Fundamental Theorem 1 & 2

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Originally Posted by Mollier

Hi, the way I understand it, the first part of the fundamental theorem of calculus says that every function that is continuous on the closed interval $[a,b]$ has an antiderivative. If we denote the antiderivative by $F$ it looks like,

$F(x) = \int^x_a f(t)dt$ .

Now, F is continuous on $[a,b]$ and differentiable on $(a,b)$ , and $F'(x)=f(x)$ for all $x$ in $(a,b)$ .
Why is $F$ not differentiable on the closed interval $[a,b]$ ?

F is not (necessarily) differentiable at the end points, a and b, because that would require that the limits of the difference quotient from both directions be the same. And we have no information about F for x> b or for x< a.

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A corollary of this is that

$\int^b_a f(x)dx = F(b)-F(a).$

From what I read, the second part is almost the same as the first part but it does not assume continuity, but it assumes that an antiderivative of $f$ exists.
Does this mean that every continuous function has an antiderivative (part 1), but if a function has an antiderivative then it must not necessarily be continuous?

Yes, if a function is continuous, then it has an anti-derivative. But that is NOT an "if and only if" theorem. There exist non-continuous functions that have anti-derivatives. In fact, take any "jump function", integrate it, and you get an example.
If, say, f(x)= 0 for x< 0, f(x)= 1 for $x\ge 0$ , then the antiderivative function is F(x)= 0 for x< 0, F(x)= x for $x\ge 0$ .

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I do not clearly see the difference between the corollary of part one, and theorem part 2.

Thanks!