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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 BoisRaymond, 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;
geomelexs . G. Darboux, Alémoire sur [es fonctions discontinues, 1875 The rigorous proof of Riemann’s theory of integrations, and its extensions byDuBoisReymond 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 subintervals, 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 subinterval 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) = illlaid" (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 subinterval (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 subinterval [xi1, 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 (DuBoisReymond, 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) (57.0: ) 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 :fa,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$ ) Ce. ;(—x)=§v).
Then Q
(3.7.14 ) “(mix = o.
—m Examples of. apylicahbns 0+ lkese browHes 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' (Frequenily 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
(3.82 ) 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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 Fall '07
 HARMSWORTH
 Math, Calculus

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