第4.1节 定积分

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4.1THE DEFINITE INTEGRAL

Weshall begin our study of the integral calculus in the same way inwhich we began with the differential calculus - by asking a questionabout curves in the plane.

Supposef is a real function continuous on an interval I andconsider the curve y=f(x). Let a<b where a,bare two points in I, and let the curve be above the x-axisfor x between a and b; that is, f(x)≥0. We then ask: What is meant by the area of the region bounded bythe curve y= f(x), the x-axis, and the lines x=aand x=b? That is, what is meant by the area of the shadedregion in Figure 4.1.1? We call this region the region under thecurve y=f(x) between a and b.

Figure4.1.1 The Region under a Curve

Thesimplest possible case is where f is a constant function; thatis, the curve is a horizontal line f(x)=k, where k is aconstant and k≥ 0, shown in Figure 4.1.2. In this case theregion under the curve is just a rectangle with height k andwidth b- a, so the area is defined as

Area=k. (b-a).

Theareas of certain other simple regions, such as triangles, trapezoidsand semicircles, are given by formulas from plane geometry.

Figure4.1.2

Thearea under any continuous curve y=f(x) will be given by thedefinite integral, which is written

Beforeplunging into the detailed definition of the integral, we outline themain ideas.

First,the region under the curve is divided into infinitely many verticalstrips of infinitesimal width dx. Next, each vertical strip isreplaced by a vertical rectangle of height f(x), base dx,and area f(x) dx. The next step is to form the sum of theareas of all these rectangles, called the infinite Riemann sum (lookahead to Figures 4.1.3 and 4.1.11). Finally, the integral ∫ba f(x)dx is defined as the standard part of theinfinite Riemann sum.

Theinfinite Riemann sum, being a sum of rectangles, has an infinitesimalerror. This error is removed by taking the standard part to form theintegral.

Itis often difficult to compute an infinite Riemann sum, since it is asum of infinitely many infinitesimal rectangles. We shall first studyfinite Riemann sums, which can easily be computed on a handcalculator.

Supposewe slice the region under the curve between a and binto thin vertical strips of equal width. If there are nslices, each slice will have width Δx=(b -a)/n. The interval[ a, b] will be partitioned into n subintervals

where

Thepoints x0,x1,……xn are called partitionpoints. On each subinterval [ xk-1, xk], we formthe rectangle of height f( xk-1). The kthrectangle will have area

f(xk-1)·Δx

FromFigure 4.1.3, we can see that the sum of the areas of all theserectangles will be fairly close to the area under the curve. This sumis called a Riemann sum and is equal to

f(x0) Δx + f( x1)Δx +… f(xn-1)Δx.

Itis the area of the shaded region in the picture. A convenient way ofwriting Riemann sums is the “∑-notation” { ∑ is the capitalGreek letter sigma},

Figure4.1.3 the Riemann Sum

Thea and b indicate that the first subinterval begins at aand the last subinterval ends at b.

Wecan carry out the same process even when the subinterval length Δxdoes not divide evenly into the interval length b-a. But then,as Figure 4.1.4 shows, there will be a remainder left over at the endof the interval [a,b], and the Riemann sum will have an extrarectangle whose width is this remainder. We let n be thelargest integer such that

a+ n Δx≤ b,

andwe consider the subintervals

[x0,x1],…，[xn-1,xn], [xn, b],

wherethe partition points are

x0=a, x1 =a+Δx,x2 =a+2Δx,……，xn=a+n Δx,b.

Figure4.1.4

xnwill be less than or equal to b but xn +Δx willbe greater than b. Then we define the Riemann sum to be thesum

____________________________

Thusgiven the function f, the interval [a,b], and the realnumber Δx> 0, we have defined the Riemann sum _____f(x) Δx. We repeat the definition more concisely.

DEFINITION

Leta<b and let Δx be positive real number. Then theRiemann sum ___ is defined as the sum

Wheren is the largest integer such that a + n Δx ≤b, and

arethe partition points.

Ifxn=b, the last term f(xn) (b -xn) iszero. The Riemann sum___f(x)Δxis a real function of three variables a, b, and Δx,

Thesymbol x which appears in the expression is called a dummyvariable (or bound variable), because the value of ______f(x) Δxdoes not depend on x. The dummy variable allows us to use morecompact notation,writing f(x)Δxjust once instead of writing f(x0)Δx, f(x1) Δx,f(x2)Δx,and so on.

FromFigure 4.1.5 it is plausible that by making Δxsmaller we can get the Riemann sum as close to the area as we wish.

Figure4.1.5

EXAMPLE1 Let f(x)=___. InFigure 4.1.6, the region under the curve from x = 0

tox=2 is a triangle with base 2 and height 1, so its area should be

Figure4.1.6

Letuscompare this value for the area with some Riemann sums. In figure4.1.7, we take Δx=_____. The interval [0, 2] divides intofour subintervals ________, ___and ____. We make a table ofvalues of f(x) at the lower endpoints.

Figure4.1.7

TheRiemannsum is then

InFigure 4.1.8, we take Δx=_____. The table of valuesis as follows.

TheRiemann sum is

Wesee that the value is getting closer to one.

Finally,let us take a value of Δxthat does not divide evenly into the interval length 2. Let Δx. We see in Figure 4.1.9 that the interval then divides into threesubintervals of length 0.6 and one of length 0.2, namely [0, 0.6],[0.6, 1.2], [1.2, 1.8], [1.8, 2.0],

Figure4.1.8

TheRiemann sum is

Example2 Let f(x) =_______,defined on the closed interval I=[-1,1]. The region under thecurve is a semicircle of radius 1. We know from plane geometry thatthe area is π/2, or approximately 3.14/2=1.57. Let us compute thevalues of some Riemann sums for this function to see how close theyare to 1.57. First take Δx=____as in Figure 4.1.10(a). We make a table of values.

TheRiemann sum is then

Nextwe take Δx=____.Then the interval [-1, 1] is divided into ten subintervals as inFigure 4.1.10 (b). Our table of values is as follows.

Figure4.1.10

TheRiemann sum is

Thuswe are getting closer to the actual area π/2 ~ 1.57.

Bytaking Δxsmall we can get the Riemann sum to be as close to the area as wewish.

Ournext step is to take Δxto be infinitely small and have an infinite Riemann sum. How can wedo this? We observe that if the real numbers a and bare held fixed, then the Riemann sum

isa real function of the single variable Δx.(The symbol x which appears in the expression is a dummyvariable, and the value of

dependsonly on Δxand not on x.) Furthermore, the term

isdefined for all real Δx>0. Therefore by the Transfer Principle,

isdefined for all hyperreal dx>0. When dx>0 isinfinitesimal, there are infinitely many subintervals of length dx,and we call

aninfinite Riemann sum(Figure 4.1.11).

Figure4.1.11 Infinite Riemann Sum

Wemay think intuitively of the Riemann sum

asthe infinite sum

whereH is the greatest hyperinteger such that a +H dx≤b, (Hyperintegers are discussed in Section 3.8.) H is positiveinfinite, and there are H +2 partition points x0,x1,…，xH,b. A typical term in this sum is the infinitely small quantityf(xK) dx where K is hyperinteger, 0 ≤K<H , and xK = a + K dx.

Theinfinite Riemann sum is a hyperreal number. We would next like totake the standard part of it. But first we must show that it is afinite hyperreal number and thus has a standard part.

THEOREM1

Letf be a continuous function on an interval I, let a<b be two pointsin I, and let dx be a positive infinitesimal. Then the infiniteRiemann sum

isa finite hyperreal number.

PROOFLet B be a real number greater than the maximum value of f on[a,b].

Considerfirsta real number Δx>0. We can see from figure 4.1.12 that the

Figure4.1.12

FiniteRiemannsum is less than the rectangular area B.(b-a);

Thereforeby the Transfer Principle,

Ina similar way we let C be less than the minimum of f on[a, b] and show that

Thusthe Riemann sum ________f(x)dx isfinite.

Weare now ready to define the central concept of this chapter, thedefinite integral. Recall that the derivative was defined as thestandard part of the quotient Δy/Δxand was written dy/dx. The “definite integral”will be defined as the standard part of the infinite Riemann sum

andis written ___ f(x) dx. Thus theΔxis changed to dx in analogy with our differential notation.The ___ is changed to the long thin S,i.e., ∫, to remind us that the integral is obtained from aninfinite sum. We now state the definition carefully.

DEFINITION

Letf be a continuous function on an interval I and let a <b be twopoints in I.

Letdxbe a positive infinitesimal. Then the definite integral offrom a to b with

respecttodx is defined to be the standard part of the infinite Riemann sumwith

respecttodx, in symbols

Wealsodefine

Bythis definition, for each positive infinitesimal dx thedefinite integral

isa real function of two variables defined for all pairs(u,w) ofelements of I. The symbol x is a dummy variable sincethe value of

doesnot depend on x.

Inthe notation ___ f(x) dx for theRiemann sum and ___ f(x) dx forthe integral, we always use matching symbols for the infinitesimal dxand the dummy variable x. Thus when there are two or morevariables we can tell which one is the dummy variable in an integral.For example, x²t can be integrated from 0 to 1 with respectto either x or t. With respect to x,

(wheredx =1/H), and we shall see later that

Withrespect to t, however,

andwe shall see later that

Thenext two examples evaluate the simplest definite integrals. Theseexamples do it the hard way. A much better method will be developedin Section 4.2.

EXAMPLE3 Given a constant C >0, evaluate the integral ___c dx.

Figure4.1.13shows that for every positive real number Δx,the finite Riemann

sumis

BytheTransfer Principle, the infinite Riemann sum in Figure 4.1.14 has thesame value,

Takingstandardparts,

Thisisthe familiar formula for the area of a rectangle.

Figure4.1.13figure 4.1.14

EXAMPLE4 Given b> 0, evaluate the integral____xdx.

Theareaunder the line y=x is divided into vertical strips of widthdx.

StudyFigure4.1.15. The area of the lower region A is the infinite Riemann

Sum

(1)areaof A = ____ xdx.

Bysymmetry,the upper region B has the same area as A;

(2)area of A = area of B.

Calltheremaining region C, formed by the infinitesimal squares along thediagonal. Thus

(3)areof A + area of B + area of C=b².

Eachsquarein C has height dx except the last one, which may be smaller,and the widths add up to b, so

(4)0≤ area of C ≤ b dx.

Putting(1)-(4)together,

Sincebdx is infinitesimal,

Takingstandardparts, we have

Figure4.1.15

PROBLEMSFOR SECTION 4.1

Computethe following finite Riemann sums. If a hand calculator is available,the Riemann sums can also be computed with ____.

1__(3x + 1)__x, _x= __2 ____(3x+1)__x, _x = ___

3__(3x + 1)__x, _x= __4 _____________, _x = ___

5___________, _x= __6 _____________, _x = 1

7__(2x - 1)__x, _x= 28 ____(x²-1)__x, _x = ___

9__(x² - 1)__x, _x= __10 ____(x²-1)__x, _x = ___

11__(5x²- 12)__x, _x= 212 ____(5x²-12)__x, _x =1

13__(1+ 1/x)__x, _x= __14 ____10-2x__x, _x = ___

15____x, _x= __16 ____2x3Δx, Δx = ___

17____Δx,Δx= __18 _____x-4Δx, Δx = 2

19__sinxΔx, Δx=__/420 ____sin²xΔx, Δx = /4

21__exΔx,Δx=1/522 ____xexΔx, Δx = 1/5

23__inxΔx, Δx= 124 _ ___, Δx=1

□25let b be a positive real number and n a positive integer.Prove that if Δx=b/n,

Usingtheformula 1 +2 …+ (n-1) = _________, prove that

□26let H be a positive infinite hyperinteger and dx = b/H.Using the Transfer Principle and Problem 25, prove that ____dx = b²/2

□27let b be a positive real number, n a positive integer,and Δx = b/n. Using the formula

Provethat

□28use problem 27 show that ___dx =b3/3

4.2FUNDAMENTAL THEOREM OF CALCULUS

Inthis section we shall state five basic theorems about the integral,culminating in the Fundamental Theorem of Calculus. Right now we canonly approximate a definite integral by the laborious computation ofa finite Riemann sum. At the end of this section we will be in aposition easily to compute exact values for many definite integrals.The key to the method is the Fundamental Theorem. Our first theoremshows that we are free to choose any positive infinitesimal we wishfor dx in the definite integral.

THEOREM1

Givena continuous function f on [a, b] and two positive infinitesimals dxand du, the definite integrals with respect to dx and du are thesame,