The Integral/chapter 13

 

1. Area on the geoboard

I start young people working on a geoboard. The geoboard I use is 1/2" plywood with a 5 by 5 square array of nails, about 2" apart with about 1" edge. I also have "spotty" paper, with dots the same distance apart as the geoboard so that my younger students can draw the shapes. I also have 1/2" spotty paper for the older students.
If it takes 1 can of paint to paint inside the square (figure A), how many cans of paint does it take to fill in figures B, C, and D?
There are many investigations one can get into: Make some shapes and find how many cans of paint it takes to paint inside them (in other words, find the area within each figure).
How many segments can you make on this 5x5 geoboard?
How many squares can you make on this 5x5 geoboard (no diagonal sides)?
What is the area of each triangle below?
                                              

You make up questions.

2. See Katy's work finding the area under a sine wave from 0 to p  (In his books, Don finds the area of a parabolic segment to find the area under a parabola, after Archimedes)

3. See Don's program to plot points on a calculator and find the area under curves

4. Finding the area under curves on 1/10" graph paper- and  seeing patterns to learn new things!

On August 28, 2001 Don started working with Grace, a 6th grader in Chicago, in his Math By Mail/Email program. On September 22, 2001, Grace's parents brought Grace and her two brothers to Champaign to work with Don. On Sunday, Dec. 2, 2001, Grace's parents brought the 3 children back to Champaign to work with Don! Grace has continued to work with Don through February, 2002.

                                                      

In November 2001 Grace started work on on the area under curves on p. 245 in Don's worksheet book. He suggested she count the squares under the parabola y = x2  from x = 0 to x = 1, (on the left above), then find the ratio of the number of squares under the curve to the number of squares in the 1x1 square. Grace found the number of squares to be about 34, so the area under the curve was 34/100. When Don asked her what simple fraction was this close to, she said 1/3. Don had Grace write "the area under the curve y = x2  from x = 0 to x = 1" as A0-1 x2 = . It would be easy to write 1 for 1*12 or 13 ,  but the square is 1 by 12  = 13 and writing it this way will show the pattern that follows. In his college math classes, Don was often confused by the NOTATION used, which kept him from really understanding the mathematical ideas.

Grace counted the squares for the area from x = 0 to x = 2, above right. She found the area to be about 1/3 of the total area (2x4). So the area under the curve y = x2 from x = 0 to x = 2 is A0-2  x2 = . Looking at what she had up to that point,


 Don asked Grace what she would get for A0-3  x2  ? She immediately said it would be

  and for the area under the curve  y = x2  from x = 0 to x = x,  she predicted correctly, . WOW!! The power of patterns! 

Then Don asked Grace to draw the line y = x and tell him what the area under it would be from x=0 to x= 1, then 2, then generalize to 0 to x. She sent Don the graph below as an attached file:

And she was able to generalize the area under the "curve"  y = x from x=0 to x=x, as  . Looking at what she had, she was able to generalize for any function xn  , the area under the curve from x=0 to x=x would be  .  Using this, Don asked Grace to do the problems on p. 246 of his worksheet book. A few days later Grace sent Don the following work as an attached file:

There were mistakes above, in #1, 2, 6, 11 and 12. Don sent the following to Grace to show her that the area under y = 4 is  A0-x  4 = 4x

Don also sent the work below:

They talked about the problems she had wrong on p.246, that of finding the area under a constant like k, which is kx0. So A0-x kx0= kx.

Grace sent her corrected versions of the problems on p. 246:

On Sunday, Dec.2, '01 when the family returned to Champaign, Don gave Grace this test to work on:

Grace finished her test at home and sent these answers to Don:

Grace is one of only a few students over the years who were able to do #4 & #5 above. Can you figure out why one needs to subtract 1 in the denominator?

Dec. 26, 2001 Grace sent the following scan to Don

which is what they did at Don's house. This shows how they got the answers to problem #1 (8/3) and #2 above (8/3 + 1/3= 9/3).

Don showed Grace the following was his notation and the "normal" calculus notation where dx is the width of the rectangles shown below:

In March 2002, see what Grace writes (unsolicited) about teaching !

As a teacher I need to try new things, do mathematics myself, and look for ways to get my students into more difficult concepts, but at their level. I treat each student as an individual; I do different things with different students. And I don't wait until I completely understand everything about an idea (which I don't think is possible anyway) before I'll get a student doing it-- that way I learn things along with my students. It's also why I encourage them to do things different ways. It makes teaching and learning enjoyable.

5. Area under curves using rectangles

Now comes a new problem, that is, the curve is the same, but the method of finding the area under it, is different. Find the area under the curve y = x2 (the dark area below), from x = 0 to 1. This time we'll make rectangles (hatched), then find the area of these, get a sequence of partial sums as we make the width of the rectangles smaller, then we'll find the limit of this sequence, which will be the area we want. The area of the hatched rectangles will approach the area under the curve. We'll start with a1  and a2  below. Notice that the height of the rectangles are square numbers because the curve is y = x2. The hatched rectangles will have an area of the height times the width.

   a1 = 0                                     

 

                                              

      

      a1 = 0 

      

    

    

    

    

     

In the Vol of Pyramid/Vol of Cube  Don shows how he finds the sum of the first n squares;

the sum of the first n-1 squares is 

The area under the curve y=x2 , from 0 to 1 is

  

because as n-> inf both the terms 1/2n and 1/6n2 ->0. The 1/3 is the same as Grace got

above. Again writing this as                 is important because it shows

 that the area is 1/3 of a 1x12 rectangle, which is what Sean, Matt, and Grace got.

 

Could you find the area under the curve y=x2 , from 0 to 2, using the rectangles under the curve? Find a1, a2, a3, and a4, using the diagrams below.  a3 is given below the diagrams.

                                                      

Generalize to an, then find the limit of an as n-> infinity.

 

This becomes  

 

Can you now find the area under y=x2 from 0-3, 0-4, 0-x?

Now find the area, using rectangles, under the curves  x0, x1 , x3 . x4 , .. xn  

From what we've done so far, what do you think would be the area under the curve xn  from x=a to x=b?

Just the difference between the areas under the curve

         

           

6. Two problems:

    a. Find the area under x2 from 0 to 1.

    b. Find the ratio of the volume of a pyramid / the volume of a cube with the same base and height

What do you notice about the answers above?

7. a. Logarithms, and 

    b. logs as area under the curves 1/x and 1/(1+x), and as infinite series

At age 11, Ian realized that he could go down, starting with y= x4 , taking the derivative until he got 0, or go up, taking the integral! He used the following notation to show this:

 y -2' = x6/30

 y -1' = x5/5

he started here y 0' = x4

 y 1' = 4x3

 y 2' = 12x2     Ian's notation for the 2nd derivative was y 2' ,                             normally written as y"

 y 3' = 24x1

 y 4' = 24

 y 5' = 0  

Jonathan was Don's student years ago and has made Don's A Map To Calculus ' clickable. See Jonathan's blog at

 http://www.technicalmisery.com/2011/02/cylinder-fun.html

for finding the volume of a cylinder.