π, Exhaustion, Method of

When you study calculus they usually do differentiation and then integration. Historically the order was different.

In school you learned what the number π stands for and you probably memorized its value to a few digits.  But could you figure out what it was if you had to?

The story of how people figured out the value of π is the story of integration.  We can start in ancient Egypt where an approximate value was got my measuring circles. It was a Greek guy, Eudoxus of Cnidus, who starting thinking about how to figure it out exactly. He developed the Method of Exhaustion to find the areas of various shapes and another Greek dude, Archimedes,  used the Method of Exhaustion to nail down the value of pie (oops, π) to a small range:

 3 ¹⁰/₇₁ < π < 3 ¹/₇.

Square-Circle-Square
(Yes it's a circle)

To understand how he did it imagine a circle with radius r. Draw a square outside it with sides of length 2r. Now draw another square  inside whose diagonal has length 2r. The sides of this square have length 2r/√2.
The area of the outer square, 4r², is obviously larger than that of the circle. The area of the inner square, 2r², is obviously smaller than the circle'In t. Therefore,

2 < π < 4.

We can narrow down the range by doing the same thing with hexagons, octagons and regular polygons with more and more sides:

Archimedes continued this until he got to a 96-sided polygon and that's how he computed the result above.  After that he was exhausted.

In the second session of Unit 4 we go through a similar argument to show that the displacement for something that's constantly accelerating can be got from the area under the velocity vs time graph just like you can do for a constant velocity. The same argument can then be used for any shape of v(t) curve.

Most of the time we just say that it's obvious. Obvious is the most dangerous word in the English language. Mathematicians teach us to be careful and to prove things using small self-evident steps so that we can be sure.

Look at these two graphs.  

You might agree that the average velocity is given by

 in both cases.  But is that true for both (a) and (b)? 

Go back to the original definition of <v>. You'll find it's only true for one case.