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# math117 lecturenotes9 - ‘ Lactate#24 I 24 grépier 3...

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Unformatted text preview: ‘ Lactate #24 I 24'! ,. grépier 3 . Integration. The integral calculus is much older than the differential calculus, because the computation of areas of surfaces and volumes of solids occupied the greatest mathematicians since antiquity: Archimedes, Kepler, Cavalieri, Viviani, Fermat, Gregory, Barrow. The decisive breakthrough came when Newton, Leibniz, and Jean Bernoulli discovered independently that integration is the inverse operation of differentiation, thus reducing all efforts of the above researchers to a couple of diﬁ‘erentiation rules, as we shall see shortly. § 3.1 Primitives (Indefinite Integral) Problem: For a given function y = f(x) we want to compute the area between the x—axis and the graph of this ﬁinction. We ﬁx a point a and denote by z = F (x) the area under f(x) between a and x. 0 Solution #1 (Newton). Newton imagines that the segment BD (Fig. 3.1) moves over the area under considerations; consequently, if x increases Fig. 3.1 by Ax the area increases by A2 = F(x + Ax) — F(x). Dividing both sides by Ax and going to limit as Ax -+O gives 21% = F '(x). On the other hand, this increase in area can also be calcufted as dz = f(x) dx (dark rectangle in Fig. 3 .1). Hence (3-1) F'(X) = f(X)~ 2.4.2 0 Solution #2 (Leibniz). Leibniz imagines the area as being a sum (later: “integral”) of small rectangles, as shown in Fig.3.2, not necessarily of the same size. If we use n rectangles then the Fig 3'2 approximate area is (3.2) 2,, = f(xl) AXl + ﬂXz) AXz + ...+f(xn-1) Axn-1+ f(xn) Axn. Hence the area of the dark rectangle is given by zn — 2.1-1 = f(xn) Axn. . . . . dz . dF . . D1v1de through by Axn and take the 11m1t as Axn —->0 to get d— = f (x); that IS d— = f (x) , which is x x the same result as Newton’s (3.1). I Remark 1: Interpretation. The area under f from a to x is obviously a function of x; call this ﬁmction (temporarily) “the area ﬁanction”. Then (3.1) says that in order to ﬁnd the area function F(x) we must ﬁnd a ﬁmction whose derivative is the given f(x). Since we know a lot of derivatives, it should be very easy to ﬁnd lots of “area ﬁmctions”. The following examples show how. I n+1 A Example 1. We know that £(x‘m) = (n + l)x“. Rewrite this as 51— x = x“. and compare dx dx n + l n+l with (3.1). Thus we can identify F(x) = X“ (with, of course, 11 at -l) and f(x) = x”, and so we n n+1 X . A n+1 conclude that the area under the graph of x“ is given by 24.3 A Example 2. We know that ad—(lnx)=% (x>0). Thus F(x) =lnxisthearea under thegraph x ofl. A x 0 Definition 1. The ﬁmction F(x) discussed above is called the primitive of f(x). Note that primitives are not unique, for to each primitive one can add an arbitrary constant c and F(x) + C Is again a pnmitive of the same function f(x). The proof lS tnvral: d—(F(x) +c) = x =9F_+\$=F(x)+o=r'(x)=f(x) (seeEq.3.1). dx dx I Remark 2. Reason for the name. The Dictionary says that the mathematical meaning of “primitive” is “a form or an expression from which another is derived”. This is a pretty good descriptive name, for F(x) is a ﬁlnction from which ﬁx) is derived (by differentiation — see Eq. (3.1). l I Remark 3. There is another name for F (x), namely “antiderivative” — but this is not as good as “primitive". I 0 Deﬁnition 2. There is a special symbol to denote the primitive F(x) of a ﬁlnction f(x), and it is (3.3) F(x) = jf(x)dx . Historically, Leibniz arrived at the adoption of this notation aﬁer a long series of attempts. He meaat the symbol I to represent the sum (3.2) in the limit as Axi —) 0 , and made it to resemble an elongated S. Leibniz wrote this (in Latin) in a letter to Jean Bernoulli on March 8, 1696, [...] notam i pro summis, ut adhibetur nota dpro differentiis [...] ire. 9“] the \$3mbol 5' {Jar swms , Mile Hana sngbol d is used- {or WMULS [mi and Bernoulli answered back immediately, on April 7, 1696, announcing that he wanted to call it “the integral”, 24.» [-51 {or which I will. mow wake. u.\$e. cg. / m ma ”meeﬁrae'IC-q [...] quod autem [. . .] vocabulum integralis etiamnum usurpaverim [. . . ]. About 18 years later, Newton — always the sore looser — wrote to Keill, on April 20 1714, And whereas M Leibnits praeﬁxes the letter I to the Ordinate of a curve to denote the Summ of the Ordinates or the arm of the Curve, I did some years before represent the same thing by inscﬁbing the Ordinate in a square [. . .] My symbols therefore [...] are the oldest in the kind. In other words, Newton would write (3.3) as F<s>= , but - as I said before — he never had the ability to invent a good notation! E Lecture—#25 I 25.1. §3.2 Integration as the inverse of differentiation. Let us compare careﬁilly Eqs. (3.1) and (3.3). It is clear that the operation denoted by I undoes what the operation d1 did; in other words, the integral isthe inverse operation ofthe x derivative, and viceversa. Following the block diagram technique of representing operators, introduced earlier, we may use the scheme f(x) ——> -———> jf(x)dx = F(x) to represent Eq. (3.3), and the scheme d F(x) —-——> dx 9 —’ F'(X) = f(X) to represent (3.1). Putting the two block diagrams together in series we see that f(x) ——> jf(x)dx =F(x)——> -—>F’(x) = ﬁx) Eq. (3.3) Eq. (3.1) so that the ﬁnal output is identical to the original input. Looking at things ﬁ'om this point of view we see that by reversing differentiation formulas we obtain formulas for primitives. Here are a few examples: n+1 X (3.4) jxndx=——+c , (ma-1) , C=arb.const. n+1 I (3.5) j—dx=1nx+c , (x>0) X (3.6) Jexdx = e" +C , (8.7) Isinx dx =—cosx+C (3.8) Icosx dx=sinx+C 25.2 1 (3.9) j1+x2dx=ian—1x+c , (3.10) J- 1 l—x 2 dx:sin‘}x+C Large tables of primitives exist, and they can be many hundreds of pages long. In recent years this knowledge has been incorporated in many symbolic computation systems, such as Maple. O The block diagrams shown above are not just theoretical schemes. They also symbolize actual operation performed by electronic circuits, the simplest of which is illustrated in Fig. 3.3. Here V, is the input voltage, the output voltage is v0, the voltage drop across the resistor Va, and the current ﬂowing in the circuit is- i. As you will learn in amulet course, the application of Ohm’s and Kirchhoﬂ‘s laws leads to the differential equation Fig. 3‘3 vi = RC%+V0. In general, vi will be an ac. signal consisting of a range of frequencies. The ac. resistance of the capacitor C (known as the impedance) is given by X0 2 36, where co = 21w is the no angular frequency of the signal. For co large, XC is very small compared with K so that most of the voltage drop takes place across the resistor. In other words, vo << vR and as a consequence the differential equation simplifies to dvO v? ——-=—1—, C st., dt RC 0" 0°” ) From which we get the primitive (3.11) v0(t)=%jvi(t)dt+C. This shows that the circuit of Fig. 3.3 is an integrator; it takes the input voltage vi(t) and outputs its integral. 2.5.3 I Remark. Modern integrator circuits do not .. Posrtive power supply have the limitations of the simple device illustrated above. Rather, they use high-gain dc. ampliﬁer circuits called operational ampliﬁers, or OP AMPS, that perform mathematical operations (hence the name). A scheme of an OP Negéﬁve PM“ supply AMP is shown in Fig. 3.4, where the ﬁinction of Fig. 8.4 the stande five terminals are labeled. By making recourse to such a device, the scheme of the modern differentiator looks as shown in Fig. 3.5, which may be symbolized by the block VéEL—q - . diagram C v°(t) vi(t)——> —> .—> vo(t) where K represents the gain (ampliﬁcation). And the very simple modiﬁcation of interchanging the resistor Fig.3.5 — An inverting differentiator R and the capacitor C produces an inverting integrator, symbolized by the block diagram vi(t) —>. ——> vo(t). I 1 L305)” #19 1 2.6.1 §3.3 The Deﬁnite Integral. (MEN , PP 537 5L“ ) Back to the area ﬁinction z = F(x) introduced in Section 1. According to Fig. 3.1 and Eq. (3.3), we may represent the area under the graph of f(x) between a and x by F(x) = If(t)dt , where t is a “dummy variable” which is used not to conﬁise it with the variable x on which the area ﬁrnction depends. Moreover, we saw there as well that the area ﬁinction F(x) is the primitive of f(x); and since one can always add the constant C to the primitive, we have C+ F(x) = Tram. Now C is arbitrary, which means we can choose any value we want for it. Therefore let us choose C = -F (a), so that we may rewrite the relation above as (3.12) F(x) — F(a) =jf(t)dt _ Two important results follow from this. First, for x = a the l.h.s. of(3. 12) vanishes — which makes perfect sense if you look at Fig. 3 .1, for the area 2 = O for x = a. At the same time we learn from the r.h.s. of (3. 12) that when the upper limit of integration is equal to the lower limit the integral vanishes. Second, if instead of a variable area ﬁinction F(x) we want the area between two ﬁxed points a and b then we pick x = b in Eq. (3.12) and get (3.13) F(b)— F(a) = Jf(t)dt. Eqs. (3.12) and (3.13) are at the foundation of all the work with integrals. The most important fact (to be put on permanent memory!) about them is the following: The integral in (3.12) — the indeﬁnite integral -is an Operation to create new functions, while the integral in (3.13) — the deﬁnite integral — is an operation to compute numbers. 26.2 I Remark. In traditional calculus textbooks the statements (3.12) and (3.13) are called the Fundamental Theorem of Calculus (part 1 and 2), or FTC for short. However, this name was introduced over a century later than the period we are considering (around 1700). Leibniz and Jean Bernoulli were so excited about the power of their discovery that they (and Euler, soon afterwards) charged on and used the new tool to solve a host of important problems that had been around since the time of Archimedes. The following examples give you an idea. I A Example 1. Area of Parabolas. The area under the nth degree parabola y = xn between a and b is given by b xn+1 b bn+l _an+l Ixndx = , n+1- n+1 b=F(b)—F(a) A where we have used the notation F(x) A Example 2. Area of a Disc. Consider the surface enclosed by the unit circle x2 + y2 = 1. Its area can be computed by computing the area of the quarter disc in the ﬁrst quadrant (Fig. 3.6) and multiplying the result by 4. The quarter circumference is given by the function f(x)=\/1—x2, Ostl, a primitive of which (as we shall see soon) is given by F(x)=12(-Jl— x2 +%sin“1(x), as is easily checked by differentiation. Thus, Fig. 5.6 applying (3 .13) we get 1 area ofunit disc = 4 [ J1 — xzdx = 4(F(l)—F(O)) = 7: A 0 26-3 A Example 3. Area of a Disc (The Sequel!) There is another elegant way of solving this problem. Consider a quarter disc of a radius r, and slice it into inﬁnitely thin sectors (Kepler, 1615, as well as Leibniz; Fig. 3.7). The area of each thin sector is given by d8 = é—rzdo, were dq) is an inﬁnitesimal increment of the angle (p identifying the sector. The sum of all these areas in the limit is (of course!) Note. These are geometric examples; however, the FTC is essential in the solution of important physics problems —- see the current assignment. ...
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