Y the FTC?

So, you’ve finally proven the Fundamental Theorem of Calculus and have written on the board:

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

And the students ask, “What good is that?” and “When are we ever going to use that?” Here’s your answer.

There are two very important uses of this theorem. Show them BOTH uses right away to help your students see why the FTC is so useful and important.

First, in words the theorem says that “the integral of a rate of change is the net amount of change.” So, if you are given a rate of change (as you are every year on the AP Calculus exam) and asked to find the amount of change (as you are every year on the AP Calculus exam), this is what you use, Show an example such as 2015 AB 1/BC1 that states,

“The rate at which rainwater flows into a drain pipe is modeled by the function R, where R\left( t \right)=20\sin \left( {\frac{{{{t}^{2}}}}{{35}}} \right) cubic feet per hour….

“(a) How many cubic feet of rainwater flow into the pipe during the 8-hour time interval ?”

The answer is of course, \displaystyle \int_{0}^{8}{{20\sin \left( {\frac{{{{t}^{2}}}}{{35}}} \right)dt}}. (Which they will soon learn how to evaluate.)

Second, a more immediate use is to avoid all that work you’ve been doing setting up Riemann sums and finding their limits. No more of that! Give them this integral to evaluate:

\displaystyle \int_{2}^{7}{{2xdx}}

Draw the trapezoid representing the area between the graph of y=2x and the x-axis on the interval [2,7] and find its area =  \displaystyle \frac{1}{2}\left( 5 \right)\left( {18+4} \right)=45

Then ask, “Does anyone know of a function whose derivative is 2x?” Let them think for a minute and someone will say, “Yeah, it’s {{x}^{2}}”  And then show them

\displaystyle \int_{2}^{7}{{2xdx}}={{7}^{2}}-{{2}^{2}}=45

Then go for a harder one:  \displaystyle \int_{0}^{{\frac{\pi }{2}}}{{\cos \left( x \right)dx}}

“Does anyone know a function whose derivative is \cos \left( x \right)?”

“Why yes, it’s \sin \left( x \right)

So, \displaystyle \int_{0}^{{\frac{\pi }{2}}}{{\cos \left( x \right)dx}}=\sin \left( {\frac{\pi }{2}} \right)-\sin \left( 0 \right)=1-0=1

That was easy!

If you want to challenge them and review some functions of the “special angles” try this one:

\displaystyle \int_{{\frac{\pi }{6}}}^{{\frac{{4\pi }}{3}}}{{\cos \left( x \right)dx}}=\sin \left( {\frac{{4\pi }}{3}} \right)-\sin \left( {\frac{\pi }{6}} \right)=\frac{{\sqrt{3}}}{2}-\frac{1}{2}

Tie the two parts together: Look at the graph of y=\sin \left( x \right). How much does it change from 0 to \frac{\pi }{2}? How much does it change from \frac{\pi }{6} to \frac{{4\pi }}{3}?

Sum up, by looking ahead:

  1. “The function whose derivative is …” is called the antiderivative.
  2. Using antiderivatives to evaluate definite integrals is easy; the hard part is finding the antiderivatives, since they are not all as straightforward as the two examples above. So, next we need to spend a few weeks learning how to find antiderivatives.[1]
  3. Given a derivative, finding its antiderivative is also the start of solving differential equations. This, too, will soon be a concern in the course.

[1] As I’ve written before, this is where it seems logical place to teach antiderivatives. Now students have a reason to find them. Teaching antidifferentiation after differentiation, before integration, seems like an intellectual exercise. Sure, it’s fun, but now we have a need for it.


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