Lots of things going on here. I immediately know that $F(x)$ does exist since $f$ is riemann integrable due to the fact that it is continuous. First I need to show that $F$ is continuous, then find $F'$ and show that it is continuous. What can I do?
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1Well I guess you know right away from the Fundamental Theorem of Calculus that $F'(x) = g'(x)f(g(x))$, right (assuming that's not a negative in front of the integral). – Jared Apr 13 '14 at 23:08
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This sounds similar to this question: http://math.stackexchange.com/questions/751340/how-to-show-integral-is-continuous#comment1564189_751340 I still think the fact that $F(x)$ is the anti-derivative means that it must be differentiable (over the interval) and therefore continuous. – Jared Apr 13 '14 at 23:16
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I can show you how to get the derivative, but I don't know about proving continuity, etc. According to the fundamental theorem of calculus, there exists a function $G(x)$ such that:
$$ F(x) = \int\limits_a^{g(x)} f(x)dx = G\left(g(x)\right) - G(a)\text{, where } G'(x) = f(x) \\ F'(x) = g'(x)G'\left(g(x)\right) - 0 = g'(x)f\left(g(x)\right) $$
I don't really like the following notation, but problems like these seem to warrant it...you could also write it like this:
$$ F'(x) = g'(x) \cdot \left(f\circ g\right)(x) $$
Jared
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Thanks! This would obviously be continuous due to composition of continuous functions though, wouldn't it? not much of a proof there required maybe. – terrible at math Apr 14 '14 at 01:05
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also, how do we know $G'(x)$ exists? We're not given that $f$ is differentiable, just continuous. – terrible at math Apr 14 '14 at 01:10
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You don't need $f$ to be differentiable since you don't take a derivative of $f$. However, you do need that $g$ is differentiable (and this is given). As for $G'$ existing, I would argue that's implied by the fundamental theorem of calculus (but I don't have anything better than that)...like I said, I'm not real sure how to prove the continuity part. – Jared Apr 14 '14 at 01:18
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ah, you're right! i was thinking of this correctly. We proved in class that a continuous function on a compact interval is riemann integrable, so I know G' exists :) – terrible at math Apr 14 '14 at 01:31
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Hint: $F = G\circ g$ where $G(x) = \int_0^x f(t)\,dt$ is a function that you can differentiate.
mookid
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Sorry! There's something wrong with the Latex in my teacher's assignments, the minus's and equals signs are reversed. – terrible at math Apr 14 '14 at 01:27
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yeah I did: it's essentially the same thing Jared did below, right? – terrible at math Apr 14 '14 at 01:54
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sounds great, thank you guys. the continuity of the derivative then comes immediately from composition of continuous functions, right? (f is given continuous so f(g(x)) is continuous, and g is cont. differentiable so g'(x)) is continuous so their product is continuous.) – terrible at math Apr 14 '14 at 02:05
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Since $f$ is continuous in $[a,b]$ then $H(x):=\int_a^x f(t) dt, x\in[a,b]$ is differentiable, by the FTC (see for instance M. Spivak, Calculus). Now, $F$ is nothing but the composition $H\circ g$, this composition being well defined since $Im g\subset Dom H$, is differentiable in $[a,b]$ since it is the composition of two differentiable functions ($H$ is so by construction, and $g$ by hypothesis). Hence, by the chain rule $F'(x)=(H\circ g)'(x)=(H(g(x))'=H'(g(x)) g'(x)=f(g(x)) g'(x)$, using the FTC ($H'(x)=f(x)$). $F'$ is a continuous function since it is the product of two continuous functions on $[a,b]$, $f(g)$ since it is the composition of continuous functions (and $Im g\subset Dom f$) and $g$. Therefore $F\in C^1([a,b])$. QED.
