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Mean Value Theorem: Derivatives vs. Integrals

Forfattertalljerome
This activity is intended to illuminate
  1. The relationship between the Mean Value Theorem (MVT) for Derivatives and the MVT for Integrals.
  2. The Fundamental Theorem of Calculus (FTC) statement involving the derivative of a definite integral. (Whether this shall be referred to as the "first" FTC or the "second" FTC is not consistently agreed upon.)
One may go through the algebraic manipulation that addresses these points yet still completely miss any geometric understanding. (Or was that just me back in school?) Interacting with this construction is intended to help foster a geometric interpretation of the MVTs and FTC. Feel free to jump right in and tinker with the construction below. See the instructions and questions below the construction to check your understanding.

MVT for Integrals

Leave the "rate f(x)" checkbox turned on and turn off the "accumulating function F(t)" checkbox. Drag point A to wherever you like along the x-axis, but then let's consider it constant for the rest of the activity. Let represent the coordinates of this point, with constant . Point B may also be dragged along the x-axis, and we will consider its position as variable throughout the activity. Let represent the coordinates of this point, with variable . Instead of dragging either point, you may alternatively select the point and then on a keyboard press the left/right arrow buttons. Holding CTRL while pressing the arrow buttons will increase the distance that the point translates. To change scaling of x vs. y axes, press Shift on keyboard and click-drag either axis, or pinch/expand either axis on a touchscreen. Let's metaphorically think of the region represented by the definite integral as a block of ice. Yes, the metaphor falls apart a bit when dips into negative values, but use your imagination to ponder the notion of "negative ice" that can cancel out "positive ice." Drag point B to lie right on top of point A. The width of the ice block has shrunk to 0, so at this moment, right?

Gradually drag point B to the right, counting the unit squares of ice that accumulate in the process. Stop at some point. Divide the ice area you've tallied by the width of the ice block. You have just calculated the _______ ________ of the function over your specified interval.

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Drag the "ice" slider all the way to the right, causing it go through the "melting" process and convert into "water." Yes, we're pretending that ice and water have the same density in this metaphor. (Relax, chemistry nerds.) Continuing the metaphor, what does the MVT for Integrals guarantee will occur for at least one location an the specified interval between points A and B?

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Turn on the "accumulating function F(t)" checkbox. Verify that it accurately tracks the "ice area" as you drag B along the horizontal axis in accordance with its definition .

MVT for Derivatives

Turning our attention to the F graph, turn on the "tangent & secant" checkbox. An unlabeled point appears on the F curve. Drag that point to a location on the F curve that satisfies the MVT for Derivatives. What does the Theorem guarantee at this particular point?

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Connect MVT for Derivatives to MVT for Integrals

Tangent line: At the same x-value(s) where you just dragged the unlabeled point on F(t) to satisfy the MVT for Derivatives, follow the vertical line that passes through that point back to the block of ice f(x). What of significance do you observe at that location on the ice?

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Secant line: Compute the slope of the secant line for F(t) on the interval between points A and B. How does this slope connect to the block of ice f(x)?

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Fundamental Theorem of Calculus

With the "tangent & secant" checkbox still on, drag the unlabeled point back and forth along the differentiable F(t) curve. What is noteworthy about the slope of the line tangent to F(t) at the unlabeled point?

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For most of this discussion we've been treating the position of point A as constant. Nonetheless, drag A left or right to a new position. While you drag A left/right: What happens to the graph of the accumulating function F(t) as you do so? How is the derivative of F(t) affected?

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Using Your Own Custom F(x) Function

In the construction above, a differentiable function is already entered for you. When point A is located at the origin, . If A is repositioned away from the origin, then is vertically translated accordingly. The input box allows you to redefine if you like, either piecewise like the default function or not. Just make sure your is differentiable in order for the MVT to apply.

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