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math117 lecturenotes11

math117 lecturenotes11 - m 3.7.Some Theoretical Questions...

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Unformatted text preview: m § 3.7..Some Theoretical Questions. The discussion of the integral so far was based on the formula (3.7.1) ' Effxrix =F<b)—F(a>, where F(x) is a primitive of f(x). We have implicitly assumed that such a primitive always exist and is unique (up to an additive constant). But what if that is not true? There are lots of fiinctions which don’t have a primitive, (perhaps the most famous one is f(x) = e”8 ), and for them the symbol fb'ffxfix is undefined. As the great B. Riemann noticed, Ourfust quesfionisdierefcpezwhatmeaningshould we b giveto j f(x)dx? a (B. Riemann, 1854, Werke p 239) A. Cauchy (1823) had made an attempt in this direction by restricting f to be a continuous fiinction on [a,b]. He asserted the following Theorem 1. If f : [a,h] —~) IR is continuous, then it is integrable. But he could not prove it —in fact, it toolt another 50 years before a rigorous proof was found (Du Bois-Raymond, 1875; Darboux, 1875). Riemann, on the other hand, bypassed this problem and, merely as a side remark in his habilitation thesis on trigonometric series, defined the integral for a much more general class of fiinctions, including discontinuous fiinctions without a primitive. As Darboux commented 21 years later In Hne pa sb By one ofthqseinsiglnsofwhichonlythegreatestmmds are mtkemafiotam capable, the famous geometer [Riemannrgeneralized the concept of definite integral, New; Gated; geomel-exs . G. Darboux, Alémoire sur [es fonctions discontinues, 1875 The rigorous proof of Riemann’s theory of integrations, and its extensions byDuBois-Reymond and Darboux is quite subtle and belongs to the area of mathematics called Real Analysis. However, we can understand Riemann’s basic ideas quite readily. We start with a fundamental assumption: 30.7. (3.7. 2) O Assumption: The fimction f: [a,b] —> R where [ab] ={x l a 5 x: b} is a bounded interval, is bounded on [ab]; that is, there exists a number M 2 0 such that, fbr all x in the interval, we have lflx)l S M‘. (Situations that violate these conditions will be considered later.) Next we make a partition of {ab}, which means that we divide the interval [2gb] into an arbitrary number of small sub-intervals, not necessarily equal. The partition is denoted by P = {x0, x1, X2, x“, xn} where a = xo< x1< X2< ...< xn.1< xn = b, and the length of a sub-interval is (5i : xi —xi_1 . Now let fl = smallest value taken by f in [19.1, xi] Fi = largest value taken by f in {xi}, xi] and use them to define the lower and upper Darboux sums s(P)=if.5,- ; S(P)=:F;51, i=l i=l which are illustrated in figure, below: ' This last statement can be written insymboiic form as 3 M 2 0 s.t V x 6 [a,b] |f(x)l S M. (3 is read ‘there exists”, and V “for all” — roughly!) 30.3 As we increase the number of points xi both fl (and s(P)) and Pi (and S(P)) will change; but it is geometrically clear from ihe Ligafiand can be proven) that all the sums s(P) will be less than the sums S(P), Now take the maximum of all the sums s(P) and define, following Darboux, ( 3 .7. 3) max s(P) = ill-laid" (the lower integral); similarly, take the minimum of all the sums S(P) and define (3.7. 4) min S(P) =1??de (the upper integral). Definition. The firnction f satisfying the assumption (Z2) is called integrable in the Riemann sense if the lowerand upper integrals (7. 3) and (7.4) areequal. In that case we remove thebars inthe integral signs and we obtain the Riemann integral ff(x)dx Geometrically, the situation is quite clear from H19, Figurefl'he lower Darboux sum, s(P), underestimates the area under the curve (panel 1); however, as the partition becomes finer and finer, the error will be less and less. Similarly, the upper Darboux sum, S(P), overestimates the area under the curve (panel 2), but the error decreases as the partition P becomes finer. The total error is given by the difference of the two areas: (Fi - fl) 6i for the ig1 partition sub-interval (panel 3), and S(P) — s(P) for the entire curve (shaded area in panel 3), and it is intuitively obvious that as the partition becomes finer and finer this error will decrease to zero. A rigorous proof, however, is far fiom obvious. The Darboux sums are very useful for theoretical reasons, but even more usefiil is the approach taken by Riemann himself. Consider a partition P and let £1, £2, ...,§n be such that XOS§ISX1S§ZSXZS...an_1S§n an; in other words, for each sub-interval [xi-1, xi] pick an arbitrary point i,- anywhere between X“ and xi (end points included). Then we call (3.7.5) o=Z:f<a.)6. image. —x.-.> the Riemann sum. A5 H12 gigom shows, we have fig f(§.) s E, and so s(P) S c S S(P). Therefore, one can show (DuBois-Reymond, Darboux) that ll Lecture, # 3) l 31.! (3.7.4,) imam—fund: as mafia 0 Note. Riemann sums are very convenient for proving properties of the integral. Although these proofs belong to Real Analysis, we shall nevertheless make good, practical use of them, and therefore list them below (for future use) in the form of Theorems. 0 Theorem 2. 0 Theorem 3. (3.7.7 ) 0 Theorem 4. (3.7.8) (5-7.0: ) 0 Theorem 5. (3.7.10) 0 Theo rem 6. Let f and g be integrable functions on [a,b], and let 7L 6 R. Then the functions f+g;xf; fg; m; germ») are also integrable. Assume that f : [a,b] —> IR is a function whose restrictions to [a,c] and [c,b], where a < c < b, are integrable. Then f is integrable on [a,b] and we have Ef(x)dx =Ef(x)dx+ff(x)dx. If f and g are integrable on [a,b] (with a < b), and if f(x) 3 g(x) for all x 6 [a,b], then ff(x)dx s fg(x)dx. As a corollary, it follows that for any integrable function we have {growls Elfixxdx . (Cauchy — Schwarz Inequality) For integrable fimctions f and 7 g on [a,b] we have 1 .L 2 1 H: f(x)g(x-)dx‘l sffffixfix) (fgflxflx) . (The Mean Value Theorem, MVT). If f : [a,b] —> R is a continuous function, then there exists a point i 6 [a,b] such that 3 l .2 (3.7.10 Hand); =f(§)(b— a). . Theorem 7' Let f : [a,h] -> R be continuous, and let g : [a,b] —> R be integrable and either always positive or always negative. Then, there a exists a g e[a,b] such that (3.7. 12) f f<x>g<x>dx =f<a)fg<x>dx. (Sometimes this result is called “the second MVT”.) 0 Th 121.8“ [,ch f :f-a,a,]—+ [K be @1311, Le. 9(4):”), -——~_. Then Q’ Q (3.7.13) f £00ch =1J§Q<lolx . -Q, o oThcorevnq . Let t :[-a,a]—-> R be 961$ ) C-e. ;(—x)=-§v). Then Q (3.7.14 ) “(mix = o. —m Examples of. apylicahbns 0+ lkese brow-Hes 9,9 H18 Riemann Maggi will be given in class and. in “H16 /—\$$ignmenl:s . l [-00th # 31. I 31' 1 § 3.9. Two Important Applications of Integration. (A) Mean (or average) value of a function. Very frequently, in scientific or engineering applications, we are not interested in the instantaneous value of a physical quantity, but rather in its mean value over a certain period of time. In such cases we may use the MVT (3.7.1I)to compute the desired average. In other words, suppose that a physical quantity is represented by the fimction fIt), integrable over [a,b]; then, we define its mean value by (5.8.1) sz—ijibfmdt' (Frequeni-ly Hoe Symboi used. is <¥>> A Example 1. Consider the sawtooth waveform 'm We figure; compute its mean value over a period. Sol 7n. Clearly, V(t + 2) = V(t),and so we shall compute its mean value overa period bythe integral A S aw ['0on signal 22-]: V(t)dt. Before we can do it though we must find V(t). That is easy, for the graph shows that this is a linear signal, and hence we may represent it by V(t) = mt + b. Now V(O) = 0 and ling. V (t) = 5 (Note that V(2) = 0). These two conditions give b = 0 and m = 3-, respectively. Therefore, I Remark 1. What would happen if we wanted the mean value over the interval [0,4]? Simple: 4V(t)dt zflf V(t)dt+L4V(t)dt) 31.2 where we have applied (3 7.7), that is Theorem 3 of the previous section. (Note that the fact that V(t) is piecewise continuous does not bother us—-the Riemann integral exists anyhow.) And since we have already V(t) over [0,2], all we need is its form over [2,4]. This is easily done, and the result is V(t) = gt—S. Thus (check it!) Nlm -_1 fi. 42__ _ V _ 400 2tdt+£(2t syn) _ , as before. Note that, although in this case the mean value is the same, sometimes if the interval of integration changes then the mean value will change. I A Example 2. A rectifier is a signal processor of the type shown on matgln. That is, it takes a sinusoidal input of amplitude VA and period To and outputs arectified sine wave; that is, a sine wave where the negative part has been “chopped off”. Compute the mean value of both the input and the output. Sol’n. For the input we have V(t) 2 VA sin (% t) , and so H10, MVT gives (check it!) ‘. —i T“ - H = = Vm—Tojo VAs1n[To)dt o. The mean output is again computed by using i‘hcz M V'T') -_LT°-_2_fi_ =LT°’2~2_1I T°.==l Vom—To L V, sm(To 1)dt T000 VAsm(Tot)dt+I Odt] VA.A To/z TC (B) Root mean square value (ms) of a fiinction. We have just seen that the average of a sinusoidal input is zero. That, of course, doesn’t mean that the sinusoidal input carries no power. By definition, the instantaneous power delivered to—say—a resistor R by a voltage V(t) is given by wefiWm, and so the average power delivered to the resistor in a time span T is, according to H12 MV'T' , 31. 3 13 = —% [01% V2(t)dt = {IB— f: V2(I)dt:l. Now notice that the units of the quantity in the square brackets is (volt)2. therefore, it is convenient to define the root mean square (rms) voltage l _ I T 2 3 (3.8-2 ) Vm —(:r—jo v (t)dt) , _ V2 so that the average power is given by P = I?“ , the same functional form as the power delivered by a static source (a battery, say). A Example 3. Calculate the rrns value of the signal f(t) = A sin 0,) . Solu'n. The signal is periodic with period T = 211:, and its amplitude is A. We apply the definition abovcand get 1! — 2 1: fm =(—1— 2 Azsinztdty =[A—j2 211: 0 1_ ___ £(t_sin2t) 2" 2 411: 2 0 Therefore, the rms value of a sinusoidal signal is given by the amplitude divided by \5 . In other words, it is about 70% of the peak value. A I Remark 2. In North America residential ac. power is supplied at 60 Hz (cycles per second) and voltage of 115 and 230 volts. These voltages are rms values of the sinusoidal waveforms supplied by the power company, and are the voltages measured by standard a.c voltmeters. The concept of ms values applies to sinusoidal currents as well, so that the power delivered to our resistor by the current i(t) through it is given by §=R(%I:iz(t)dt)=RI;w and most ac. ammeters are calibrated to display the rms value of a current. I ...
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