設(shè)f(x)在(-∞,+∞)內(nèi)連續(xù),且f(x)>0,證明F(x)=[∫(0-x)tf(t)dt]/[∫(0-x)f(t)dt]在(0,+∞)單調(diào)增加
∫(0-x)表示下標(biāo)為0 上標(biāo)為x,
此問(wèn)題的核心是求該函數(shù)的導(dǎo)數(shù),然后證明其導(dǎo)數(shù)大于0(我想難點(diǎn)可能在導(dǎo)數(shù)分析上).
對(duì)F(x)關(guān)于x求導(dǎo)
![](http://c.hiphotos.baidu.com/zhidao/wh%3D600%2C800/sign=500095d0d9f9d72a1731181be41a040c/f9198618367adab400a1eaac8bd4b31c8601e4e8.jpg)
對(duì)F(x)的表達(dá)式, 可知其分母大于0, 對(duì)其分子項(xiàng)進(jìn)行分析, f(x)是大于零的,由因?yàn)榉e分項(xiàng)里面x>t,故積分項(xiàng)也是大于零的,故
![](http://d.hiphotos.baidu.com/zhidao/wh%3D600%2C800/sign=f01bbc2ff736afc30e5937638329c7fc/4034970a304e251f669e1517a786c9177e3e53f0.jpg)
從而證得F(x)>0.亦即F(x)在區(qū)間(0,+\infty)是單調(diào)遞增的.