Supplement for Math 121, fall 2001
Problem 1. Height/Area
We are making a garden in the shape shown in the figure. In order to find an approximation of the area of the garden we stretch strings to measure the distances across the garden at intervals of ½ yard (or meter) as shown:
a. Find an approximation of the area.
b. Assume that between each successive pair of measurements the distances are between the distances of the two measurements. With this assumption what is the largest and smallest values that the area could be?
c. How could we make the approximation of the area more accurate?
Problem 2. Speed/Distance
We are traveling in a car and the odometer is broken but we record the speed at hour intervals as follows (note that 0.1 hr = 6 min):
| time | 0.0 hr | 0.1 hr | 0.2 hr | 0.3 hr | 0.4 hr | 0.5 hr | 0.6 hr | 0.7 hr |
| speed | 40 mph | 60 mph | 55 mph | 50 mph | 45 mph | 40 mph | 45 mph | 35 mph |
a. Find an approximation for the distance traveled.
b. Assume that between each successive pair of speedometer readings the speed is between the speeds of the two readings. With this assumption what is the largest and smallest values that the distance could be?
c. How could we make the approximation of the distance more accurate?
Further examples are in Section 4.3 and its Exercises: 1-6, 17-20, 21-24.
Introduction
Problem 3. What do we mean by area?
Here is an irregular region. Approximate its area (in cm2) Use any method that you can think of.
What properties of area are you assuming in making your approximation?
How could you make your approximation more accurate?
Axioms for Area
You have a good intuitive understanding of area of many regions in the plane. In particular, you know the following, which we will accept as axioms for the notion of area:
A straight line has no area. 
.
If a region is cut into two parts by a straight line, then the area of the whole region is the sum of the areas of the two parts: 
.
If the 
region R lies completely inside the region S, then the area of R is less than the area of S. 
.
Translations, rotations, reflections do not change areas:
.
Note that none of the above axioms require units. They are totally independent of what units one wants to use; in fact, there is no need to choose any units at all for these four axioms to make sense. Now the fifth axiom requires units.
The area of a rectangle is (length) x (width).
.
Examples.
We can use these properties to show that the area of a parallelogram is:
And: 
.
In a similar way it is possible to find the areas of all polygons. We can find the area of any polygon by dividing it into a finite number of triangles, but what about the area of regions with curved boundaries? For example, what is the area of a circle?
History of Area of a Circle
We will show two methods for showing that the area of a circle is (1/2) x R x C, where R is the radius and C is the circumference. (Check that this is the same as pR2.)
Archimedes' Method
In the first method, imagine that the circle is divided into many, many wedges (like pieces of pie). This method has been attributed to Archimedes:
Now each of these wedges is very close to being a triangle and the more wedges there are, the more they become indistinguishable from true triangles. Now imagine infinitely many wedges, each infinitesimally thin. Each of these wedges is a "triangle" with height equal to the radius R. If we let b be the base of the wedges then area of the circle is
Sum (1/2) R b), summed over all the infinitesimal triangles.
Since they all have the same height we can factor that out, so:
Area of circle = (1/2)R (Sum( b )) = (1/2)R C,
since the sum of the bases of all the triangles is the circumference.
We can use our limit techniques to prove this precisely by first assuming that the circle is divided into n equal wedges.
Now, by looking at the two true triangles, one contained in the wedge and one containing the wedge, we see that:
Then, by summing over all the wedges we get:
.
Then in the limit as n goes to infinity the sum of the cords and the sum of the tangents both converge to the length of the circumference C and thus we can conclude by the squeeze theorem that the area of the circle is ½ R C.
The Talmud Method:
The second method is from an 9th century commentary to the Jewish Talmud:
Imagine the circle consisting of very very thin concentric threads. Now, as indicated in the drawing, glue the threads along the dotted radius and cut them along the solid radius.
Now comb out the threads to form a triangle. (Why is a triangle formed?)
This triangle has the same area as the circle and it has base C and height R and thus the area of the triangle and circle is (1/2) R C .
Ancient Chinese "Comb" Method:
In this method, divide the circle into 2n wedges as in Archimedes' Method and then separate the wedges except along the circumference so that they form two "combs". Then put the two combs together as shown:
As we let n go to infinity, the union of the two combs approaches a rectangle of sides R and (1/2)C.
Issues with Area -- Infinitesimal method that does not work.
But we must be careful with infinitesimal arguments such as those above. For example:
Notice that both triangles is this figure can be considered as made up of a collection of (infinitely many) vertical line segments and that each vertical segment (for example, s, t) in the left triangle has a corresponding vertical segment (s*, t*) of the same length in the right triangle, and vice versa. Thus we can not think of adding up line segments to get area but rather must think of adding up infinitesimal rectangles. However, the above example shows that we must be careful when comparing infinitesimal rectangles. We can think of the vertical segments, s and s*, as infinitesimal rectangles each with infinitesimal area but clearly, if we are going to get the usual notion of area for the triangles, we must think of the infinitesimal rectangle s* as having 3 times the area of the infinitesimal rectangle s. We will return to this when we talk about the change of variable formula later.
Problem 4: Area of the Circle.
Which of the above methods for finding the area of a circle do you like the best? Explain why and fill in any details left in the above descriptions.
Explain why the method you picked avoids the errors described in the last section.
Which regions have well-defined areas?
This is the subject matter of the branch of mathematics that is called "Measure Theory" which we will not go into in this course. If the boundary of the region is irregular enough then the region may not have a well-defined area. But, on the other hand, if the boundary of the region is regular enough then there is a well-defined area which we can approximate as closely as we wish.
There are many non-calculus ways in which people approximate areas. For example, the amount of light that is reflected from shiny foil is proportional the area of the foil. Also, the weight of a uniform quality of paper is proportional to its area. So, if you cut the area out of foil or paper, you can approximate its area accurately by measuring the amount of light reflected or its weight.
A very popular method used by computers is the Monte Carlo Method is which the computer "throws" random "darts" at a rectangle that contains the region whose area is desired. The computer then counts the number of the random darts that land in the region. After throwing a very large number of darts the ratio of (the number of darts that hit to region) to (the total number of darts) is a good approximation to the ratio of (the area of the region) to (the area of the rectangle).
Problem 5: Area Under a Graph of a Monotone Function.
Let f be a function that is non-negative and monotone increasing {that is, c < d implies that 0< f(c)< f(d)} on the interval [a,b] . Let F be any antiderivative of f.
a. Let LHA(D) {or RHA(D) } denote the approximation of the area between the graph of f and the x-axis using rectangles with width D and height determined by the value of f at the left-hand {or right-hand} endpoints. Show that
RHA(D) - LHA(D) = ( f(b) - f(a) ) D .
[Hint: Follow the ideas from P-Problem 8.]
b. Use a and the Squeeze Theorem to argue that the area under the graph of f is equal to the limits of either LHA(D) or RHA(D) as D goes to zero.
[Hint: Again follow your ideas from P-problem 8.]
c. Show that, for any two points, c < d, in [a,b] it is always true that
.
{Hint: Divide thru by (d-c) and apply the Mean Value Theorem to the function F.}
d. Using the above, show that 
equals the limits of the right-hand and left-hand approximations and thus that 
is equal to the area between the graph of f and the x-axis.
[Hint: Use Part c in each rectangle from Parts a and b.]
Definition of Definite Integral
For a < b, we define the definite integral 
to be the signed area under the graph of f(x) from x = a to x = b, where by "signed area" we mean the area above the x-axis is considered positive and the area below the x-axis is considered negative. Define 
.
e. Let f be any function which has only finitely many local maxima and local minima on the interval [a,b]. If F is any antiderivative of f on the interval [a,b] then show that
and thus that
.
Numerical Approximations
We can numerically approximate the definite integral by using rectangles whose heights are determined by left endpoints:
or right endpoints:
or midpoints:
Problem 6: Rules for the Definite Integral
Prove the "Rules for definite integrals" listed on page 347 of Thomas' Calculus (10th edition) using either:
a. the 5 properties of area that we started with,
b. the approximation methods by rectangles, or
c. the Problem 5e.