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Let $f(x) = f(c) + \int_c^x h(t) \ dt$ where $f,h:[a,b] \to \mathbb{R}$ and $h$ is riemann integrable, and $c \in [a,b]$. The book I'm looking at does this:

$$f'(x) = \frac{d}{dx}\left(f(c) + \int_c^x h(t) \ dt \right)$$ $$f'(x) = \frac{d}{dx}\left(f(c) + \int_a^x h(t) \ dt + \int_c^a h(t) \ dt \right)$$ $$f'(x) = h(x)$$

I am assuming they used the fundamental theorem of calculus, but I do not know how, or why the split of the integral was necessary

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  • $\begingroup$ It's not necessary if you apply it to $h_{\mid[c,b]}$, but the author apparently chose to apply it to $h$ (which is defined in $[a,b]$). $\endgroup$
    – Git Gud
    Apr 21, 2015 at 18:02

1 Answer 1

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Notice that the function $h$ is defined at $[\color{red}a,b]$, and the Theorem is applied at $$g(x) = \int_a^x h(t)dt \underbrace{\implies}_{FTC} g'(x) = h(x) $$

Then as $f(c)$ and $\int_c^ah (t) dt$ are constants, it follows your result.

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  • $\begingroup$ You're welcome!! Glad I could help. $\endgroup$ Apr 21, 2015 at 18:09

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