Riemann Integral Property for Continuous, Monotonic, Non-negative Function

Question

If $f$ is continuous, non-negative, and monotonically increasing function on $[0,∞)$, then prove that $\int^{x}_{0} f(t)dt\leq xf(x)$ $\forall x ≥ 0$

My attempt:

Define $F(x)=\int^{x}_{0} f(t)dt$. Since $f$ is continuous, then $F$ is differentiable in $[0,∞)$.

Choose $x\in [0,∞)$

By Mean Value Theorem, $F^{'}(x_{0})=[F(x)-F(0)]/x$ for some $x_{0}\in [0,x]$

$\implies x.F^{'}(x_{0})=[F(x)-F(0)]$. Since $F(0)=0$, we have $x.F^{'}(x_{0})=F(x)$. Also, $F^{'}(x_{0})=f(x_{0})$. So, $x.f(x_{0})=\int^{x}_{0} f(t)dt $.

Since $f(x_{0})\geq 0, x\geq 0$ and $f$ is monotonically increasing, then $\forall x>x_{0}$,

$x.f(x)\geq \int^{x}_{0} f(t)dt$.

Since $x$ is arbitrary, we can choose $x$ arbitrarily close to $0$ and hence the corresponding $x_{0}$ can be found arbitrarily close to $0$. So, for all $x$, we have $x.f(x)\geq \int^{x}_{0} f(t)dt$

Is the proof correct and rigorous enough? If not, then please suggest an alternative proof in full.

Answer 1

Yes, it is correct, except with the following minor points:

• Handle the case where $x=0$ separately as Neeraj said, because you want to avoid dividing by 0.
  • You can instead use directly the version of mean value theorem $\exists c \in [a,b]$ s.t. $(b-a)f'(c)=f(b)-f(a)$.
  • However, the "stronger" mean value theorem says when $a<b$, $\exists c \in (a,b)$ s.t. $(b-a)f'(c)=f(b)-f(a)$.
• You should immediately define $F(x)=\int_0^x f(t) dt$, and not say that $F$ is the indefinite integral earlier.
• Edit: I didn't see the last line earlier. I think it was meant to handle the case $x=0$, but it is not complete unless you say that both sides are continuous functions. As I said in my comments, why don't you just do $x=0$ separately?

Also, there is a simple proof.

$\int_0^x f(t) dt\le\int_0^x f(x) dt=xf(x)$.

(since $\forall t\in [0,x], f(t)\le f(x)$)