Lecture for Week 14 (Secs. 6.3–4)Definite Integralsand the Fundamental Theorem1Integral calculus — the second half of thesubject, after differential calculu s — has two as-pects:1. “Undoing” differentiation. This is the prob-lem of finding antiderivatives, which we’vealready discuss ed.2. The study of “adding things up” or “accu-mulating” something. This is very closelyrelated to the area topic of last week.2The fundamental theorem of calculus showsthat these two things are essentially the same.To reveal the basic idea, conside r a speed-distance problem: We know that if an objectmoves at a constant speed for a certain period oftime, then the total distance travel ed isdistance = speed × time.Suppose in s t ead that the speed is a varying func-tion of time. If we conside r a very short time in-terval, [ti−1, ti], then the speed is approximate ly3constant (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, oranything in between — say f (wi) for some wi∈[ti−1, ti]. The choice won’t make any difference i nthe end.(There should be a picture here, but I don’thave time to draw it now.)4So the total distance traveled be tween t = aand t = b is approximatelynXi=1f(wi)∆ti, where ∆ti− ti−1.As we let n → ∞, the approximation shouldbecome e xact.In effect, we have con cluded t hat:5(A) The distance traveled between times a andb is the area un der the graph of the speedfunct i on between t he vertical lines t = a andt = b (if area is defined in appropriate units,and if the speed is always nonnegative).On the other hand,(B) The distance function is an antid erivative ofthe speed function.6Conclusion : Areas and antiderivatives arevery closel y related. In some sense , they are thesame thing!Before conti nuing we need a more precisedefini t ion of t he definit e integral as a limit of asum,Zbaf(x) dx = limkP k→0nXi=1f(x∗i)∆xi.That occupies th e first half of Sec . 6.3, and it7involves all t he issues about varying the size s∆xiof t he strips , et c., th at I discussed last weekabout are a.One important difference between areas andgeneric defini t e integrals: Th e integrand functionis al lowed 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 below the graph an d abovethe x axis, and A−is t he area above the graphand below the x axis. ( In Fig. 3 on p. 379, A+isyellow and A−is blue.)Note also that the “dummy variable” i n anintegral can be any lett er that doesn’t cause confu-sion:Z62x3dx =Z62t3dt.Also, this thing is not a function of the variable ofintegration, i t is just a number.9Here is an example where a bad choice of l et -ter would cause confusion:x2− xZx22t3dt (∗)must not be written 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 obvi 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 need to liebetwe en a and b.Perhaps more important (from the point ofview of avoiding mistakes) is an identity th at isnot in the list:13Zf(x)g(x) dx =Zf(x) dx ×Zg(x) dxFALSE!The integral of a product is not the product ofthe inte grals, just as (and because) th e derivativeof a product is not the product of the de rivat ives.Roughly s peaking, every derivati ve formu la tu rn saround to give an integral formu l a. Th e integralformula correspond ing to the product ru le forderivatives is integration by parts (Sec. 8.1).14Order properties of the integral1. If f( x) ≥ g(x) for all x in [a, b], thenZbaf(x) dx ≥Zbag(x) dx.(Here we assume a < b, of cou rse.) In partic-ular, if f is nonnegative, thenZbaf(x) dx ≥ 0 also.152. If f is continuous and f(x) ≥ 0, thenZbaf(x) dx (with a < b) is 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, the n there is a z in ( a, b)such thatRbaf(x) dx = f(z) (b − a).5. If m ≤ f(x) ≤ M in [a, b], the nm (b − a) ≤Zbaf(x) dx ≤ M (b − a).17Now back to the “fundamental theorem”.Roughly s peaking, it s ays t hat differentiation“undoes” i ntegration, and vice versa. They areinverse ope rati ons (al most) , like squ aring andtaking the sq uare root, except that they operateon functi ons instead of numbers.The two f unctions involve d are related asare the two readings on a car’s speedomet er-odometer panel. The odometer re ading is theintegral of the speed omet er reading. The speedo-18meter reading is the derivative of the odometerreading. That is the e s s ence of calc ulus!The theorem has two parts, one for e ach or-der of th e operati ons. And I state each part intwo versions, depen ding on which functi on (theintegral or the derivative) take s center s tage.In stating the theorem, we assume for simplic-ity that f (the derivative) is continuous.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 apply the theorem:ddxZx102u du = 2x,ddyZy31tdt =1y.We know these facts w ithout necessarilyknowing what the integrals themselves are. Youmay know that the first one isZx102u du = x2− 100,22either (the hard way) in analogy to Ex er cise 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, us e th e ch ai n rule:ddvZv2102u du = 2v2× 2v = 4v3.(Make sure you und erstand this. The i ntegral isa 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−
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