# Fundamental Theorem of Calculus Perplexing 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!

• Your inequality should flip if x < 0. Commented Oct 1, 2018 at 13:36

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, $$x<0$$.
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 $$f(x)>Cx$$.