by:

Heather Bridges

Note: The class needs to be familiar with the expanding of expressions
of the form ( a+b)^n using the binomial theorem.

Isaac Newton developed a way of estimation the area under a curve back in
the late 1600's. His ideas were discovered many years later in De Analysi.

Newton found that if y = a x ^ (m/n) was the form of the curve then

Have the students to check to see if this works using a familiar curve
lik y = x. If we put y = x in the form as stated above then a = 1, m=1,
and n=1. So after plugging in the values into Newtons formula we see that
the area under the curve is A = 1/2 x^2.

The students should then use another way to find that same area using the
formula for a triangle. The will discover that the area is still the same
so Newton's formula passes their test.

Newton then went on to add that if the curve is made up of several terms
likewise the area will be made up of several terms using the same methods
as above.

The students could us a few examples of this like given the curve

the area under the curve is .

Now , the whole idea is for the students to use the binomial theorem and
Newton's formula for estimation the area under the curve to develop the
approximation for pi.

Prompt the students to name all the things they know about the semicircle
given only the information above. The should be able to come up with the
radius, the center, the equation, and the area. Divide the students into
two groups. Half of the students work to find the area of ABD by Newton's
ideas and the other using straight geometry while the teacher provides assistance.

Area by Newton:

Beginning with the equation for the semicircle, adjust it so it is in the
form of Newton's or in other words solve for y. The students should then
use the binomial theorem to expand the equation. Then using Newton's method
they can find a formula for the area under the curve without substituting
in a value for x. Then since we know that x = .25 they can plug in the
value using a spreadsheet software to find that the Area is approximately
equal to .0767.

Area by geometry:

The first step is to just focus on the triangle ADC. The students can use
the pythagorean theorem to solve for the length of AD. Then they can find
the area of ADC. The tricky part is for the students to pick up on the
fact that the portion ABC of the semicircle is one third of it. Using this
they can discover that the area = 1/24 *pi. THen they can subtract the
area of triangle ADC from the portion ABC to get the area of the portion
ABD.

Then come back together to use all the students information

Finally we can solve this to find that pi = 3.142.

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