# Do all continuous functions have antiderivatives?

If not all continuous functions are differentiable, so how is it that all continuous functions have anti-derivatives?

• yes, this is stated in the fundamental theorem of calculus. – Masacroso Dec 10 '17 at 8:48
• Because $g(x)=\int_0^x f(t)\,dt$ is differentiable and $g'(x)=f(x)$. – user228113 Dec 10 '17 at 8:48
• Not all natural numbers can be divided by 2, how is it that all can be multiplied by 2? – Professor Vector Dec 10 '17 at 8:49
• Also @ProfessorVector: both facts from number theory are almost trivial to prove. The case here is that it is difficult to show that continous functions possess anti-derivatives. I suppose your analogy is more to illustrate that such contrasts should not be hard to believe. – Paramanand Singh Dec 10 '17 at 9:46
• @ProfessorVector : if it were that trivial most calculus books would have proved it. It is trivial only when you are already aware of it. The fact that continuous functions are integrable is a typical example of "theorems whose proofs are beyond the scope of the introductory books" but somehow their statements are in their scope. – Paramanand Singh Dec 10 '17 at 10:51

Let $I\subset\mathbb R$ be an interval with more than one point. If $f\colon I\longrightarrow\mathbb R$ is a continuous function, the existence of an anti-derivative of $f$ can be proved as follows: take $a\in I$ and define$$\begin{array}{rccc}F\colon&I&\longrightarrow&\mathbb R\\&x&\mapsto&\displaystyle\int_a^xf(t)\,\mathrm dt.\end{array}$$Then $F$ is a primitive of $f$, by the Fundamental Theorem of Calculus.