+sinxf'(x)[0,π] -∫[0,π] f'(x)dsinx
=∫[0,π]f(x)sinxdx-∫[0,π] f'(x)cosxdx
=∫[0,π]f(x)sinxdx-∫[0,π] cosxdf(x)
=∫[0,π]f(x)sinxdx-cosxf(x)[0,π] +∫[0,π] f(x)dcosx
=∫[0,π]f(x)sinxdx-cosxf(x)[0,π] -∫[0,π] f(x)sinxdx
=-cosxf(x)[0,π]
=f(π)+f(0)
=3
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