Question

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=?

Answer 1

∫[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