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This section is aimed at students in upper secondary education in the Danish school system, some objects will be simplified some details will be omitted.

Integrals

The so called definite integral of a function f on an interval \([a,b]\) is defined as the area under the graph of f in that interval. There are several ways to define this and one is to divide the interval into sub-intervals \(a=x_0< x_1< x_2 < \cdots< x_N=b\) and then add up the areas of the rectangles that have the function value \(f(t_n)\) as its height and the width \(\Delta x_n=x_n-x_{n-1}\). If we then make the sub-intervals narrower and narrower, we will approach the true area under the graph. This yields the following definition for the definite integral $$\int_a^b f(x)dx=\lim_{N\to\infty}\sum_{n=1}^Nf(t_n)\Delta x_n$$

The Fundamental Theorem of Calculus

The importance of the theorem is that it relates derivatives and integrals in the sense that they are somehow "inverse" operations, i.e. for functions with the necessary characteristics, we have $$f(x)=\frac{d}{dx}\int_a^xf(t)dt$$ and $$\int_a^x\frac{d}{dt}f(t)dt=f(x)-f(a)$$
The second of these is immediately useful for integrating a broad class of functions, namely functions that are the derivative of another. This other function \(F(x)\) is called an antiderivative of \(f(x)=F'(x)\) and every other antiderivative will differ from it by a constant.

Indefinite Integrals

The class of antiderivatives for a function \(f(x)\) is called the indefinite integral and is written as a definite integral without the limits, i.e. $$\int f(x)dx=F(x)+C$$
The first result of the main theorem is traditionally a bit more technical, but I've tried coming up with a practical use-case. Imagine you are building a "rectangular" fence along a river bank that can be described by the graph of some function \(f(x)\), and you have a limited amount of fencing materials, lets say enough for \(l\) m. The fenced-in area can then be described by $$A(h)=b\cdot h-\int_x^{x+h}f(t)dt$$ where b is the "width" of the area, from the "x-axis" to the horizontal section, \(x\) and \(h\) the horizontal position and distance along the river-bank, respectively. The task is then to determine the optimal dimensions, to this end lets relate them to one-another by $$l=h+b-f(x)+b-f(x+h)\implies b=(l-h+f(x)+f(x+h))/2$$ This yields $$A(h)=(l-h+f(x)+f(x+h))\cdot h/2-\int_x^{x+h}f(t)dt$$ To maximize this function we find the stationary point by differentiating with respect to h and setting it to zero. I.e. \begin{align} 0=&(l-h+f(x)+f(x+h)+h\cdot(f'(x+h)-1))/2-f(x+h)\\ 0=&l-2h+f(x)-f(x+h)+hf'(x+h) \end{align} Here the derivative and integral cancel and if I divide by \(h\) we get that $$f'(x+h)-\frac{f(x+h)-f(x)}{h}=2-\frac{l}{h}$$ which means that the difference between the tangent- and secant slope at the end-point, needs to have the value \(2-l/h\).
We will need two versions of the so-called mean value theorem for our proof.

The Mean Value Propositions

The first states that if \(f(x)\) is differentiable in the interval \(]a,b[\) then there exists a \(c\in]a,b[\) such that $$f'(c)=\frac{f(b)-f(a)}{b-a}$$ which in essence means that there is a value in the domain where the slope is the same as the average slope on the entire interval.
The second theorem states that if \(f(x)\) is continuous on \([a,b]\), then we can find a \(c\in[a,b]\) such that $$f(c)(b-a)=\int_a^bf(x)dx$$ which means that the area under the graph is equal to the area of a rectangle with the interval length as its width and height corresponding to the function value at some point in the interval.
Being propositions, I will not prove these, but here is an example of their direct use, before I delve into the proof of the FToC.
In this example I will investigate the so-called Gustav's law, which states that the constants from the two versions of the mean value propositions is actually the same constant for exponential functions. Consider the function \(f(x)=de^{kx}\) then the first proposition says that $$dke^{kc}=\frac{de^{kb}-de^{ka}}{b-a}$$ and the second states that $$de^{kc}(b-a)=\frac{de^{kb}-de^{ka}}{k}$$ Which are equivalent expressions by swapping the \(b-a\) and \(k\). The value of the constant turns out to be $$c=\frac{1}{k}\log\left(\frac{e^{kb}-e^{ka}}{dk(b-a)}\right)$$

Proof

For the first part we want to differentiate the integral, so lets write it as the function $$F_a(x)=\int_a^xf(x)dx$$ and start by considering the rise of the secant \begin{align} &&F_a(x+h)-F_a(x)=&\int_a^{x+h}f(x)dx-\int_a^xf(x)dx\\ &&=&\int_x^{x+h}f(x)dx\\ &&=&(\cancel{x}+h\cancel{-x})f(c)\\ \implies&&\frac{F(x+h)-F(x)}{h}\to&f(x) \end{align} since \(c\in[x,x+h]\to[x,x]\) by the second mean-value proposition.
For the second statement, lets seperate the interval \([a,x]=[x_0,x_1]\cup [x_1,x_2]\cup\cdots\cup[x_n,x_{n+1}]\) where \(a=x_0\) and \(x=x_{n+1}\). Then I can create the following telescopic sum \begin{align} f(x)-f(a)=&f(x_{n+1})-f(x_n)+f(x_n)-f(x_{n-1})+\\ &\cdots+f(x_2)-f(x_1)+f(x_1)-f(x_0)\\ =&\sum_{k=0}^n f(x_{k+1})-f(x_k)\\ =&\sum_{k=0}^n \frac{f(x_{k+1})-f(x_k)}{x_{k+1}-x_k}(x_{k+1}-x_k)\\ =&\sum_{k=0}^n f'(c_k)\Delta x_k\\ \to&\int_a^bf'(x)dx \end{align} where \(\Delta x_k=x_{k+1}-x_k\). Since the left-hand-side is a constant, the limit must be that same constant.

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