Question

高数积分题

设f(x)在[0,2π]上具有二阶连续导数，且f''(x)≥0证明：
∫(0→2π)f(x)cosxdx≥0

Answer 1

∫(0,2π)f(x)cosxdx
=∫(0,2π)f(x)d(sinx)
=f(x)sinx|(0,2π)-∫(0,2π)sinxf'(x)dx
=-∫(0,2π)sinxf'(x)dx
=-∫(0,π)sinxf'(x)dx-∫(π,2π)sinxf'(x)dx

对于后一项，令t=x-π，则x=t+π，dx=dt
∫(0,2π)f(x)cosxdx
=-∫(0,π)sinxf'(x)dx-∫(0,π)sin(t+π)f'(t+π)dt
=-∫(0,π)sinxf'(x)dx+∫(0,π)sintf'(t+π)dt
=-∫(0,π)sinxf'(x)dx+∫(0,π)sinxf'(x+π)dx
=∫(0,π)sinx[f'(x+π)-f'(x)]dx
因为f''(x)>=0，所以f'(x)单调递增，即f'(x+π)-f'(x)>=0
又因为sinx在(0,π)上>0，所以sinx[f'(x+π)-f'(x)]>=0
即∫(0,π)sinx[f'(x+π)-f'(x)]dx>=0
即∫(0,2π)f(x)cosxdx>=0，证毕