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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!

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    $\begingroup$ Your inequality should flip if x < 0. $\endgroup$ Commented Oct 1, 2018 at 13:36

2 Answers 2

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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, $x<0$.

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  • $\begingroup$ Hi thanks for your explanation! It didnt occur to me that the integral properties demand that b>a. Thanks alot! $\endgroup$
    – HK Tan
    Commented Oct 1, 2018 at 13:44
  • $\begingroup$ I'm glad I could help. $\endgroup$ Commented Oct 1, 2018 at 13:44
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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 $f(x)>Cx$.

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