Lecture for Week 14 (Secs. 6.3–4)Definite Integralsand the Fundamental Theorem1Integral calculus — the second half of thesubject, after differential calculus — has two as-pects:1. “Undoing” differentiation. This is the prob-lem of finding antiderivatives, which we’vealready discusse d.2. The study of “adding things up” or “acc u-mulating” something. This is very closel yrelated to the area topic of last we ek.2The fundamental theorem of calculus showsthat these two things are essentially the same.To reveal the basic idea, consider a speed-distance problem: We know th at if an objectmoves at a constant speed for a certain period oftime, then the total distance traveled isdistance = speed × time.Suppose instead that the speed is a varying func-tion of time. If we con si der a very short time in-terval, [ti−1, ti], then the speed is approximately3constant (at least if f is continuous) and hencewe can approximate the distance bydistance = speed in that interval × (ti− ti−1).As the speed we may choose the maximumspeed in the short interval, or the minimum, oranyt hing in between — say f (wi) for some wi∈[ti−1, ti]. The choice won’t make any difference inthe end.(There should be a picture here, but I don’thave time to draw it now. )4So the total distance travele d between t = aand t = b is approximatelynXi=1f(wi)∆ti, where ∆ti− ti−1.As we let n → ∞, the approximation sh ou l dbecome exact.In effect, we h ave concluded that:5(A) The distance traveled betwee n times a andb is the area under the graph of the speedfunction between the vertical lines t = a andt = b (if area is defined in appropriate units,and if the speed is alway s nonnegative).On the other hand,(B) The distance function is an antiderivative ofthe speed f unction.6Conclusion: Areas an d antiderivatives arevery closely related. In some sense, the y are thesame th i ng!Before continui ng we need a more prec i sedefinition of the definite integral as a limit of asum,Zbaf(x) dx = limkP k→0nXi=1f(x∗i)∆xi.That occ upies the first half of Sec. 6.3, and it7involves all the issues about varying the sizes∆xiof the strips, etc., that I discussed last weekabout area.One important difference between areas andgeneric definite i ntegrals: The integrand functi onis al l owed to be negative in some (or all) places).Areas must always be positive (or zero), but inte-grals can be negative. In general,Zbaf(x) dx = A+− A−,8where A+is t he area bel ow the graph and abovethe x axis, an d A−is t he area above the graphand below the x axis. (In Fig. 3 on p. 379, A+isyell ow an d A−is blue.)Note also that the “dummy variable” in anintegral can be any letter that doesn’t cause confu-sion:Z62x3dx =Z62t3dt.Also, this thing is not a function of the variable ofintegration, it is just a number.9Here is an example where a bad choice of let-ter would cause confusion:x2− xZx22t3dt (∗)must not be writt en asx2− xZx22x3dx.(The integral in (∗) is a function of x, though notof t.)10Algebraic properties of integrals1.Zba[f(x) + g(x)] dx=Zbaf(x) dx +Zbag(x) dx.2.Zbacf(x) dx = cZbaf(x) dxfor a constant c.113.Zbaf(x) dx =Zcaf(x) dx +Zbcf(x) dx.4.Zbadx meansZba1 dxand equals b − a.Remarks: Formula (4) is just the area of arectangle. The others are also rather obv i ous forareas. Recall that making (3) obvious was one ofthe reasons for allowing strips of varying widths∆xi.125.Zabf(x) dx means −Zbaf(x) dxif b > a.It follows that in (3), c does not ne ed to liebetween a and b.Perhaps more important (from the point ofview of avoiding mistakes ) is an identity that isnot in the list:13Zf(x)g(x) dx =Zf(x) dx ×Zg(x) dxFALSE!The integral of a p roduc t is not the product ofthe integrals, just as (and because) the derivativeof a product i s n ot the product of the derivatives.Roughly speaking, every derivative formula turnsaround to give an integral formula. The inte gralformula c orrespond i ng to the product rule forderivative s is integration by parts (Sec. 8.1).14Order properties of the integral1. If f(x) ≥ g(x) for all x i n [a, b] , thenZbaf(x) dx ≥Zbag(x) dx.(Here we assume a < b, of course.) In partic-ular, if f is nonnegative, thenZbaf(x) dx ≥ 0 also.152. If f is continuous an d f (x) ≥ 0, thenZbaf(x) dx (with a < b) i s strictly greaterthan 0 unless f(x) = 0 for all x in [a, b].3.Zbaf(x) dx≤Zba|f((x)| dx.164. Mean value theorem for integrals: Iff is continuous, then there is a z in (a, b)such thatRbaf(x) dx = f (z) (b − a).5. If m ≤ f(x) ≤ M in [a, b], thenm (b − a) ≤Zbaf(x) dx ≤ M (b − a).17Now back to the “fundamental theorem”.Roughly speaking, it says that differentiation“undoes” integration, and vice ve rsa. The y areinverse operations (almost), l i ke squaring andtaking the square root, ex cept that they operateon functions instead of numbers.The two functions involved are related asare th e two readings on a car’s speedometer-odometer pan el. The odometer reading is theinte gral of the speedometer reading. The speedo-18meter read i ng is the derivative of the odometerreading. That is the essence of c al culus!The theorem has two parts, one for each or-der of the operations. And I st ate each part intwo versions, d epending on which f unction (theinte gral or the derivative) takes center stage.In stating the theorem, we assume for simplic-ity that f (the derivative) is conti nuous.19Fundamental Theorem, Part 1:ddxZxaf(t) dt = f (x).That is,G(x) ≡Zxaf(x) dtis an antiderivative of f.20ExerciseEvaluateddxZx102u du,ddyZy31tdt,ddvZv2102u du.21For the first two, just appl y the the orem:ddxZx102u du = 2x,ddyZy31tdt =1y.We know these facts without necessarilyknowing what the i ntegrals t hemselves are. Youmay k now that the fir st one isZx102u du = x2− 100,22either (the hard way) in analogy to Exercise 6.3.16or (the easy way) peeking ahead to Part 2 of thetheorem. The second integral requires a logarithmfunction (see Sec. 6.6).For the third one, use the chain rule:ddvZv2102u du = 2v2× 2v = 4v3.(Make sure you understand this. The integral i sa function of v2and therefore of v.)23Fundamental Theorem, Part 2:ZbaH′(x) dx = H(b) − H(a).That is,Zbaf(x) dx = H(b) − H(a),where H is any antiderivative of f.24Now we can fill in the remarks I made onthe previous example:ExerciseFindZx102u du,Zy31tdt.25Sinceddxx2= 2x,Zx102u du = u2u=x− u2u=10≡ u2x10= x2− 100.Certainly much easier than “Use
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