You might want to look for a treatise by Archimedes on "the method", in which he does this computation using Cavalieri's principle (more or less) and a rather clever mechanical argument. There's a good case to be made that in doing so, he was inventing some of the key ideas of calculus, and the proofs he offers don't really meet the modern standard of rigor, but nonetheless, it's a remarkable achievement.
Let $(x_1,y_1) = (0,0)$ and $(x_2,y_2) = (\pi,2\pi).$
Let $f(x) = 2x + \sin x.$
Then $f'(x) = 2 + \cos x$ and the maximum value of $f'(x)$ on $[x_1,x_2]$ is
$f'(0) = 3.$
Now let $g(x) = 2x + \sin^2 x.$
Then $g'(x) = 2 + \sin (2x)$ and the maximum value of $g'(x)$ on $[x_1,x_2]$ is
$g'(\frac\pi4) = 3.$
These are two functions between the same two endpoints on the same interval with the same maximum value of the derivative.
But the area between the segment and $f(x)$ is $2$ while the area between the segment and $g(x)$ is $\frac\pi2.$
Let $h(x) = 2x + \frac54 \sin^2 x.$ Now $h'(\frac\pi4) = \frac{13}4,$ which is greater than the maximum value of $f'.$ But the area between the line segment and $h(x)$
is just $\frac{5\pi}{8},$ which is less than the area between the segment and $f(x)$ (which is $2$).
So no, you cannot tell much about the area under the curve just by looking at the greatest value of $f'(x).$
In particular it is not true that a greater maximum value of $f'(x)$ means a greater area bounded by the curve and the line segment.
Of course, there is a relationship between $f'$ and the area enclosed by the graph of $f$ and the segment.
The relationship is guaranteed by the Fundamental Theorem of Calculus.
Find the indefinite integral of $f'$ and set the constant of integration to select the antiderivative that passes through $(x_1,y_1).$
Now you have recovered $f$ and can use the integral of $f$ to find the area enclosed between the segment and the graph of $f.$
The key thing is that the value of a function (such as $f'$) at a single point gives you only limited information about the integral of the function, even if that single point is chosen to be the the maximum value of that function.
And when you ask about the area enclosed by $f$ and a line segment,
you're asking something about the integral of the integral of $f'.$
Best Answer
I'll try to give a proof which should be on the right track.
Let $P$ and $Q$ be two points on the parabola, $M$ and $N$ their projections on the directrix, $F$ the focus. When $Q$ approaches $P$ the area of trapezoid $PQNM$ becomes the double of the area of triangle $PQF$, because bases $PM$ and $PF$ are equal, $QN$ approaches $PM$, and the ratio of altitudes $QH$ and $QK$ tends to $1$, because line $PQ$ becomes by definition the tangent at $P$, which is the bisector of angle $\angle FPM$ (focal property of parabola).
The area of the parabolic sector is the sum of triangles like $PQF$, while the area of the parabolic rectangle is the sum of trapezoids like $PQNM$, hence the thesis follows.