Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

So I was given $f(x)$ continuous and positive on $[0,\infty)$, and need to show that $g(x)$ increasing on $(0,\infty)$

And $g(x)={\int_0^xtf(t)dt\over \int_0^xf(t)dt} $

So my approach is I want to show that $g'(x)>0$, so I used FTC and quotient rule to take the derivative of $g'(x)$, but then I got suck at midway because I cannot simplify it.

share|improve this question

2 Answers 2

up vote 4 down vote accepted

We have \begin{align*} g'(x) &= \frac{xf(x)\cdot \int_0^x f(t)\,dt - \int_0^x tf(t)\,dt \cdot f(x)}{(\int_0^x f(t)\, dt)^2} \end{align*} Now the denominator is positive, we look at the numerator \begin{align*} xf(x)\int_0^x f(t)\,dt - \int_0^x tf(t)\, dt \cdot f(x) &= \int_0^x xf(x)f(t)\, dt - \int_0^x tf(x)f(t)\, dt\\ &= \int_0^x (x-t)f(x)f(t)\, dt \end{align*} Now $f(t) > 0$ for $t > 0$, $f(x) > 0$ and $(x-t)> 0$ for $t > 0$. So the numerator is positive for $x > 0$, therefore $g' > 0$ on $(0,\infty)$ as wished.

share|improve this answer

You want to show that for $x>0$, \begin{equation}g^{\prime}(x)=\frac{xf(x)\int_0^xf(t)dt-f(x)\int_0^xtf(t)dt}{(\int_0^xf(t)dt)^2} >0\end{equation} which implies since $f$ is positive, $0<x\int_0^xf(t)dt-\int_0^xtf(t)dt=\int_0^x(x-t)f(t)dt$ which is true since $x>t>0$ and $f$ is positive.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.