Fundamental Theorem of Calculus Perplexing Question?

Question

Suppose $f'(x) \geq C \geq 0$ for some constant $C$. Assume that $f'$ is differentiable everywhere. Suppose $f(0) = 0$. Can I then conclude that:

$\int_0^x f'(x) dx \geq \int_0^x C dx$ and hence $f(x) \geq Cx$?

Here, what I am applying is a property of integrable functions from my notes, which says that if we have two functions $f,g$ which are integrable, then we have $f \geq g$ => $\int f \geq \int g$.

And FTC(II) which states that if $f$ is differentiable and $f'$ is integrable on [a,b], then $\int^b_a f' = f(b) - f(a)$

However, the statement I've deduced is actually false. Take $f(x) = x^3/3+2x$ for example. We know that $f'(x) = x^2 + 2 \geq 2$ $\forall x$. However, we cannot deduce that $f(x) \geq 2x$ $\forall x$ as this is false for $ x < 0$

Can anyone tell me what has went wrong here? Am stuck here for quite some time. Thanks in advance!

Answer 1

The implication$$\bigl((\forall x\in[a,b]):f(x)\geqslant g(x)\bigr)\implies\int_a^bf(x)\,\mathrm dx\geqslant\int_a^bg(x)\,\mathrm dx$$holds indeed, but note that $a\leqslant b$. That's the problem with what you did: in that situation, $x<0$.

Answer 2

You've "centered" your problem at $x=0$. And your conclusion is true for $x\geq 0.$ Note that for $x< 0$, the fact that $f'(x)>C$ doesn't imply that $f(x)>Cx$.