令f(x)=sinx/x，(π/2<=x<=π)，则f'(x)=(xcosx-sinx)/x^2<=0所以f(x)在[π/2,π]上单调递减所以0=sinπ/π<=sinx/x<=sin(π/2)/(π/2)=2/π根据积分中值定理，存在k∈[π/2,π]，使得∫(π/2,π)sinx/xdx=(π/2)*sink/k所以0<=(π/2)*sink/k<=1即0<=∫(π/2,π)sinx/xdx<=1
