The Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus or FTC, as its name suggests, is a very important idea. It is not sufficient to present the formula and show students how to use it. Show them where it comes from.

Here is an approach to demonstrate the FTC. I try to sneak up on the result by proposing a problem and then solving it. Here is the outline.

    1.  Suppose we have a differentiable function f that goes from \left( a,f\left( a \right) \right) to \left( b,f\left( b \right) \right). What is the net change in f over this interval? Easy it’s f\left( b \right)-f\left( a \right).  No problem, but way too easy for a calculus class. So let’s try a harder way!

 

    1. Partition the interval [a, b] as you would for a Riemann sum, and calculate the change in f on each subinterval. The subintervals may be the same width or not. The change in y on the general subinterval [xi-1, xi] is f\left( {{x}_{i}} \right)-f\left( {{x}_{i-1}} \right).

 

    1. Approximate the net change over the whole interval by adding these \displaystyle \sum\limits_{i=1}^{n}{\left( f\left( {{x}_{i}} \right)-f\left( {{x}_{i-1}} \right) \right)} .
      Is this a Riemann sum? No it is not! There is no \Delta x part. What to do?

 

    1. The expression f\left( {{x}_{i}} \right)-f\left( {{x}_{i-1}} \right) looks familiar.
      It is part of the equation for the Mean Value Theorem: \displaystyle {f}'\left( c \right)=\frac{f\left( b \right)-f\left( a \right)}{b-a} or {f}'\left( c \right)\left( b-a \right)=f\left( b \right)-f\left( a \right).
      If we adapt this to the subinterval letting ci be the number guaranteed by the MVT on each subinterval  [xi-1, xi], then f\left( {{x}_{i}} \right)-f\left( {{x}_{i-1}} \right)={f}'\left( {{c}_{i}} \right)\left( {{x}_{i}}-{{x}_{i-1}} \right)

 

    1. We can rewrite the sum in step 3 as \displaystyle \sum\limits_{i=1}^{n}{\left( f\left( {{x}_{i}} \right)-f\left( {{x}_{i-1}} \right) \right)}=\sum\limits_{n=1}^{n}{\left( {f}'\left( {{c}_{i}} \right)\left( {{x}_{i}}-{{x}_{i-1}} \right) \right)} .

 

  1. This is a Riemann sum and therefore, \displaystyle \underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{n=1}^{n}{\left( {f}'\left( {{c}_{i}} \right)\left( {{x}_{i}}-{{x}_{i-1}} \right) \right)}=\int_{a}^{b}{{f}'\left( c \right)dx} .
  2. So what is this equal to? We have already found what this limit is in step 1, so we now have:

\displaystyle \int_{a}^{b}{{f}'\left( x \right)dx}=f\left( b \right)-f\left( a \right).

This is called the FTC. And it is important.

The first thing it tells us is that the integral of a rate of change is the (net) amount of change. This will help us do variety of problems.

The second thing it tells us is that, if we can find a function of which the integrand is the derivative (i.e. its antiderivative), then we can find the value of a definite integral by evaluating an antiderivative at the endpoints and subtracting. No more struggling with trying to find the limit of Riemann sums or graphing the function and hoping you can break it into regions with easy to find areas. All we need is an antiderivative and then one quick computation will do the trick from now on.

There is more to the FTC. This will be the subject of the next post.

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