2. 考研
  3. 设f(x),g(x)在[0,1]上的导数连续，且f(0)=0，f’(x)≥0,g’(x)≥0.证明：对任何a∈[0,1]有 ∫0ag(x)f’(x)dx+∫01f(x)g’(x)dx≥f(a)g(1)

设f(x),g(x)在[0,1]上的导数连续，且f(0)=0，f’(x)≥0,g’(x)≥0.证明：对任何a∈[0,1]有 ∫0ag(x)f’(x)dx+∫01f(x)g’(x)dx≥f(a)g(1)

admin2022-09-05  49
问题 设f(x),g(x)在[0,1]上的导数连续，且f(0)=0，f’(x)≥0,g’(x)≥0.证明：对任何a∈[0,1]有
∫0ag(x)f’(x)dx+∫01f(x)g’(x)dx≥f(a)g(1)
选项
答案证法一 设F(x)=∫0xg(t)f’(t)dt+∫01f(x)g’(x)dx－f(x)g(1),x∈[0,1] 则F(x)在[0,1]上的导数连续性，并且 F’(x)=g(x)f’(x)－f’(x)g(1)=f’(x)[g(x)－g(1)], 由于X∈[0,1]时,f(x)≥0, g’(x)≥0,因此F’(x)≤0,即F(x)在[0.1]上单调递减. 注意到 F(1)=∫01g(t)f’(t)dt+∫01f(t)g’(t)dt－f(1)g(1) 而∫01g(t)f’(t)dt=∫01g(t)df(t)=g(t)f(t)|01－∫01f(t)g’(t)dt =f(1)g(1)－∫01f(t)g’(t)dt 故F(1)=0 因此x∈[0,1]时，F(x)≥0由此可得对任何a∈[0,1]时有 ∫0ag(x)f’(x)dx+∫01f(x)g’(x)dx≥f(a)g(1) 证法二 ∫0ag(x)f’(x)dx=g(x)f(x)|0a－∫0af(x)g’(x)dx =f(a)g(a)－∫0af(x)g’(x)dx ∫0ag(x)f’(x)dx+∫01f(x)g’(x)dx=f(a)g(a)－∫0af(x)g’(x)dx+∫01f(x)g’(x)dx =f(a)g(a)+∫a1f(x)g’(x)dx 由于x∈[0,1]时，f’(x)≥0,因此f(x)在[0,1]内单调递增 f(x)≥f(a),x∈[a,1] 又由于x∈[0,1]时，g’(x)≥0因此 f(x)g’(x)≥f(a)g’(x),x∈[a,1] ∫a1f(x)g’(x)dx≥∫a1f(a)g’(x)dx=f(a)[g(1)－g(a)] 从而∫0ag(x)f’(x)dx+∫01f(x)g’(x)dx≥f(a)g(a)+f(a)[g(1)－g(a)]=f(a)g(1)
解析
转载请注明原文地址:https://kaotiyun.com/show/XwR4777K
0

考研数学三

相关试题推荐
随机试题
最新回复(0)