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I have a line segment $\overline{AB}$, with endpoints $(x_1,y_1)$ and $(x_2,y_2)$. Drawing any function $f(x)$through those 2 points, the function must have an average velocity over $[x_1,x_2]$ of $\frac{y_2-y_1}{x_2-x_1}$. However, the area bounded by the line and $f(x)$ can vary greatly, and what I'm trying to find is the relationship between $f'(x)$ and that area. I'm not sure about how to formally express this next part, but as the average velocity of both the line and function over $[x_1,x_2]$ is the same, the average of the sum of the values of $f'(x)$ at all the infinite points on $f(x)$ over the domain $[x_1,x_2]$ will be the same as the average of the sum of the slope/value of the line's derivative at all the infinite points the line in the domain $[x_1,x_2]$. However, the actual values of $f'(x)$ could vary significantly. Visually, it appears the greater the greatest y-value of $f'(x)$ over $[x_1,x_2]$, the greater the area if $f(x)$ has no stationary points in the domain $[x_1,x_2]$. For example, with the line segment with endpoints $(0,0)$ and $(2,4)$, and the function $f(x)=x^\frac{3}{2}*\sqrt{2}$, the area bounded by $y=2x$ and $f(x)$ is $\frac{16-2\sqrt{2}}{5}$, and the greatest y-value of $f'(x)$ over $[1,2]$ is 3. If $f(x)=\frac{x^3}{2}$, then the area bounded is 4, and the greatest y-value of $f'(x)$ over $[0,2]$ is 6.

Am I correct in saying that the greater the greatest y-value of $f'(x)$ over $[x_1,x_2]$, the greater the area bounded, if $f(x)$ does not have stationary points in the domain $[x_1,x_2]$? And if so, does there exist a relation that, given a line segment over $[x_1,x_2]$, can give the area bounded by the line segment and $f(x)$, given only the greatest y-value of $f'(x)$ over $[x_1,x_2]$?

Thanks to David K for answering the above questions. I only have one last question: I am aware of how to actually find the exact area, but, is there some general correlation that can be drawn between some property of the derivative, and the area? Which is to say, using no calculus methods, can one draw any conclusions about the area given the equation of the derivative? Or even given the actual function?

I also have no idea what tags I should use on this, so I would appreciate it too if someone could put on the right tags.

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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 $f'.$

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    $\begingroup$ Thank you for your answer! I have changed my question based on your answer. $\endgroup$ – H Huang Feb 12 at 7:21
  • $\begingroup$ Whether there are stationary points or not is irrelevant. I have edited them out of the answer. (Note: if the function is continuously differentiable over the entire interval, there will be a point at which $f'$ equals the slope of the line segment. That's the Mean Value Theorem, and it applies to the examples in the question as well.) $\endgroup$ – David K Feb 12 at 13:12

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