This section will show how the concept of the integral can be used to calculate the area of many plane regions. This problem has been important in economics for over 4000 years.
Like all major rivers, the Tigris and Euphrates in Mesopotamia (now part of Iraq) and the Nile in Egypt would occasionally change course as a result of severe floods. Some farmers would gain new land from the river, while others would lose land. Since taxes were often assessed on land area, it became necessary to re-calculate the area of a parcel of land whose boundary might be an irregularly shaped river bank.
Rather later, but still around 360 B.C., the Greek mathematician Eudoxos developed a generalmethod of exhaustionfor determining the areas of irregularly shaped plane regions.
The idea was to exhaust the area by inscribing within it an expanding sequence of polygonal regions, whose area can be calculated exactly by summing the areas of a finite collection of triangles. Provided this sequence does indeed “exhaust” the area by including every point in the limit, we can define theareaof the region as the limit of the increasing sequence of areas of the inscribed polygonal regions.
Eudoxos and Archimedes, amongst others, used the method of exhaustion in order to determine quite accurate approximations to the areas of a number of specific plane regions, especially for a circular disk. (See Example 7.11.1 for an illustration of how this might work.) The method, however, turned out to work only in some special cases, largely because of the algebraic problems encountered. Nearly 1900 years passed after Eudoxos before an exact method could be devised, combining what we now call integration with the new differential calculus due to Newton and Leibniz. Besides allowing areas to be measured with complete accuracy, their ideas have many other applications. Demonstrating the precise logical relationship between differentiation and integration is one of the main achievements of mathematical analysis. It has even been argued that this discovery is the single most important in all of science.
The problem to be considered and solved in this section is illustrated in Fig. 1:How do we compute the areaAunder the graph of a continuous and nonnegative functionf over the interval[a, b]?
y ⫽ f(x)
A ⫽ ? y
a b x
y ⫽ f(x)
A(t) y
a t b x
Figure 1 Figure 2
Lettbe an arbitrary point in [a, b], and letA(t )denote the area under the curvey =f (x) over the interval [a, t], as shown in Fig. 2. Clearly,A(a)=0, because there is no area from a toa. On the other hand, the area in Fig. 1 isA=A(b). It is obvious from Fig. 2 that, because f is always positive,A(t )increases ast increases. Suppose we increaset by a positive amountt. ThenA(t+t )is the area under the curvey =f (x)over the interval [a, t+t]. Hence,A(t+t )−A(t )is the areaAunder the curve over the interval [t, t+t], as shown in Fig. 3.
y ⫽ f(x) y
a t t ⫹Δt b x ΔA
t t ⫹Δt ΔA f(t)
f(t ⫹Δt)
Figure 3 Figure 4
In Fig. 4, the areaAhas been magnified. It cannot be larger than the area of the rectangle with basetand heightf (t+t ), nor smaller than the area of the rectangle with baset and heightf (t ). Hence, for allt >0, one has
f (t ) t≤A(t+t )−A(t )≤f (t+t ) t (∗) Becauset >0, this implies
f (t )≤ A(t+t )−A(t )
t ≤f (t+t ) (∗∗)
Let us consider what happens to(∗∗)ast → 0. The interval [t, t+t] shrinks to the single pointt, and by continuity off, the valuef (t+t )approachesf (t ). The Newton quotient [A(t+t )−A(t )]/t is squeezed betweenf (t )and a quantityf (t+t )that tends tof (t ). This quotient must therefore tend tof (t )in the limit ast →0.
So we arrive at the remarkable conclusion that the functionA(t ), which measures the area under the graph off over the interval [a, t], is differentiable, with derivative given by A(t )=f (t ) for alltin(a, b) (∗∗∗) This proves thatthe derivative of the area functionA(t )is the curve’s “height” function f (t ), and the area function is therefore one of the indefinite integrals of f (t ).1
1 The functionfin the figures is increasing in the interval [t, t+t]. It is easy to see that the same conclusion is obtained whenever the functionfis continuous on the closed interval [t, t+t]. On the left-hand side of(∗), just replacef (t )byf (c), wherecis a minimum point of the continuous functionf in the interval; and on the right-hand side, replacef (t +t )byf (d), wheredis a maximum point off in [t, t+t]. By continuity, bothf (c)andf (d)must tend tof (t )as t→0. So(∗∗∗)holds also for general continuous functionsf.
S E C T I O N 9 . 2 / A R E A A N D D E F I N I T E I N T E G R A L S 301 Let us now usex as the free variable, and suppose thatF (x)is an arbitrary indefinite integral off (x). ThenA(x) = F (x)+C for some constantC. Recall thatA(a) = 0.
Hence, 0=A(a)=F (a)+C, soC= −F (a). Therefore, A(x)=F (x)−F (a) where F (x)=
f (x) dx (1)
SupposeG(x)is another function withG(x)=f (x). ThenG(x)=F (x)+C for some constantC, and soG(x)−G(a)=F (x)+C−(F (a)+C)=F (x)−F (a). This argument tells us that the area we compute using (1) is independent of which indefinite integral off we choose.
E X A M P L E 1 Calculate the area under the parabolaf (x)=x2over the interval [0,1].
Solution: The area in question is the shaded region in Fig. 5. The area is equal toA = F (1)−F (0)whereF (x)is an indefinite integral ofx2. Now,
x2dx= 13x3+C, so we chooseF (x)= 13x3. Thus the required area is
A=F (1)−F (0)= 13 ã13−13ã03 =13
Figure 5 suggests that this answer is reasonable, because the shaded region appears to have roughly 1/3 the area of a square whose side is of length 1.
1
1 y ⫽ x2 y
x
f(x) ⫽ px ⫹ q pb ⫹ q
pa ⫹ q
a b
y
x
Figure 5 Figure 6
The argument leading to (1) is based on rather intuitive considerations. Formally, mathema- ticians choose todefinethe area under the graph of a continuous and nonnegative function fover the interval [a, b] as the numberF (b)−F (a), whereF(x)=f (x). The concept of area that emerges agrees with the usual concept for regions bounded by straight lines. The next example verifies this in a special case.
E X A M P L E 2 Find the areaAunder the straight linef (x)=px+qover the interval [a, b], wherea, b,p, andq are all positive, withb > a.
Solution: The area is shown shaded in Fig. 6. It is equal toF (b)−F (a)whereF (x)is an indefinite integral ofpx+q. Now,
(px+q) dx=12px2+qx+C. The obvious choice of an indefinite integral isF (x)= 12px2+qx, and so
A=F (b)−F (a)=1
2pb2+qb −1
2pa2+qa =12p(b2−a2)+q(b−a) As Fig. 6 shows, the areaAis the sum of a rectangle whose area is(b−a)(pa+q), and a triangle whose area is12p(b−a)2, which you should check gives the same answer.
The Definite Integral
Letf be a continuous function defined in the interval [a, b]. Suppose that the functionF is continuous in [a, b] and has a derivative withF(x)=f (x)for everyxin(a, b). Then the differenceF (b)−F (a)is called thedefinite integraloff over [a, b]. We observed above that this difference does not depend on which of the indefinite integrals off we choose asF. The definite integral off over [a, b] is therefore anumberthat depends only on the functionf and the numbersaandb. We denote this number by
b
a
f (x) dx (2)
This notation makes explicit the functionf (x)we integrate and the interval of integration [a, b]. The numbersaandbare called, respectively, thelowerandupper limit of integra- tion. The variablexis adummyvariablein the sense that it could be replaced by any other variable that does not occur elsewhere in the expression. For instance,
b
a
f (x) dx= b
a
f (y) dy= b
a
f (ξ ) dξ are all equal (toF (b)−F (a)). But do not write anything likey
a f (y) dy, with the same variable as both the upper limit and the dummy variable of integration, because that is meaningless. The differenceF (b)−F (a)is denoted byb
a F (x), or by F (x)b
a. Thus:
D E F I N I T I O N O F T H E D E F I N I T E I N T E G R A L b
a
f (x) dx= b
a
F (x)=F (b)−F (a)
whereF is any indefinite integral off over an interval containing bothaandb.
(3)
E X A M P L E 3 Evaluate (a) 5
2
e2xdx (b) 2
−2
(x−x3−x5) dx Solution:
(a) Because
e2xdx= 12e2x +C, 5
2
e2xdx= 5
2 1
2e2x = 12e10−12e4 =12e4(e6−1).
(b) 2
−2(x−x3−x5) dx= 2
−2(12x2−14x4−16x6)=(2−4−646)−(2−4−646)=0.
After reading the next subsection and realizing that the graph off (x) = x−x3−x5 is symmetric about the origin, you should understand better why the answer must be 0.
Definition (3) does not necessarily requirea < b. However, ifa > bandf (x)is positive throughout the interval [b, a], thenb
a f (x) dxis a negative number.
Note that we have defined the definite integral without necessarily giving it a geometric interpretation as the area under a curve. In fact, depending on the context, it can have different
S E C T I O N 9 . 2 / A R E A A N D D E F I N I T E I N T E G R A L S 303 interpretations. For instance, iff (r)is an income distribution function, thenb
a f (r) dris the proportion of people with income betweenaandb. (See Section 9.4.)
Although the notation for definite and indefinite integrals is similar, the two integrals are entirely different. In fact, b
a f (x) dx denotes a single number, whereas
f (x) dx represents any one of the infinite set of functions all havingf (x)as their derivative.
Area when f(x) Is Negative
Iff (x)≥0 over [a, b], then b
a
f (x) dx is the area below the graph off over [a, b] (4) Iff is defined in [a, b] andf (x)≤0 for allxin [a, b], then the graph off, thex-axis, and the linesx =aandx =bstill enclose an area. This area is−b
a f (x) dx, with a minus sign before the integral because the area of a region must be positive (or zero), whereas the definite integral is negative.
E X A M P L E 4 Figure 7 shows the graph off (x)=ex/3−3. Evaluate the shaded areaAbetween the x-axis and this graph over the interval [0,3 ln 3]. (Note thatf (3 ln 3)=0.)
Solution: Becausef (x)≤0 in the interval [0,3 ln 3], we obtain A= −
3 ln 3 0
ex/3−3 dx= −
3 ln 3 0
(3ex/3−3x)
= −(3eln 3−3ã3 ln 3)+3e0= −9+9 ln 3+3=9 ln 3−6≈3.89
Is the answer reasonable? Yes, because the shaded set in Fig. 7 seems to have an area somewhat less than that of the triangle enclosed by the points(0,0),(0,−2), and(4,0), whose area is 4, and a little more than the area of the inscribed triangle with vertices(0,0), (0,−2), and(3 ln 3,0), whose area is 3 ln 3≈3.30.
1
⫺1
⫺2
1 2 3 4
f(x) ⫽ ex兾3⫺ 3 y
x c1
a
b
c2 c3
y ⫽ f(x) y
x
Figure 7 Figure 8
Suppose the functionf is defined and continuous in [a, b], and that it is positive in some subintervals, negative in others, as shown in Fig. 8. Letc1,c2,c3 denote three roots of the equationf (x) =0—that is, three points where the graph crosses thex-axis. The definite integralb
a f (x)dx is the sum of the two shaded areas above thex-axis, minus the sum of the two shaded areas below thex-axis. The total area bounded by the graph off, the x-axis, and the linesx =a andx =b, on the other hand, is calculated by computing the
positive areas in each subinterval [a, c1], [c1, c2], [c2, c3], and [c3, b] in turn according to the previous definitions, and then adding these areas. Specifically, the shaded area is
− c1
a
f (x) dx+ c2
c1
f (x) dx− c3
c2
f (x) dx+ b
c3
f (x) dx
In fact, this illustrates a general result: the area between the graph of a functionf and the x-axis is given by the definite integralb
a |f (x)|dx of the absolute value of the integrand f (x), which equals the area under the graph of the nonnegative-valued function|f (x)|. P R O B L E M S F O R S E C T I O N 9 . 2
1. Compute the areas under the graphs of (a)f (x)=x3and (b)f (x)=x10over [0,1].
2. Compute the area bounded by the graph of the function over the indicated interval. In (c), sketch the graph and indicate by shading the area in question.
(a) f (x)=3x2 in [0,2] (b) f (x)=x6 in [0,1]
(c) f (x)=ex in [−1,1] (d) f (x)=1/x2 in [1,10]
3. Compute the areaAbounded by the graph off (x)=1/x3, thex-axis, and the two linesx= −2 andx= −1. Make a drawing. (Hint:f (x) <0 in [−2,−1].)
4. Compute the areaAbounded by the graph off (x)= 12(ex+e−x), thex-axis, and the lines x= −1 andx=1.
⊂SM⊃5. Evaluate the following integrals:
(a) 1
0
x dx (b)
2 1
(2x+x2) dx (c) 3
−2
1
2x2−13x3 dx (d)
2 0
(t3−t4) dt (e) 2
1
2t5− 1
t2
dt (f)
3 2
1 t−1+t
dt
⊂SM⊃6. (a) Letf (x)=x(x−1)(x−2).Calculatef(x). Where isf (x)increasing?
(b) Sketch the graph and calculate 1
0
f (x)dx.
7. (a) The profit of a firm as a function of its outputxis given by f (x)=4000−x−3 000 000/x, x >0 Find the level of output that maximizes profit. Sketch the graph off.
(b) The actual output varies between 1000 and 3000 units. Compute the average profitI = 1
2000 3000
1000
f (x) dx.
8. Evaluate the integrals (a)
3 1
3x
10dx (b)
−1
−3
ξ2dξ (c) 1
0
αeβτdτ (β=0) (d) −1
−2
1 ydy
S E C T I O N 9 . 3 / P R O P E R T I E S O F D E F I N I T E I N T E G R A L S 305