More about the FTC

In keeping with my idea from the last post of sneaking up on ideas, here is a way to sneak up on the other part of the FTC.

Consider these three functions F1(t) = 3, F2(t) = 2t and F3(t) = 2t + 3

For each of these three functions do the following:

  1. Graph each function on the interval [0, t] for t\ge 0.
  2. Let x be a value of t and let x move to the right along the t-axis starting at zero. Using area formulas, write functions, A1(x), A2(x), and A3(x), for the areas of the region between the graph and the t-axis. Make a table of values for x = 1, 2, 3, …, 10. Do not use any “calculus”; the area expressions are easy to find from geometry.  (A1(t) = 3t; A2(t) = t2; A3(t) = t2 + 3)
  3. Consider what happens as the as x moves from zero to the right.  The expressions are a function that gives the area for each x. Now write expressions for the three areas with an integral that gives the same area. The lower limit will be 0 and the upper limit will be x.
  4.  These are examples of functions defined by integrals. Each different value of x as the upper limit of integration, will give a unique value (area).
  5. Differentiate the area functions and compare them to the functions from which they came. What do you notice?

Consider the FTC where the integral has an upper limit of x. That is, a function defined by an integral. (The lower limit does not have to be zero.)

\displaystyle A\left( x \right)=\int_{a}^{x}{{f}'\left( t \right)dt}=f\left( x \right)-f\left( a \right)

What is {A}'\left( x \right)?

\displaystyle {A}'\left( x \right)=\frac{d}{dx}\int_{a}^{x}{{f}'\left( t \right)dt}=\frac{d}{dx}\left( f\left( x \right)-f\left( a \right) \right)={f}'\left( x \right)-0

or

\displaystyle \frac{d}{dx}\int_{a}^{x}{{f}'\left( t \right)dt}={f}'\left( x \right)

The is the other part of the FTC which says that the derivative of a function defined by an integral is the integrand, f (x) evaluated at x.

A good way to demonstrate this to your students is to do a simple integral like

\displaystyle \int_{\pi /4}^{x}{\cos \left( t \right)dt}=\sin \left( x \right)-\tfrac{\sqrt{2}}{2},

and then differentiate to show that

\displaystyle \frac{d}{dx}\int_{\pi /4}^{x}{\cos \left( t \right)dt}=\frac{d}{dx}\left( \sin \left( x \right)-\tfrac{\sqrt{2}}{2} \right)=\cos \left( x \right)

This part of the FTC says, roughly, very roughly, says that the derivative of an integral is this integrand. The other part, discussed in the last post, just as roughly, says the integral of the derivative is the antiderivative.

Keep you notes on this activity. In my next post I will use these functions to demonstrate some of the important properties of integrals.

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