# Can part 1 of the fundamental theorem of calculus be stated without using the area function concept?

Being new to calculus, I'm trying to understand Part 1 of the Fundamental Theorem of Calculus.

Ordinarily, this first part is stated using an " area function" F mapping every x in the domain of f to the number " integral from a to x of f(t)dt".

However, I encounter difficulties to understand what is the status of this area function, being apparently neither an indefinite integral , nor a definite integral( for, I think, a definite integral is a number, not a function); if this " area function" is not an " integral " ( of some sort), I do not understand in which way asserting that F'=f amounts to saying " integration and differentiation are inverse processes" as it is said informally.

Hence my question : is there an easier to understand version of FTC Part 1 that does not make use of the area function concept?

Note : I think I understand in which way the area function is a function and what it " does". What I do not understand is the role it plays in proving that " integration and differentiation a reverse processes" ( being given this function is neither a definite integral, nor an indefinite integral, as MSE answers I got previously tend to show).

• "Ordinarily"? I don't think I have ever seen the Fundamental Theorem stated in terms of an "area function". Who says "area function"? – David K Feb 13 '20 at 13:38
• @DavidK. - some also say " accumulation function" ; the function F that maps every x in dom(f) to the number " integral from a to x of f(t)dt". – user655689 Feb 13 '20 at 13:50
• Sorry for misrepresenting you in a (now deleted) comment, I lost track of the train of thought for a moment. – David K Feb 13 '20 at 14:02
• I would say $F$ is a function that is defined in terms of definite integrals. We get into the question of "what's the definition of a definite integral," but if you want to stick to a definition in which if you get two different numerical values you must have two definite integrals (rather than one definite integral with a variable bound), we can define $F$ in terms of a family of definite integrals in which everything is the same except the upper bound. – David K Feb 13 '20 at 14:05
• BTW you're not wrong to be concerned about the exact definitions of these things. It's a point that often gets glossed over. – David K Feb 13 '20 at 14:09

I think the key issue here is that you are unable to understand how integration and differentiation are reverse processes.

In order to understand and appreciate it fully you need to know the definition of derivative (easy) and that of integral (difficult and mostly avoided in beginner's calculus texts).

Just as derivative is defined as a limit, the integral $$\int_{a} ^{b} f(x) \, dx$$ is also defined as a complicated limit based on $$a, b, f$$. There are some technicalities involved here and you can have a look at this answer for more details.

The link between derivatives and integrals is then understood by analyzing the integral $$\int_{a} ^{x} f(t) \, dt$$. The idea is to understand how the integral varies as the interval of integration varies. And there you have the Fundamental Theorem of Calculus part 1 which says that

FTC Part 1: Let $$f:[a, b] \to\mathbb {R}$$ be Riemann integrable on $$[a, b]$$. Then the function $$F:[a, b] \to \mathbb {R}$$ defined by $$F(x) =\int_{a} ^{x} f(t) \, dt$$ is continuous on $$[a, b]$$. And further if $$f$$ is continuous at some point $$c\in[a, b]$$ then $$F$$ is differentiable at $$c$$ with $$F'(c) =f(c)$$.

In simpler terms if the function $$f$$ being integrated is continuous on entire interval of integration then $$F'(x) =f(x)$$ in entire interval. Thus we are able to figure out the rate at which the integral varies as the interval of integration varies.

And this gives us a way of evaluating integrals without using the complicated definition of integral. Rather one hopes to find an anti-derivative and just subtract its values at end points of the interval. More formally we have

FTC Part 2: Let $$f:[a, b] \to\mathbb {R}$$ be Riemann integrable on $$[a, b]$$ and further let's assume that $$f$$ possesses an anti-derivative $$F$$ on $$[a, b]$$ ie there exists a function $$F:[a, b] \to \mathbb {R}$$ such that $$F'(x) =f(x)$$ for all $$x\in[a, b]$$. Then $$\int_{a} ^{b} f(x) \, dx=F(b) - F(a)$$

Yes, $$\int_a^bf(t)\,\mathrm dt$$ is a number. But if you change $$a$$ or $$b$$ (or both), you usually get a different number. So, $$(a,b)\mapsto\int_a^bf(t)\,\mathrm dt$$ is a function of $$a$$ and $$b$$ (and $$f$$). And, in particular, for $$a$$ (and $$f$$) fixed, $$x\mapsto\int_a^xf(t)\,\mathrm dt$$ is a function. And the Fundamental Theorem of Calculus states that, if $$f$$ is continuous, then $$F$$ is differentiable and $$F'=f$$.

• By the same token, if we define some function $f$ from real numbers to real numbers, then $f(0)$ is a number, and so is $f(17)$, and so is $f(a)$ if $a$ is a number. The fact that you can look at an application of a function and say it is a number does not negate the fact that you had a function to apply. – David K Feb 13 '20 at 13:45