Is there any layman proof or explanation for area as an definite integral? I am reading a chapter on integrals from a School textbook. In the chapter, a function called the area function is defined based on the definite integral. It is as follows
$$A(x) = \int\limits_{a}^{x} f(x) dx$$
The following are two theorems are given in the textbook
Theorem 1

Let $f$ be a continuous function on the closed interval $[a, b]$ and
let $A(x)$ be the area function. Then $A'(x) = f(x)$ for all $x \in
 [a, b]$

Theorem 2

Let $f$ be a continuous function on the closed interval $[a, b]$ and
$F$ be an anti-derivative function of $f$. Then $\int\limits_{a}^{b} f(x) dx = [F(x)]_{a}^{b} = F(b)-F(a)$.

but it is mentioned that the proof of the theorems stated is beyond the scope of the book I am studying.
Is there any elementary or layman proof/explanation for understanding how the definite integral quantifies the area under the curve?
 A: As you have written, the first fundamental theorem of calculus states that if $f$ is continuous on $[a,b]$, then for all $x\in[a,b]$,
$$
\frac{d}{dx}\int_{a}^{x}f(t) \, dt=f(x) \, .
$$
To understand why, note that by the definition of the derivative,
$$
\frac{d}{dx}\int_{a}^{x}f(t) \, dt= \lim_{h\to0}\frac{\int_{a}^{x+h}f(t) \, dt-\int_{a}^{x}f(t) \, dt}{h}=\lim_{h\to0}\frac{\int_{x}^{x+h}f(t) \, dt}{h} \, .
$$
In a real analysis course, you will be given the tools to prove rigorously that this limit is equal to $f(x)$. However, if you are just looking for intuition, then consider how if $h$ is very small, then $\int_{x}^{x+h}f(t) \, dt$ is the area of an approximately rectangular shape, with width $h$ and $f(x)$:

As $h\to0$, any difference between $\int_{x}^{x+h}f(x) \, dx$ and the area of a rectangle becomes negligible. Hence,
$$
\lim_{h\to0}\frac{\int_{x}^{x+h}f(t) \, dt}{h}=\lim_{h\to0}\frac{h\cdot f(x)}{h}=f(x)
$$
This result is called "fundamental" for two reasons. First, there is no reason, a priori, to think that integration (finding the area under the curve) has anything to do with differentiation (finding the tangent to a curve); the first fundamental theorem of calculus shows that they are in fact intimately related. Second, the first fundamental theorem of calculus guarantees that if $f$ is continuous on a closed interval $[a,b]$, then it must be the derivative of some function on that interval, namely $F(x)=\int_{a}^{x}f(t) \, dt$. This is crucial in proving the second fundamental theorem of calculus.
The second fundamental theorem of calculus is now a simple corollary of the first. Suppose that $S$ and $F$ are both antiderivatives of $f$ on $[a,b]$, where $F(x)=\int_{a}^{x}f(t) \, dt$. Then, since $S'(x)=F'(x)$, there must be a number $C$ such that
$$
S(x)=F(x)+C=\int_{a}^{x}f(t) \, dt+C
$$
for all $x\in[a,b]$. To find the value of $C$, set $x=a$. Then,
$$
S(a)=F(a)+C=\int_{a}^{a}f(t) \, dt + C= C
$$
Hence, $S(x)-S(a)=F(x)$. Setting $x=b$, this becomes $S(b)-S(a)=F(b)$, or
$$
\int_{a}^{b}f(t) \, dt = S(b)-S(a) \, .
$$
NB the term "second fundamental theorem of calculus" is also sometimes used to describe a stronger result, where we don't even need to assume that $f$ is continuous. Again, this would be covered in a real analysis course.
