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第4.3节 不定积分
4.3INDEFINITE INTEGRALS
TheFundamentalTheorem of Calculus shows that every continuous function fhasat least one antiderivative, namely F(x)=___f(t)dt. Actually,fhasinfinitely many antiderivatives, but any two antiderivatives of fdiffer only by a constant. This is an important fact aboutantiderivatives, which we state as a theorem.
THEOREM1
Letf be a real function whose domain is an open interval I.
(i)If F(x) isan antiderivative of f(x), then F(x) + C is an antiderivative of f(x)for every real number C.
(ii)If F(x) andG(x) are two antiderivatives of f(x), then F(x) - G(x) is constantfor all x in I. That is ,
G(x)= F(x) +C
forsome real number C.
DiscussionParts (i) and (ii) together show that if we can find oneantiderivative F(x)off(x),then the family of functions
F(x)+C,C=a real number
givesall antiderivatives of f(x).We see from Figure 4.3.1 that the graph of F(x)+C isjust the graph of F(x)movedvertically by a distance C.Thegraphs of F(x)andF(x)+Chavethe same slopes at every point x.For example, let f(x)=3x².Then F(x)=x3is an antiderivative of 3x²because
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Butx3+6 and x3-___arealso antiderivatives of 3x². In fact, x3+Cisan antiderivative of 3x²foreach real number C.Theorem1 shows that 3x²hasno other antiderivatives.
Figure4.3.1
PROOFWe prove (i) by differentiating,
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Part(ii) follows from a theorem in Section 3.7 on curve sketching. If afunction has derivative zero on I,then the function is constant on I.Thedifference F(x)-G(x) hasderivative f(x)- f(x)=0and is therefore constant. We used this fact in the proof of theFundamental Theorem of Calculus.
Incomputing integrals of f,we usually work with the family of all antiderivatives of f.Weshall call this whole family of functions the indefiniteintegral off.The symbol for the indefinite integral is∫ f(x)dx.IfF(x)isone antiderivative of f,the indefinite integral is the set of all functions of the form F(x)+C0, C0constant.
Weexpress this with the equation
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Itis an equation between two families of functions rather than betweentwo single functions. C is called the constantof integration.To illustrate the notation,
∫ 3x²dx=x3+C.
Werepeat the above definitions in concise form.
DEFINITION
Letthe domain of f be an open interval I and suppose f has anantiderivative. The family of all antiderivatives of f is called theindefinite integral off and is denoted by ∫ f(x)dx.
Givena function f, the family of all functions which differ from F only bya constant is written F(x) +C. Thus if F is an antiderivative of f wewrite
∫ f(x)dx=F(x) +C.
Whenworking with indefinite integrals, it is convenient to usedifferentials and dependent variables. If we introduce the dependentvariable ubyu=F(x),then
du=F'(x)dx = f(x)dx.
Thusthe equation ∫ f(x)dx=F(x) +C.
canbe written in the from ∫ du=u +C.
Thedifferential symbol dandthe indefinite integral symbol ∫behaveas inverses to each other. We can start with the family of functionsu+C,form du,and then form ∫du=u+ C toget back where we started. Some of the rules for differentiationgiven in Chapter 2 can be turned around to give a set of rules forindefinite integration.
THEOREM2
Letu and v be functions of x whose domains are an open interval I andsuppose du and dv exist for every x in I.
(i)∫ du=u+C
(ii)Constant Rule ∫cdu=c∫du.
(iii)Sum Rule ∫du + dv = ∫du + ∫dv.
(iv)Power Rule ∫ur du = __________+C, where r is rational, r ≠-1,
andu>0 on I
(V)∫sinu du =-cosu +C.
(vi)∫cosu du =-sinu +C.
(vii)∫ eudu =eu+ C.
(viii)_______________=1n|u|+C(u≠ 0).
DiscussionThePower Rulergivesthe integral of urwhen r≠-1,while Rule (viii) gives the integral of urwhen r= -1.When we put u= f(x) andv=g(x), the Constant and Sum Rules take the form
ConstantRule
∫ cf(x)dx=c
∫f(x)d(x).
SumRule ∫(f(x)+g(x))dx=∫ f(x)d(x)+∫ g(x)d(x).
Inthe Constant and Sum Rules we are multiplying a family of functionsby a constant and adding two families of functions. If we do eitherof these two things to families of functions differing only by aconstant, we get another family of functions differing only by aconstant. For example,
7(3x4+C)=21x4+ 7C=21x4+C '
isthe family of all functions equal to 21x4plusa constant. Similarly,
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isthe family of all functions equal to 5x + 2_____ plus a constant.
PROOFOF THEOREM 2
(i)This is just a short form of the theorem that u+Cisthe family of all functions which have
thesamederivative as u.
(ii)We have cdu =d(cu),whence
∫ cdu =∫d(cu)=cu+C=c(u+C')=c∫du.
(iii)du + dv =d(u + v),
∫du+dv = ∫d(u+v) =u+v+C=∫ du+ ∫ dv.
(iv)___________________
Rules(v)-(viii)are similar. Only the last formula, (viii), requires an explanation.The absolute value in 1n |u|comesabout by combining the two cases u>0and u<0. When u >0,
u=| u|and
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Whenu <0, 1n u isundefined, but |u|=-uand 1n| u|=1n(-u).Thus
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Thus,in both cases, when u≠0,
d(1n|u|=____du,
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EXAMPLE1 ∫ (2x-1+ 3sin x )dx=21n | x|-3 cos x+C.
Wecan use the rules to write down at once the indefinite integral ofany polynomial.
EXAMPLE2 ∫(4x3- 6x-²+2x+1)dx=x4-2x3+x²+x+C.
EXAMPLE3
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Indefiniteintegration is much harder than differentiation, because there are norules for integrating the product or quotient of two functions. Itoften requires guesswork. The short list of rules in Theorem 1 willhelp, and as this course proceeds we shall add many more techniquesfor finding indefinite integrals.
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Ourrules give no hint on finding this integral. However, once the answeris given to us we can easily prove that it is correct bydifferentiating,
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Hereis warning that may prevent some common mistakes.
Warning:the integral of the product of two functions is not equal to theproduct of the integrals.
Thesamegoes for quotients. That is,
Wrong:
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Forexample,
Wrong:
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Correct:
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Wrong:
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Forexample,
Wrong:
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Correct:
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Theindefinite integral can be used to solve problems of the followingtype. Given that a particle moves along the y-axiswith velocity v=f(t),and that at a certain time t=t0itsposition is y=y0. Find the position yasa function of t.
EXAMPLE5 A particle moves with velocity v=1/t²,t >0.At time t=2 it is at position y=1.Find the position yasa function of t.We compute
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Sincedy/dt=v, yisone of the functions in the family -1 /t+C. We can find the constant Cbysetting t=2and y=1,
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Thenthe answer is
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Thenext theorem shows that in such a problem we can always find theanswer if we are given the position of the particle at just one pointof time.
THEOREM3
Supposethe domain of f is an open interval I and f has an antiderivative.Let P(x0, y0) be any point with x0 in1. Then f has exactly one antiderivative whose graph passes throughP.
PROOFLet Fbeany antiderivative of f.Then F(x)+ Cisthe family of all antiderivatives. We show that there is exactly onevalue of Csuchthat the function F(x)+Cpassesthrough P(x0,y0)(Figure 4.3.2). We note that all of the following statements areequivalent:
(1)F(x)+ Cpassesthrough P(x0,y0).
(2)F(x0)+ C = y0.
(3)C = y0-F(x0).
Thusy0-F(x0)is the unique value of Cwhichworks.
Figure4.3.2
TheFundamental Theorem of Calculus, part (ii), may be expressed brieflyas follows, where fiscontinuous on I.
If∫ f(x)dx = F(x) +C, then
_________f(x)d(x)=F(b) - F(a).
Forevaluating definite integrals we introduce the convenient notation
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Itis read “F(x) evaluatedfrom atob.”
TheConstant and Sum Rules hold for definite as well as indefiniteintegrals:
ConstantRule
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SumRule
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TheConstant Rule is shown by the computation
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TheSum Rule is similar.
EXAMPLE6 Evaluatethe definite integral of y=(1+t)/t3fromt =1to t=2
(seeFigure 4.3.3).
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Thusthe area under the curve y=(1+t)/t3fromt=1to t=2 is _______
Figure4.3.3
EXAMPLE7 Findthe area of the region under one arch of the curve y=sin x
(seeFigure 4.3.4).
Onearch of the sine curve is between x=0and x =π.The area is the definite integral
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Thearea is exactly 2.
Figure4.3.4
EXAMPLE8 Findthe area under the curve y=-2x-1fromx=-5to x=-1.
(SeeFigure 4.3.5.)
Thearea is given by the definite integral
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Firstcompute the indefinite integral
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Nowcompute the definite integral.
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Figure4.3.5
Thisexample illustrates the need for the absolute value in theintegration rule
∫ x-1dx = 1n| x|+ C.
Thenatural logarithm 1n x is undefined at x = -5and x= -1,but 1n | x| is defined for all x ≠0.The absolute value sign is put in when integratingx-1 and removed when differentiating 1n | x|.
EXAMPLE9 In computing definite integrals one must first make surethat the function to be integrated is continuous on the interval. Forinstance,
Incorrect:
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Thisis clearly wrong because 1/x² >0 so the area under the curvecannot be negative. The mistake is that 1/x² is undefined atx=0 and hence the function is discontinuous at x=0.Therefore the area under the curve and the definite integral
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areundefined (Figure 4.3.6).
Figure4.3.6
PROBLEMSFOR SECTION 4.3
Evaluatethe following integrals.
1
3
5
7
9
11
13
15
17
19
21
23
25
27
29
31
InProblems 32 -36, find the position y asa function of t giventhe velocity v = dy/dt andthe value of y atone point of time.
32v = 2t+3,y = 0 whent =0
33v = 4t² -1,y = 2 whent =0
34v = 3t4 ,y = 0 whent =-1
35v = 2sint ?,y =10 whent =0
36v = 3t-1,y = 1 whent =1
InProblem37 -42, find the position yandvelocity v?asa function of tgiventhe acceleration a andthe values of yandv att =0or t=1.
37a =t,v =0and y=1when t=0
38a =-32,v =10and y=0when t=0
39a =3t²,v =1and y=2when t=0
40a =1-____,v =-2and y=1when t=0
41a =t-3,v =1and y=0when t=1
42a =-sint,v =0and y=4when t=0
43Whichof the following definite integrals are undefined?
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44Find the function fsuchthat f `isconstant, f(0)= f `(0)and f(2)=f`(2).
45An object moves with acceleration a=6t.Findits position y asa function of t,given that
y=1when t=0and y =4 when t=1.
46Find the function hsuchthat h''isconstant, h(1)=1, h(2)=2, and h(3)=3.
□47Suppose that F''existsfor all x.and let (x0, y0) and (x1, y1) be two given points.
Provethatthere is exactly one function G(x)suchthat
G(x0)= y0
G'(x1)= y1
G''(x)= F ''(x)for all x.
□48Assume that F ''(x)exists for all x,and let (x1,y1)and (x2,y2)be two points
withx1 ≠ x2.
Provethatthere is exactly one function G(x)such that G ''(x)= F ''(x)for all x,and the graph of Gpassesthrough the two points (x1,y1)and (x2,y2).
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