y = f(x), below by y = -1 and on the sides by x = 1/2n and x = 1. This area n~x

Transcription

1 124 CHAP. II. AREAS Show that bcf (x) dx = Clb f (x) dx where C is a costat. Assume J thatf(x) is such that the defiitio of the itegral i (11.34) applies, ad use this defiitio to prove the above formula Cosider the fuctio y =f(x) which is represeted i Fig (Chapter I, Sectio 13). Fid the area of the regio which is bouded above by y = f(x), below by y = -1 ad o the sides by x = 1/2 ad x = 1. This area A depeds o. Does lim A,, exist? If so, what is this limit ad what does it represet? ~x 10. THE TRAPEZOIDAL RULE We have see i the precedig sectios that the area uder a curve ca be approximated by lower sums or upper sums, depedig o whether we wat a approximatio from below or above. The error at each subiterval is idicated i Fig. II.28 by the shaded area i case we take the lower sum, ad by the dotted area i case we use the upper sum, as a approximatio. Eve though this error becomes small ad ultimately vaishes as Ax = xk+l - xx. -> 0 for all k, i case the fuctio is well behaved i the iterval i which the area is sought, there is o reaso why we should't look for a better approximatio, i.e., a approximatio that yields for the same subdivisio a smaller error tha either the upper or the lower sum. Let us cosider Fig. II.29. If we joi the begiig poit Pl of the curve i the iterval xk < x < xk+i with its edpoit P2 by a straight lie ad approximate the area uder the curve by the shaded trapezoid, the we ca expect to obtai i geeral a more accurate result tha the oe obtaiable from a approximatio by a upper or a lower sum. Of course, there are cases where upper or lower sums do yield better results for a give subdivisio tha the trapezoidal approximatio would yield. We depict two such cases i Fig Y Xk Fig xk+1 x

2 TRAPEZOIDAL RULE 125 y Fig I Fig (a) the shaded area represets the error that is committed if the trapezoidal approximatio is used, ad the dotted area represets the error resultig from a approximatio by the lower sum (iscribed rectagle). I Fig. II.30(b) the dotted area represets the error due to a upper sum approximatio ad the shaded area agai idicates the error produced by a trapezoidal approximatio. It is obvious that, i both cases, the dotted area is smaller tha the shaded area, i.e., the error due to the trapezoidal approximatio is larger tha the error committed by the lower ad upper sum approximatios, respectively. However, the itroductio of additioal divisio poits will, i geeral, reverse this situatio. y y xk Xk+1 Xk k+1 x (a) (b) Fig

3 126 CHAP. If. AREA; Fig Let us ow derive a geeral formula for the approximatio of a area uder a curve by a sum of areas of trapezoids. First, we ote that the area of the trapezoid ABCD i Fig is give by (11.38) Area of trapezoid = b a + C (see Problem 11.12). This formula is quite obvious if we otice that the shaded triagles i Fig are cogruet. Now we cosider a fuctio y = f(x) of which we assume that it has o discotiuities i a S x S b (see Fig ). We divide the iterval a< x s b ito equal subitervals of legth Ox=b -a 2

4 TRAPEZOIDAL RULE 127 ad call the divisio poits x1i x2, x3,, x_1, puttig x0 = a, x,, = b. Clearly, xk=a+kb a fork=0,1,2,. At each oe of these divisio poits we erect the ordiate Yk = f (xk), joi each pair of adjacet poits (xk, yk), (xk+1, yk+1) for all k = 0, 1, 2, - 1, by a straight lie ad compute the area of every oe of the trapezoids thus obtaied (see Fig ). Clearly, we obtai for the area Ak of the (k + 1)th trapezoid accordig to (11.38) Ak=Axyk+1+yk 2 for all k = 0, 1, 2, f(b) =f(x) = y,. Now we sum all these areas ad obtai, - 1, where we agree that f (a) = f (xo) = yo ad Ox( Ax(Zo+y1+y2+...+y-1+ 2") Sice Ax = b - a, we obtai the followig approximatio for the area Fie

5 128 CHAP. II. AREAS uder the curve y = f (x) betwee x = a ad x = b : _ (I1.39) Jf(x) dx - b a (yo + 2y1 + 2y y-1 + y). This formula is called the trapezoidal rule for quite obvious reasos. I Sectio 8 we approximated the area uder the curve y=x(1-x)+1 betwee x = 0 ad x = 1 by a lower sum with 16 subitervals ad obtaied $16 = Let us ow use the trapezoidal rule with the same umber of subitervals for the same problem. We have 1 1 _3 _1 _`16 5 xo0,x116,x28,x3 16'x4 4'x5 x6 ad, accordigly, YO _11 8, x7 16 ' x$ 2 ' x9 16 ' x1o 8 ' x11 _3 13 _7 _15 x12 4,x13 16,x14 8,x15 16,x16= ,y1=256y2-256'y3=256'y4=256y5= Y6 256 ' y' ' YS Hece, we obtai rl Jo [x(1 - x) + 1] dx 1 32 = z for k = ',k - y16-k,,, We will see i Sectio 13, Problem 11.81, that the true value of this itegral is foud to be Thus we see that the approximatio which we just obtaied is cosiderably better tha the approximatio as obtaied from the lower sum i Sectio 8. Let us ow fid as aother applicatio of the trapezoidal rule a approximatio to the value of 7r, i.e., the area of the uit circle. The uit circle with the ceter i the origi has the equatio y=1'1 -x2. 256

6 II.10. TRAPEZOIDAL RULE 129 We cosider oe quadrat of this circle above the x-axis betwee x = 0 ad x = 1 ad choose = 32. We have ad obtai accordigly the followig values for yk. 1/0 = 1 21/9 = /17 = y25 = y1 = /10 = /18 = y26 = y2 = y11 = y19 = y27 = /3 = y12 = /20 = /28 = y4 = y13 = y21 = /29 = /5 = /14 = /22 = y30 = /6 = /15 = /23 = y31 = /7 = y16 = /24 = /32 = 0. 21/s = Hece, we obtai with (11.39) ad, therefore, 7r 1 4 =J Ji - x2 dx_ = r ti This result deviates from the true value of IT = by which is ot as accurate as we might have expected it to be. The reaso is that the trapezoidal approximatios to the circle do ot yield very good results i the eighborhood of the poit (1, 0). Problems (' Fid a approximatio to J x10 dx by usig the trapezoidal rule with 0 = 4 ad compare it to the result that is obtaied from a approximatio by the lower sum S4. (True value of the itegral is II.61. A fuctio f (x) is said to be mootoically icreasig if, for ay pair of values x1 < x2, the relatio f(xl) < f(x2) holds. Show that the trapezoidal rule applied to a mootoically icreasig fuctio is obtaied by takig the arithmetic mea of lower sum ad upper sum. (Hit: Note that, if f (X) is mootoically icreasig, the max f (x) = 1/k+1 i Xk < X < xk+l, mi f (x) _ 1/i i xk < x < xk+l.) 1'1 II.62. Fid a approximatio to J x2 dx by usig the trapezoidal rule. Start 0 with = 2 ad cotiue doublig the umber of subitervals util o more chages of the result i the third decimal place ca be expected.

7 130 CHAP. IT. AREAS Prove that the trapezoidal rule yields the exact result if applied to the itegratio of all liear fuctios, i.e., fuctios of the type y = ax + b. ri Use the trapezoidal rule to fid a approximatio to = 4J V! - x2 dx 0 takig 16 subitervals betwee 0 ad j ad 32 subitervals betwee j ad 1. Compare your result with the oe obtaied i this sectio ad the oe obtaied i Sectio INTEGRATION BY A LIMIT OF A SUM PROCESS We will devote this sectio to the discussio of two very simple examples, i which we will compute the area of a rectagle ad a triagle by a itegratio process, i.e., by the evaluatio of the appropriate limits of lower sums, ad thus demostrate how the formulas for the area of these basic geometric cofiguratios ow appear as special cases of our geeral defiitio of area as give i (11.34) (Sectio 8). (It would be awkward, ideed, if this were ot so. After all, the geeral defiitio of area as give i Sectio 8 was based o the formula for the area of a rectagle, ad the formula for the area of a triagle, i tur, was developed from the formula for the area of a rectagle-see Sectio 2.) First, let us cosider a rectagle with the dimesios a ad b. We choose our coordiate system so that the rectagle takes the positio as idicated i Fig We ca iterpret this problem as the problem of fidig the area of a regio which is bouded above by y = b (the lie parallel to the x-axis at distace b from the x-axis) betwee x = 0, x = a ad above the x-axis. If we subdivide the iterval 0 < x < a ito subitervals of legth Ax = a, we obtai i every oe of these subitervals mi f (x) =f(k) = b ad y maxf(x) = f(y7k) = b, y=b -> x Fig

8 INTEGRATION BY LIMIT PROCESS 131 (a, b) a Fig i.e., f Therefore, (11.40) f (Y7k) = b ad, cosequetly, S=gbAx=ball=ab. k=1 k=1 f0a b dx =lims =lims =limab = ab, - oo as was to be expected. Next, we cosider a right triagle where the two sides formig the right agle have the dimesios a ad b. We choose our coordiate system i such a way that the triagle will take the positio as idicated i Fig. II.35. The lie through (0, 0) ad (a, b), formig the hypotheuse of the triagle, has the equatio Thus we face the problem of fidig the area uder the curve y = b x a betwee x = 0, x = a ad above the x-axis. A subdivisio of the iterval 0 < x < a ito equal subitervals will yield a udersum (lower sum), as idicated for = 6 i Fig I geeral, we have clearly 1=o, ad, cosequetly, J 0,.f 2=Th b a 2a -1 =-,..., = a, f 2b ( - 1)b..., f () _.

9 2-F Thus with Ox = a we have CHAP. II. AREAS s-a[,+b+2b+3b+..,+(- 1)b LL =ab[ (-1)]. I order to discover the limit of this expressio as -* oo, we have to evaluate the sum (-1) for ay ad represet this sum i terms of. For this purpose, we rewrite the sum i the more elaborate form 1 1). If we add the first ad the last term, we obtai. If we add the secod ad the ext to the last term, we obtai. If we add the third term ad the term precedig the ext to the last, we obtai agai, etc... Now, if is odd, we obtai exactly 2 1 such sums, i.e, the sum of this series is give by y 1..(-1) (-1) 2 Fig

10 II.11. INTEGRATION BY LIMIT PROCESS 133 Suppose is eve. The we obtai 2 2 such sums ad the term 2 i the middle is left out. Thus we obtai altogether (-1)= =(2 1) which is the same formula as the oe obtaied above for odd. Thus we ca state that (-1)=(-1) 2 for ay iteger. Hece, we obtai S _ ad we see easily that _ab(-1)ab2-=ab(1-1) ab (11.41) f " b ab(1 1) ab. 1) x dx lim S lim lim (1 p a -fcc 2 2 as we expected to obtai. Next we cosider the geeral case where the triagle is ot ecessarily a right triagle. We choose our coordiate system so that the origi coicides with the vertex A at which the triagle has a obtuse agle, if it has a obtuse agle at all (ote that a triagle has at most oe obtuse agle). If there is o obtuse agle, the we let the origi coicide with just ay vertex. Further, we choose the x-axis so that it coicides with the height of the triagle through A (see Fig ). With the otatio itroduced i Fig , we have p + q = a. The equatio of the lie through AC is give by ypx where h = b by the theorem of Pythagoras ad the equatio of the lie through AB is give by Therefore, the area of the triagle (ABC) is give by It h fo hxdx+ fo J- hxj dx.

11 134 CHAP. If. AREAS Y Fig I view of (11.41), where we ow write h istead of a ad p istead of b i case of the first itegral, ad h istead of a ad q istead of b i case of the secod itegral, the area of the triagle (ABC) is give by hp + hq = h(p + q) = ha Sice a is the base of our triagle, we thus obtai the well kow formula Area of triagle = 2 base x height. Problems Evaluate the followig sums: (a) I k (b) k k=1 k= (c) 1 k (d) 1 (k + k k1 k Fid the area of a right triagle with sides 3, 4, 5 by a limit of a sum process Fid the area of a triagle with sides 3, 5, 7 by a limit of a sum process.

12 1I.12. FINITE SUMS Fid the area of a regio bouded by y = [x], the x-axis, x = 0 ad x = 10 by a limit of a sum process Cosider f (x) = x2 i the iterval 0 <_ x < 1. Fid S., for the area uder y = x2, betwee x = 0, x = 1 ad above the x-axis. Use equal subitervals ad evaluate the lower sum for = 4, 8, 32. Make a cojecture about the limit. 12. FINITE SUMS I the precedig sectio we foud that -1 ( - k=1 2 by a method which C. F. Gauss ( ) purportedly iveted i grade school whe his icompetet teacher required the class to add up all the umbers from 1 to 100 i the hope that this would keep the class busy for a while. It is almost eedless to say that Gauss had the last laugh. I the followig sectio we will eed a formula for the sum -1 k2= (-1)2. k=1 I order to evaluate this sum, we cosider the followig two sums, which appear to pose a more complicated problem (ad do, as a matter of fact, if we were out to evaluate them), amely, ad -1 k=1-1 k=1 '10= (-1)3 Clearly, -1-1 (11.42) k3 - (k - 1)3 = ( - 1)3. k=1 k=1 O the other had (11.43) 1 k3 - I (k - 1)3 = 1 [k3 - (k - 1)3] k=1 k=1 k=1 Sice 1(k-1)3= (-2)3. -1 (k3-k3+3k2-3k+1) k=1-1 _ (3k2-3k + 1). k= (3k2-3k+1)=3k2-31k+I1 k=1 k=1 k=1 k=1

13 136 CHAP. II. AREAS ad -1 I k = ( - 1) k=1 2 k=1 we have i view of (11.42) ad (11.43) I1=(-1), (See Problem 11.23), 3 I k2-3 ( - 1) + ( - 1) = ( - i)3 k=1 2 ad, cosequetly, after shiftig of terms ad dividig by 3, k=1 ) + (- )-- (- 3 r =(-1)[2(- 1) L 6 J =(- 1)r LL 6 = ( - 1)(2-1) Let us list all the summatio formulas which we obtaied so far for later referece purposes: -1 (11.44) 57.1=(-1) k=1-1 ( - 1) (11.45) _7 k = k=1 2 (11.46) -1 2 = ( - 1)(2-1) k=1 6 Problems II Evaluate 101 k 1000 (a) A (b) A(k + 1) I k2 (d) I k2 k=1 k= Fid a formula for k3. [Hit: Cosider the sums 10 ad -1 k=1 k=1 A (k - 1)4 ad proceed as i the text.] b:= Fid a formula for (2m+1). [Hit: Observethat m =2( m).] 6

14 AREA UNDER PARABOLA Evaluate Evaluate ()2 + (2a) r37r oo. (( lea I2 8r3,T AREA UNDER A PARABOLIC ARC Let us cosider the problem of fidig the area A which is bouded by a ope rectagle with a parabolic arc beig the supplemetary boudary. First, we cosider the problem of fidig the area uder the parabola y=x2 betwee x = 0, x = a ad above the x-axis (see Fig. II.38). We subdivide the iterval 0 < x < a ito equal subitervals of legth a. coordiates of the divisio poits are xl, a x2=-1 2a x3=_i 3a...,x-1= ( - 1)a ad for the sake of uiformity we put y 0 = x0 ad a=x. The the x1 X2 X3 Fig

15 138 We ote that mi f (x) = f (xk) i xk C x C xk+l CHAP. II. AREAS ad obtai, therefore, for the lower sum i view of / f(xk) = { the followig expressio 2 2 J (k) =,l2 s=0+a( +_ - + a ( - 1)a 12 )2 (a)2 11 fl a3 a3 ( - 1)(2-1) =3[ (-1)2]=3 6 3 ( ) = 63 or, as we may write, Sice we have S 3 ( lim(1-3 + )=1 - co 2 22 (11.47) A= x2 dx = It follows from (11.47) right away that 0 a f x2dx 3 which represets the area uder the parabola y = x2, betwee x = 0, x = b, ad above the x-axis. Hece, we obtai for the area uder the parabola y = x2 betwee x = a, x = b, ad above the x-axis (11.48) x2dx 0 a3 = a 3 (see Fig ). This formula, together with the formulas (11.40) ad (11.41) of Sectio 11, will eable us to fid the area uder a parabolic are (11.49) y = Ax2 + Bx + C b3 a

16 AREA UNDER PARABOLA 139 Y x = a x=b Fig betwee ay poit x = a ad ay other poit x = b, if we also assume that the fuctio i (11.49) is positive betwee a ad b. (Otherwise, we would have to divide the itegratio iterval as we have demostrated i Sectio 9.) We eed for this purpose a property of the defiite itegral which ca be easily established, amely, that (11.50) J [f(x) + g(x)] dx =J f(x) dx + bg(x) dx. b b a a a We will prove this relatio for mootoically icreasig fuctios f (x) ad g(x) oly, but we wish to poit out that it is true i geeral-provided that the two itegrals o the right exist. If f (x) ad g(x) are mootoically icreasig (for defiitio of mootoically icreasig fuctios, see Problem 11.61), it is clear that mi [f(x) + g(x)] = f(xk) + g(xk) [= mi f (x) + mi g(x)] i ay iterval xk < x < xk+1. Therefore, --`1-1 / -1 s = Ox G [f (xk) + g(xk)] = Ox.1 f (xk) + Ax Y_ g(xk) k=0 k=0 k=0 If the limits of both sums o the right side of this relatio exist as -> cc, which we will assume, the ivokig our rule accordig to which the limit

17 140 CHAP. If. AREAf of a sum of two sequeces is equal to the sum of the limits of the twc sequeces, provided that both limits exist (see Sectio 4), we have that -1-1 b b lim S = lim Ax I f (xk) + I'M Ox I g(xk) = f f (x) dx + f g(x) dx. -+oo -oo k=0 k=0 Ja Ja O the other had, lim S = f b[f (x) + g(x)] d x 'm Ja ad (11.50) follows. Now, if we wish to itegrate the fuctio i (11.49), we make use of the relatio (11.50) as follows: f Jb(Ax2 + Bx + C) dx = f Ax2 dx + f t(bx + C) dx. a a a We apply (11.50) agai to the last itegral o the right ad obtai b b b b (Ax2+Bx+C)dx=J Ax2dx+ f Bxdx+rCdx. a a a Ja We have see i Problem that where k is a costat. Hece f bkf(x) dx = kb f(x) dx a a f(ax2 + Bx + C) dx = A fbx2 dx + B f bx dx + C f.bdx..' a a a From (11.48) it follows that From (11.40) ad (11.41) i Sectio 11 it follows that b L 62 a2 f xdx= ad f dx=b - a a 2 a (see also Problems ad 11.76). Thus we fially arrive at the formula (I1.51) f. b x2 dx = b3 - a3 b 3 f (Ax2+Bx+C)dx=Ab -a3 b 2 -a 2 + B +C(b-a). a 3 2 3

18 AREA UNDER PARABOLA 141 Problems We have see i Sectio 11 that b fj 0 b dx = ab. Derive from this formula that f dx = b ad, i tur, dx = b - a. 0 Ja a We have see i Sectio 11 that x dx = ab I. Derive from this result that 0bx dx = b 2 2 ad, i tur, b x dx = Evaluate : a b a 2 b2 - a2 2. ('1 0 (a) J (4x2 + 6x + 3) dx (b) I (x2-12x + 1) dx 0 g (c) (3x2 + 2x + 4) dx (d) fo I101(x2-500) dx Fid the area of the regio betwee the curve y = f(x), the x-axis, ad x=a,x=b: (a) f(x) = x2-1, a = -2, b = 2 (b) f (x) = x2-3x + 2, a = 0, b = 4 (c) f (X) = x2 + 3x + 2, a = -5, b = -1 (d) f(x) = x2 - x, a = -1, b = 1 Note that these fuctios chage sig i the itegratio iterval Determie A, B, C so that the parabola loo y=axe+bx+c passes through the poits (1, 1), (2, 2) ad (3, 5) Fid the area uder the parabola i Problem betwee x = 1 ad x=3. Cl Fid J [x(1 - x) + 1] dx ad compare the result with the values which 0 we obtaied for the udersums i Sectio 8 ad from the trapezoidal rule i Sectio Fid the area of the regio which is ecompassed by the parabola y = x2 ad the lie y = x. (Sketch the regio.) Fid the area of the regio which is ecompassed by the two parabolas (Sketch the regio.) y=x2ad y=4x2+1.

19 142 CHAP. II. AREAS 14. SIMPSON'S RULE I Sectio 10 we derived the so-called trapezoidal rule which yields for a give subdivisio a approximate value to the defiite itegral. We arrived at this rule by approximatig the fuctio y = f (x) to be itegrated by a polygo ad the, i tur, by approximatig the area uder the curve y = f (x) by a sum of areas of trapezoids. It seems quite obvious that we ca obtai better approximatios to the area, if we use a closer approximatio to the curve tha the oe furished by a polygo. This is exactly what we are goig to do ow. Agai we subdivide the iterval a < x < b by divisio poits x1, x2, x3,, x,,_1 ito equal subitervals of legth Ax = b a but we will assume ow that is a eve umber for reasos which will soo become obvious. Now we erect the ordiates at all divisio poits (see Fig ). The ordiates at divisio poits with a eve subscript we represet by a solid lie ad the ordiates at divisio poits with a odd subscript we represet by a broke lie. Now, istead of replacig the curve withi each subiterval by a straight lie, we will replace the curve betwee each two cosecutive divisio poits with a eve subscript x2k, x2k+2 by a parabola y = Axe + Bx + C which passes also through the poit [x2k+l,f(x2k+1)). It seems quite apparet that a much better approximatio is obtaied i this maer. Now all we have to do is fid the equatio of the 2 parabolas ivolved, apply formula Y -x Fig. H.40

20 SIMPSON'S RULE 143 (11.51) from Sectio 13 to fid the areas uder those parabolas, ad form the sum of all these areas. Agai, we let a = x0, b = x,, to uify the otatio. Further, we let f(xk) = y7,, k = 0, 1, 2, 3,...,. Now let us cosider the subiterval x2k < x < x2k+2 (see Fig. II.41). We have to fit a parabola y = Ax2 + Bx + C through the three poits (x2k, y2k), (x2k+1, Y2kt1), (x2k+2+ Y2k+2) ad the fid the area uder the parabolic are. We ote that the area uder this parabolic arc depeds oly o the distace Ax betwee cosecutive divisio poits ad the legths of y2k, Y27,t1 ad y2k+2 It is quite immaterial where x2k+1 is located relative to the origi. We ca simplify our problem to some extet if we assume that x2k+1 = 0, i.e., shift the etire cofiguratio i Fig by x27,+1 uits to the left (see Fig ). Sice Ax = b - a we have ow x2k = -Ax = - b a ad x27,+2 = Ax = b a Thus we obtai for the area uder the parabolic arc y = Ax2 + Bx + C betwee X2k ad x2k+2 i view of (11.51), (11.52) Ox (Ax2 + Bx + C) dx = A (Ax)3 J-oX + C[Ax - (-Ax)] 3 = 2A (Ox) + 2CAx. 3 3(-Ax)3 + B (6.x)2 -( -0x)2 We ote that B has dropped out etirely due to the shift of x2k+1 ito the origi of the coordiate system ad thus oly A ad C are left to be deter. mied so that y = Ax2 + Bx + C passes through the three poits (-Ax, y2k), (0, Y2k+1), (Ax, y2k+2) Y2k = A(-Ax)2 + B(-Ax) + C y2k+1 = C Y2k+2 = A(0x)2 + B(Ax) + C. Additio of the first ad the last equatio yields (11.53) Y2k+2 + Y2k = 2A(0x)2 + 2C.

21 144 CHAP. II. AREAS x 2k x2k+1 Fig X2k+2 From the secod equatio we have (11.54) C = Y2k+1 ad hece, i view of (11.53), (11.55) A =?/2k+2 +?12k - 2Y2k+1 2(Ax)2 Fig

22 SIMPSON'S RULE 145 If we substitute the values for A ad C from (11.55) ad (11.54) ito (11.52), we see that the area A2k uder the parabolic are betwee x2k ad x2k+2 is (11.56) A2k = Y2k+2 + 1/2k - 2y2k+1 2( x)3 + 2y2k+1Ax 2(0x)2 3 = (Y27,+2 + y2k - 2y2k+1 + 2y2k+1 Y2k+2 + 4y2k+1 + y2kax r Ax = 3 / 3 Now we sum all these areas A0, A2, A4, A-2 ad obtai -1 2 Ox L,o A2k = 3 [(Yo + 4y1 + y2) + (Y2 + 4y3 + Y4) + (y4 + 4y5 + y6) (y-4 + 4y-3 + y-2) + (y-2 + 4y-1 + y)] Ax = 3 (yo+4y1+21/2+4y3+2y4+4y5+2ys+...+2y-4 + 4y-3 + 2y /-1 + 1/). Thus we obtai the followig approximatio to the defiite itegral of the fuctio f (x) betwee a ad b, if we also substitute for Ax its value Ax=b -a ('b (11.57) J.f(x)dx-b3a(yo-f 41/1-F 2y2+4y3-f a + 2y y-1 + y) This is called Simpso's rule. This rule is easy to remember: the first ad the last ordiate yo ad y are each take oce, all other eve-umbered ordiates are take twice ad all the odd-umbered ordiates are take four times. I order to obtai some isight ito the effectiveess of this formula, we will try it out i computig a approximatio of the area of oe quadrat of the uit circle, which will yield a approximatio to 4. 2y4+... We will see that the approximatio which we will thus obtai is better tha the oe we arrived at by usig the trapezoidal rule i Sectio 11, but ot as good as the oe obtaied i Sectio 5. The uit circle has the equatio y= 1 -x2. We subdivide the iterval 0 < x S 1 ito 32 subitervals ad compute the values of yk by makig use of square root tables. * * Stadard Mathematical Tables, CRC, 12th ed. pp

23 146 CHAP. II. AREAS We obtai : 1/o = 1 32y1 = Y2 = /3 = y4 = /5 = /6 = y7 = /8 = y9 = y10 = y11 = /12 = /13 = y14 = /15 = /16 = /17 = y18 = y19 = Y20 = /21 = /22 = /23 = /24 = y25 = /26 = /27 = Y28 = /29 = /30 = y31 = Thus, 1/32 = 0 k=15 k= Y2k = , 4 1 y2k+1 = k=1 k=0 ad cosequetly, after elemetary operatios 7r ti as compared to the value which was obtaied by the trapezoidal rule usig the same umber of subitervals. Problems Use Simpso's rule to fid f 1 (2x2+3x+1)dx. Take = 2, 4, 8. How do you always obtai the same result? Use Simpso's rule to fid a approximatio for dx.

24 U. 15. INTEGRAL WITH VARIABLE LIMIT Use Simpso's rule to fidjsixdx. a approximatio to Take = 4. (For values of si x, see Appedix III, Sectio 2.) Use the trapezoidal rule for the itegral i Problem with the same umber of divisio poits. The true value of this itegral is 2. Compare the results. 15. DEFINITE INTEGRAL WITH A VARIABLE UPPER LIMIT We have see i Sectio 13 that the area uder a parabolic arc y = x2 betwee x = a ad x = b is give by fbx2 dx = b3 - a3. a 3 3 Let a = 0 i the followig discussio for reasos of coveiece. Now let us cosider various values of b ad evaluate this area i every istace. First, let b = 0, the we obtai Next we let b = J. The 0 x2dx=0. f0 For b = 1, we obtai f x2dx = 1. Jo 3 2 Forb=2, f x2dx= Forb=3, f x2dx= For b = 9, fo x2dx= For b = 12, f x2 dx = 576, etc. 0i 24 f x2 dx 1 b We see that the value of J x2 dx depeds o the value which we assig to rb 0 b, or as we ca say: J x2 dx is a fuctio of b. This is also quite clear from 0 lookig at Fig. II.43, where we have draw three vertical lies at three