Bounds and the Fundamental Theorem of Calculus

Question

Suppose $f : \mathbb{R} \to \mathbb{R}$ is continuous. Fix $a \in \mathbb{R}$ and define $$ F(x) := \int_a^x f(t) \, \mathrm{d}t. $$ Every version of the Fundamental theorem of calculus (FTC) I've seen tells us that $F$ is differentiable for $x \geq a$ and that $F'(x) = f(x)$ for all $x \geq a$.

My question is : Is the above result also true for $x < a$ ?

My guess : I think it holds for $x < a$, since in that case I believe we have $$ F(x) =\int_a^x f(t) \, \mathrm{d}t = - \int_{-a}^{-x} f(-t) \, \mathrm{d}t $$ and by the FTC and the chain rule it follows that $$ F'(x) = - f(-(-x)) \cdot (-1) = f(x). $$ Is this correct ?

Yes. Another way to think of it is to pick some $b<x$, then $F(x) = \int_b^x f -\int_b^a f$. — copper.hat, Jan 28 '14 at 16:46
@copper.hat I tried to get the $x$ as the upper bound but did not succeed ! Thanks. — Amateur, Jan 28 '14 at 16:51

score 2 · Accepted Answer · answered Jan 28 '14 at 16:54

Well, you've developed a somewhat circular argument because you want to show $F$ is differentiable for $x < a$, but then you use that in your proof.

But, $a$ was chosen arbitrarily. So, if you choose a different starting value, some $\tilde a < x$, then it will be true that

$$ \tilde F (x) := \int_{\tilde a}^x f(t)\,dt $$ is differentiable for $x > \tilde a$, $\tilde F' = f$, etc.

Then for $\tilde a < x < a$

$$ \tilde F(x) = \int_{\tilde a}^a f(t)\,dt - \int_{x}^a f(t)\,dt = C + \int_a^x f(t)\,dt = C + F(x) $$

so $\tilde F$ and $F$ only differ by a constant, and everything follows from there.

Thank you for pointing out that my argument is circular. – Amateur Jan 28 '14 at 17:00 — Amateur, Jan 28 '14 at 17:00

Bounds and the Fundamental Theorem of Calculus

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