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Ignore the technique details such as continuity or existence of derivatives etc, the Fundamental Theorem of Calculus can be put as

There are two parts to the theorem. The first part asserts that the derivative of a definite integral of a function, equals the original function; while the second part asserts that the definite integral of the derivative of a function, equals the original function.

  1. List item Let $f$ be a function defined on a closed interval $[a, b]$. Let $F$ be the function defined, for all $x$ in $[a, b]$, by $$F(x)=\int_{a}^{x}f(t) \ dt$$ Then for all $x$ in $(a, b)$ $$F'(x)=f(x)$$
  2. Let $F$ be a function on $[a,b]$, and $f$ a function on a closed interval $[a,b]$ which is a derivative of $F$ in $(a,b)$: $$F'(x)=f(x)$$ Then $$\int_{a}^{b}f(x) \ dx=F(b)-F(a)$$

Part 2 seems is the one get used everyday. Without part 2, it would be rather hard to compute any definite integral, e.g. $\int_0^1 \sqrt{x} \ dx$, based on the definition of definite integral, i.e. the definite integral is the limit of the sum

$$\int_a^b f(\tau) \ d\tau := \lim_{\Delta \tau \rightarrow 0} \sum_{i=0}^{n} f(\tau_i) \Delta \tau$$ , where $\Delta \tau = \frac{b - a}{n}$, and $\tau_i = a + i \Delta \tau$ ($i$ from $0$ to $n$) are the time moments evenly split $[a, b]$.

However, is there any application of the first half of the theorem?

I understand as the symmetric point of view, or completeness of the theorem, part 1 is necessary. Just wonder, if there's some case that part 1 come into save?

athos
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  • The first part only establishes the symmetry between integrating and differentiating. It is not actually needed in applications where integrals have to be calculated. It is only of theoretical interest. – Peter Dec 05 '22 at 10:46
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    There are plenty of applications. You just don't notice them because they are given without explicit comment. For example, it is a common practice to define the natural logarithm as $\ln x = \int_1^x \frac{dx}x$, from which it follows that $\frac{d}{dx} \ln x = \frac 1x$, by the first FTC. And from there that $\frac {d}{dx}e^x = e^x$. In ordinary differential equations, solutions are often given by similar integral formulas. – Paul Sinclair Dec 06 '22 at 02:45
  • @PaulSinclair Indeed, I didn't notice this way of defining $\ln x$ and $e^x$, though I passed by it in https://math.stackexchange.com/a/834547/26632 . Could you pls elaborate a bit more on the ODE examples? – athos Dec 06 '22 at 09:35

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I believe the validity of the first part is required in order to be able to differentiate under the integral sign (aka 'Feynman's Trick').

PMar
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  • As it’s currently written, your answer is unclear. Please [edit] to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. – Community Dec 05 '22 at 13:25
  • pls kindly elaborate? I know Feynman's trick but don't get your point. – athos Dec 05 '22 at 17:53