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# Integration, Average Behavior: Area of Regions in the Plane

Module by: Lawrence Baggett. E-mail the author

Summary: This module contains an axiom of choice and also includes various theorems and exercises related to the usage of integrals in finding the area of regions in the plane.

It would be desirable to be able to assign to each subset SS of the Cartesian plane R2R2 a nonnegative real number A(S)A(S) called its area. We would insist based on our intuition that (i) if SS is a rectangle with sides of length LL and WW then the number A(S)A(S) should be LW,LW, so that this abstract notion of area would generalize our intuitively fundamental one. We would also insist that (ii) if SS were the union of two disjoint parts, S=S1S2,S=S1S2, then A(S)A(S) should be A(S1)+A(S2).A(S1)+A(S2). (We were taught in high school plane geometry that the whole is the sum of its parts.) In fact, even if SS were the union of an infinite number of disjoint parts, S=n=1SnS=n=1Sn with SiSj=SiSj= if ij,ij, we would insist that (iii) A(S)=n=1A(Sn).A(S)=n=1A(Sn).

The search for such a definition of area for every subset of R2R2 motivated much of modern mathematics. Whether or not such an assignment exists is intimately related to subtle questions in basic set theory, e.g., the Axiom of Choice and the Continuum Hypothesis. Most mathematical analysts assume that the Axiom of Choice holds, and as a result of that assumption, it has been shown that there can be no assignment SA(S)SA(S) satisfying the above three requirements. Conversely, if one does not assume that the Axiom of Choice holds, then it has also been shown that it is perfectly consistent to assume as a basic axiom that such an assignment SA(S)SA(S) does exist. We will not pursue these subtle points here, leaving them to a course in Set Theory or Measure Theory. However, Here's a statement of the Axiom of Choice, and we invite the reader to think about how reasonable it seems.

AXIOM OF CHOICE Let SS be a collection of sets. Then there exists a set AA that contains exactly one element out of each of the sets SS in S.S.

The difficulty mathematicians encountered in trying to define area turned out to be involved with defining A(S)A(S) for every subset SR2.SR2. To avoid this difficulty, we will restrict our attention here to certain “ reasonable” subsets S.S. Of course, we certainly want these sets to include the rectangles and all other common geometric sets.

Definition 1:

By a (open) rectangle we will mean a set R=(a,b)×(c,d)R=(a,b)×(c,d) in R2.R2. That is, R={(x,y):a<x<bandc<y<d}.R={(x,y):a<x<bandc<y<d}. The analogous definition of a closed rectangle[a,b]×[c,d][a,b]×[c,d] should be clear: [a,b]×[c,d]={(x,y):axb,cyd}.[a,b]×[c,d]={(x,y):axb,cyd}.

By the area of a (open or closed) rectangle R=(a,b)×(c,d)R=(a,b)×(c,d) or [a,b]×[c,d][a,b]×[c,d] we mean the number A(R)=(b-a)(d-c).A(R)=(b-a)(d-c). .

The fundamental notion behind our definition of the area of a set SS is this. If an open rectangle R=(a,b)×(c,d)R=(a,b)×(c,d) is a subset of S,S, then the area A(S)A(S) surely should be greater than or equal to A(R)=(b-a)(d-c).A(R)=(b-a)(d-c). And, if SS contains the disjoint union of several open rectangles, then the area of SS should be greater than or equal to the sum of their areas.

We now specify precisely for which sets we will define the area. Let [a,b][a,b] be a fixed closed bounded interval in RR and let ll and uu be two continuous real-valued functions on [a,b][a,b] for which l(x)<u(x)l(x)<u(x) for all x(a,b).x(a,b).

Definition 2:

Given [a,b],l,[a,b],l, and uu as in the above, let SS be the set of all pairs (x,y)R2,(x,y)R2, for which a<x<ba<x<b and l(x)<y<u(x).l(x)<y<u(x). Then SS is called an open geometric set. If we replace the << signs with signs, i.e., if SS is the set of all (x,y)(x,y) such that axb,axb, and l(x)yu(x),l(x)yu(x), then SS is called a closed geometric set. In either case, we say that SS is bounded on the left and right by the vertical line segments {(a,y):l(a)yu(a)}{(a,y):l(a)yu(a)} and {(b,y):l(b)yu(b)},{(b,y):l(b)yu(b)}, and it is bounded below by the graph of the function ll and bounded above by the graph of the function u.u. We call the union of these four bounding curves the boundary of S,S, and denote it by CS.CS.

If the bounding functions uu and ll of a geometric set SS are smooth or piecewise smooth functions, we will call SS a smooth or piecewise smooth geometric set.

If SS is a closed geometric set, we will indicate the corresponding open geometric set by the symbol S0.S0.

The symbol S0S0 we have introduced for the open geometric set corresponding to a closed one is the same symbol that we have used previously for the interior of a set. Study the exercise that follows to see that the two uses of this notation agree.

## Exercise 1

1. Show that rectangles, triangles, and circles are geometric sets. What in fact is the definition of a circle?
2. Find some examples of sets that are not geometric sets. Think about a horseshoe on its side, or a heart on its side.
3. Let ff be a continuous, nonnegative function on [a,b].[a,b]. Show that the “region” under the graph of ff is a geometric set.
4. Show that the intersection of two geometric sets is a geometric set. Describe the left, right, upper, and lower boundaries of the intersection. Prove that the interior (S1S2)0(S1S2)0 of the intersection of two geometric sets S1S1 and S2S2 coincides with the intersection S10S20S10S20 of their two interiors.
5. Give an example to show that the union of two geometric sets need not be a geometric set.
6. Show that every closed geometric set is compact.
7. Let SS be a closed geometric set. Show that the corresponding open geometric set S0S0 coincides with the interior of S,S, i.e., the set of all points in the interior of S.S. HINT: Suppose a<x<ba<x<b and l(x)<y<u(x).l(x)<y<u(x). Begin by showing that, because both ll and uu are continuous, there must exist an ϵ>0ϵ>0 and a δ>0δ>0 such that a<x-δ<x+δ<ba<x-δ<x+δ<b and l(x)<y-ϵ<y+ϵ<u(x).l(x)<y-ϵ<y+ϵ<u(x).

Now, given a geometric set SS (either open or closed), that is determined by an interval [a,b][a,b] and two bounding functions uu and l,l, let P={x0<x1<...<xn}P={x0<x1<...<xn} be a partition of [a,b].[a,b]. For each 1in,1in, define numbers cici and didi as follows:

c i = sup x i - 1 < x < x i l ( x ) , and d i = inf x i - 1 < x < x i u ( x ) . c i = sup x i - 1 < x < x i l ( x ) , and d i = inf x i - 1 < x < x i u ( x ) .
(1)

Because the functions ll and uu are continuous, they are necessarily bounded, so that the supremum and infimum above are real numbers. For each 1in1in define RiRi to be the open rectangle (xi-1,xi)×(ci,di).(xi-1,xi)×(ci,di). Of course, didi may be <ci,<ci, in which case the rectangle RiRi is the empty set. In any event, we see that the partition PP determines a finite set of (possibly empty) rectangles {Ri},{Ri}, and we denote the union of these rectangles by the symbol CP.=i=1n(xi-1,xi)×(ci,di).CP.=i=1n(xi-1,xi)×(ci,di).

The area of the rectangle RiRi is (xi-xi-1)(di-ci)(xi-xi-1)(di-ci) if ci<dici<di and 0 otherwise. We may write in general that A(Ri)=(xi-xi-1)max((di-ci),0).A(Ri)=(xi-xi-1)max((di-ci),0). Define the number APAP by

A P = i = 1 n ( x i - x i - 1 ) ( d i - c i ) . A P = i = 1 n ( x i - x i - 1 ) ( d i - c i ) .
(2)

Note that APAP is not exactly the sum of the areas of the rectangles determined by PP because it may happen that di<cidi<ci for some ii's, so that those terms in the sum would be negative. In any case, it is clear that APAP is less than or equal to the sum of the areas of the rectangles, and this notation simplifies matters later.

For any partition P,P, we have SCP,SCP, so that, if A(S)A(S) is to denote the area of S,S, we want to have

A ( S ) i = 1 n A ( R i ) = i = 1 n ( x i - x i - 1 ) max ( ( d i - c i ) , 0 ) i = 1 n ( x i - x i - 1 ) ( d i - c i ) = A P . A ( S ) i = 1 n A ( R i ) = i = 1 n ( x i - x i - 1 ) max ( ( d i - c i ) , 0 ) i = 1 n ( x i - x i - 1 ) ( d i - c i ) = A P .
(3)
Definition 3:

Let SS be a geometric set (either open or closed), bounded on the left by x=a,x=a, on the right by x=b,x=b, below by the graph of l,l, and above by the graph of u.u. Define the areaA(S)A(S) of SS by

A ( S ) = sup P A P = sup P = { x 0 < x 1 < ... < x n } i = 1 n ( x i - x i - 1 ) ( d i - c i ) , A ( S ) = sup P A P = sup P = { x 0 < x 1 < ... < x n } i = 1 n ( x i - x i - 1 ) ( d i - c i ) ,
(4)

where the supremum is taken over all partitions PP of [a,b],[a,b], and where the numbers cici and didi are as defined above.

## Exercise 2

1. Using the notation of the preceding paragraphs, show that each rectangle RiRi is a subset of the set SS and that RiRj=RiRj= if ij.ij. It may help to draw a picture of the set SS and the rectangles {Ri}.{Ri}. Can you draw one so that di<ci?di<ci?
2. Suppose S1S1 is a geometric set and that S2S2 is another geometric set that is contained in S1.S1. Prove that A(S2)A(S1).A(S2)A(S1). HINT: For each partition P,P, compare the two APAP's.

## Exercise 3

Let TT be the triangle in the plane with vertices at the three points (0,0),(0,H),(0,0),(0,H), and (B,0).(B,0). Show that the area A(T),A(T), as defined above, agrees with the formula A=(1/2)BH,A=(1/2)BH, where BB is the base and HH is the height.

The next theorem gives the connection between area (geometry) and integration (analysis). In fact, this theorem is what most calculus students think integration is all about.

## Theorem 1

Let SS be a geometric set, i.e., a subset of R2R2 that is determined in the above manner by a closed bounded interval [a,b][a,b] and two bounding functions ll and u.u. Then

A ( S ) = a b ( u ( x ) - l ( x ) ) d x . A ( S ) = a b ( u ( x ) - l ( x ) ) d x .
(5)

### Proof

Let P={x0<x1<...<xn}P={x0<x1<...<xn} be a partition of [a,b],[a,b], and let cici and didi be defined as above. Let hh be a step function that equals didi on the open interval (xi-1,xi),(xi-1,xi), and let kk be a step function that equals cici on the open interval (xi-1,xi).(xi-1,xi). Then on each open interval (xi-1,xi)(xi-1,xi) we have h(x)u(x)h(x)u(x) and k(x)l(x).k(x)l(x). Complete the definitions of hh and kk by defining them at the partition points so that h(xi)=k(xi)h(xi)=k(xi) for all i.i. Then we have that h(x)-k(x)u(x)-l(x)h(x)-k(x)u(x)-l(x) for all x[a,b].x[a,b]. Hence,

A P = i = 1 n ( x i - x i - 1 ) ( d i - c i ) = a b ( h - k ) a b ( u - l ) . A P = i = 1 n ( x i - x i - 1 ) ( d i - c i ) = a b ( h - k ) a b ( u - l ) .
(6)

Since this is true for every partition PP of [a,b],[a,b], it follows by taking the supremum over all partitions PP that

A ( S ) = sup P A P a b ( u ( x ) - l ( x ) ) d x , A ( S ) = sup P A P a b ( u ( x ) - l ( x ) ) d x ,
(7)

which proves half of the theorem; i.e., that A(S)abu-l.A(S)abu-l.

To see the other inequality, let hh be any step function on [a,b][a,b] for which h(x)u(x)h(x)u(x) for all x,x, and let kk be any step function for which k(x)l(x)k(x)l(x) for all x.x. Let P={x0<x1<...<xn}P={x0<x1<...<xn} be a partition of [a,b][a,b] for which both hh and kk are constant on the open subintervals (xi-1,xi)(xi-1,xi) of P.P. Let a1,a2,...,ana1,a2,...,an and b1,b2,...,bnb1,b2,...,bn be the numbers such that h(x)=aih(x)=ai on (xi-1,xi)(xi-1,xi) and k(x)=bik(x)=bi on (xi-1,xi).(xi-1,xi). It follows, since h(x)u(x)h(x)u(x) for all x,x, that aidi.aidi. Also, it follows that bici.bici. Therefore,

a b ( h - k ) = i = 1 n ( a i - b i ) ( x i - x i - 1 ) i = 1 n ( x i - x i - 1 ) ( d i - c i ) = A P A ( S ) . a b ( h - k ) = i = 1 n ( a i - b i ) ( x i - x i - 1 ) i = 1 n ( x i - x i - 1 ) ( d i - c i ) = A P A ( S ) .
(8)

Finally, let {hm}{hm} be a nondecreasing sequence of step functions that converges uniformly to u,u, and let {km}{km} be a nonincreasing sequence of step functions that converges uniformly to l.l. See part (d) of (Reference). Then

a b ( u - l ) = lim m a b ( h m - k m ) A ( S ) , a b ( u - l ) = lim m a b ( h m - k m ) A ( S ) ,
(9)

which proves the other half of the theorem.

OK! Trumpet fanfares, please!

## Theorem 2

(A=πr2.A=πr2.) If SS is a circle in the plane having radius r,r, then the area A(S)A(S) of SS is πr2.πr2.

### Proof

Suppose the center of the circle SS is the point (h,k).(h,k). This circle is a geometric set. In fact, we may describe the circle with center (h,k)(h,k) and radius rr as the subset SS of R2R2 determined by the closed bounded interval [h-r,h+r][h-r,h+r] and the functions

u ( x ) = k + r 2 - ( x - h ) 2 u ( x ) = k + r 2 - ( x - h ) 2
(10)

and

l ( x ) = k - r 2 - ( x - h ) 2 . l ( x ) = k - r 2 - ( x - h ) 2 .
(11)

By the preceding theorem, we then have that

A ( S ) = h - r h + r 2 r 2 - ( x - h ) 2 d x = π r 2 . A ( S ) = h - r h + r 2 r 2 - ( x - h ) 2 d x = π r 2 .
(12)

We leave the verification of the last equality to the following exercise.

## Exercise 4

Evaluate the integral in the above proof:

h - r h + r 2 r 2 - ( x - h ) 2 d x . h - r h + r 2 r 2 - ( x - h ) 2 d x .
(13)

Be careful to explain each step by referring to theorems and exercises in this book. It may seem like an elementary calculus exercise, but we are justifying each step here.

REMARK There is another formula for the area of a geometric set that is sometimes very useful. This formula gives the area in terms of a “double integral.” There is really nothing new to this formula; it simply makes use of the fact that the number (length) u(x)-l(x)u(x)-l(x) can be represented as the integral from l(x)l(x) to u(x)u(x) of the constant 1. Here's the formula:

A ( S ) = a b ( l ( x ) u ( x ) 1 d y ) d x . A ( S ) = a b ( l ( x ) u ( x ) 1 d y ) d x .
(14)

The next theorem is a result that justifies our definition of area by verifying that the whole is equal to the sum of its parts, something that any good definition of area should satisfy.

## Theorem 3

Let SS be a closed geometric set, and suppose S=i=1nSi,S=i=1nSi, where the sets {Si}{Si} are closed geometric sets for which Si0Sj0=Si0Sj0= if ij.ij. Then

A ( S ) = i = 1 n A ( S i ) . A ( S ) = i = 1 n A ( S i ) .
(15)

### Proof

Suppose SS is determined by the interval [a,b][a,b] and the two bounding functions ll and u,u, and suppose SiSi is determined by the interval [ai,bi][ai,bi] and the two bounding functions lili and ui.ui. Because SiS,SiS, it must be that the interval [ai,bi][ai,bi] is contained in the interval [a,b].[a,b]. Initially, the bounding functions lili and uiui are defined and continuous on [ai,bi],[ai,bi], and we extend their domain to all of [a,b][a,b] by defining li(x)=ui(x)=0li(x)=ui(x)=0 for all x[a,b]x[a,b] that are not in [ai,bi].[ai,bi]. The extended functions lili and uiui may not be continuous on all of [a,b],[a,b], but they are still integrable on [a,b].[a,b]. (Why?) Notice that we now have the formula

A ( S i ) = a i b i ( u i ( x ) - l i ( x ) ) d x = a b ( u i ( x ) - l i ( x ) ) d x . A ( S i ) = a i b i ( u i ( x ) - l i ( x ) ) d x = a b ( u i ( x ) - l i ( x ) ) d x .
(16)

Next, fix an xx in the open interval (a,b).(a,b). We must have that the vertical intervals (li(x),ui(x))(li(x),ui(x)) and (lj(x),uj(x))(lj(x),uj(x)) are disjoint if ij.ij. Otherwise, there would exist a point yy in both intervals, and this would mean that the point (x,y)(x,y) would belong to both Si0Si0 and Sj0,Sj0, which is impossible by hypothesis. Therefore, for each x(a,b),x(a,b), the intervals {(li()x),ui(x))}{(li()x),ui(x))} are pairwise disjoint open intervals, and they are all contained in the interval (l(x),u(x)),(l(x),u(x)), because the SiSi's are subsets of S.S. Hence, the sum of the lengths of the open intervals {(li(x),ui(x))}{(li(x),ui(x))} is less than or equal to the length of (l(x),u(x)).(l(x),u(x)). Also, for any point yy in the closed interval [l(x),u(x)],[l(x),u(x)], the point (x,y)(x,y) must belong to one of the SiSi's, implying that yy is in the closed interval [li(x),ui(x)][li(x),ui(x)] for some i.i. But this means that the sum of the lengths of the closed intervals [li(x),ui(x)][li(x),ui(x)] is greater than or equal to the length of the interval [l(x),u(x)].[l(x),u(x)]. Since open intervals and closed intervals have the same length, we then see that (u(x)-l(x)=i=1n(ui(x)-li(x)).(u(x)-l(x)=i=1n(ui(x)-li(x)).

We now have the following calculation:

i = 1 n A ( S i ) = i = 1 n a i b i ( u i ( x ) - l i ( x ) ) d x = i = 1 n a b ( u i ( x ) - l i ( x ) ) d x = a b i = 1 n ( u i ( x ) - l i ( x ) ) d x = a b ( u ( x ) - l ( x ) ) d x = A ( S ) , i = 1 n A ( S i ) = i = 1 n a i b i ( u i ( x ) - l i ( x ) ) d x = i = 1 n a b ( u i ( x ) - l i ( x ) ) d x = a b i = 1 n ( u i ( x ) - l i ( x ) ) d x = a b ( u ( x ) - l ( x ) ) d x = A ( S ) ,
(17)

which completes the proof.

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