5.1 - 5 Cl INTEGRALS 5.1 Areas and Distances 1. (a) Since f...

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Unformatted text preview: 5 Cl INTEGRALS 5.1 Areas and Distances 1. (a) Since f is increasing, we can obtain a lower estimate by using left endpoints. We are instructed to use five rectangles. so 71 : 5. 5 L5 : Elf($i—1)Az [A302 17;; 2 105—0 :2] =f($0)‘2+f($1)'2+f(332)'2+f($3)‘2+f($4)‘2 =2lf(0)+f(2)+f(4)+f(6)+f(8)l m2(1+3+4.3+5.4+6.3) 22(20) =40 Since f is increasing, we can obtain an upper estimate by using right endpoints. R5 : :flxifllm : 2lf($1) + f($2) + f(963) + flu) + f($5)] : 2[10(2) + NH + f(6) + f(8) + f(10)] z 2(3 + 4.3 + 5.4 + 6.3 i 7) = 2(26) 2 52 Comparing R5 to L5, we see that we have added the area of the rightmost upper rectangle. f (10) - 2. to the sum and subtracted the area of the leftmost lower rectangle, f (0) ~ 2. from the sum. 10 (b) L10 = f(:6i_1)Am [Ax = 10130 :1] I1lf($0)+f($1)+“'+f(w9)l :f(0)+f(1)+m+f(9) z1+2.1+3+3.7+4.3+4.9+5.4+5.8+6.3+6.7 2432 R10 : iflxflAm:f(1)+f(2)+~--+f(10) _ L + 1 f (10) 1 flo) add rightmost upper rectangle. — 10 ‘ 7 ~ subtract leftmost lower rectangle =43.2+7~1:49.2 433 434 U CHAPTERS INTEGRALS 6 6 2- (a) L6 = Z f(:c;,1)Aa: [Ax : 12—0 : 2] 1:1 : 2lf($0) + f($1)+ “302) + f($3) + 1%“) + f($5)l 2 man + M) + M) + H6) + M) + f(10)l z 2(9 + 8.8 + 8.2 + 7.3 + 5.9 + 4.1) : 2(43.3) : 86.6 (ii) R6 : L6 + 2 ~ f(12) — 2 - f(0) m 86.6 + 2(1) 7 2(9): 70.6 [Add area of rightmost lower rectangle and subtract area of leftmost upper rectangle] 6 (iii) M6 = Z fact) Am i=1 : 2[f(1) + f(3) + f(5) + f(7) + MD + NH] z 2(8.9 + 8.5 + 7.8 + 6.6 + 5.1 + 2.8) = 2(397) : 79.4 (b) Since f is decreasing, we obtain an overestimate by using lefl endpoints; that is. L6. (c) Since f is decreasing. we obtain an underestimate by using right endpoints; that is. R5. (d) MG gives the best estimate. since the area of each rectangle appears to be closer to the true area than the overestimates and underestimates in L6 and R6. || | ,_. ._. 4 3. (a) R4 : Z f(:c,)A:c [Am 511 7 11:1 =f($1)'1+f($2)'1+f(5€3)'1+f(934)'1 :f(2)+f(3)+f(4)+f(5) :g+§+i+%:%%=1-28§ Since f is decreasing on [1. 5]. an underestimate is obtained by using the right endpoint approximation. R4. 4 L4 : Z f(£l?1_1)A£C 1:1 : f(1) + f(2) + f(3) + f(4) :1+§+§+§=%:2.08§ L4 is an overestimate. Alternatively. we could just add the area of the leftmost upper rectangle and subtract the area of the rightmost lower rectangle; that is. L4 2 R4 + f(1)~17 f(5) - 1. SECTION 5.1 AREAS AND DISTANCES C1 435 5 570 4. (a) R5 2 ;f(:l)l)AZII [13:10: —5— :1] = f($1)-1+f(r2)~1+f(933)~1+f($4)-1+f(w5)-1 :f(1)+f(2) +f(3)+f(4)+f(5) :24+21+16+9+0=70 Since f is decreasing on [0. 5]. R5 is an underestimate. 5 (13) L5 : Z f($z>1)AI i=1 : f(0) +f(1) +f(2) +f(3) +f(4) :25+24+21+16+9:95 L5 is an overestimate 5. (a)f(x):1+x2andAx=2;§E_—l) =1 : R3:1-f(0)+11f(1)+1-f(2):1v1+1~2+1-5:8. Ax:%fl:0.5 => Rs 2 0-5[f(—0‘5) + W) + f(0.5) + W) + f(1~5) + f(2)] : 0.5(125 +1+1.25 + 2 + 3.25 + 5) : 0.5(13.75) : 6.875 (b)L3 :1~f(—1)+1-f(0)+1-f(1)=1~2+111+11225 L6 : 0~5[f(—1)+ f(—0-5) + NH +f(0.5) +f(1) +f(1.5)] : 0.5(2 +1.25 + 1 +1.25 + 2 + 3.25) :0.5(10.75) : 5.375 (c) M3 : 1~f(70.5)+1-f(0.5)+ 1 ~ f(1.5) :1-1.25+1~1.25+1~3.25:5.75 [We : 0.5[f(—0.75) + f(—0.25) + f(0.25) + No.75) + f(1.25) + f(1~75)] : 0.5(1.5625 + 1.0625 + 1.0625 + 1.5625 + 2.5625 + 4.0625) : 0.5(11.875) : 5.9375 (d) M6 appears to be the best estimate. 436 l: CHAPTERS INTEGRALS 6. (a) 2 -- *1 (b) f(:c) : €42 and Ax = 2—_(‘—2) :1 => (1) R4 :1-f(71)+1~f(0) (ii) M4 :1-f(—1.5)+ 1~f(—0.5) +1 ~ f(1) + 1 -f(2) +1 - f(0.5) +1 -f(1.5) : e—l +1+e—1 +e‘4 : €72.25 +e~0125 +e—O.25 +6—2.25 m 1.754 m 1.768 (c) (1)128 : 0-5lf(—1-5) + f(—1)+ f(*0~5) + MD + mm + f(1) + f(1-5) + f(2)] : 6—2125 +671 +e—O.25 + 1 +67025 +e—1 +6—225 +6 2 1.761 (ii) Due to the symmetry of the figure. we see that Ms : (0.5)(2)[f(0.25) + f(0175)+ f(1.25) + f(1.75)] : e7010625 +e—O.5625 +e—l,5625 +63.0625 m 1.766 7. Here is one possible algorithm (ordered sequence of operations) for calculating the sums: 1 Let SUM : 0. XiMIN : 0. X_MAX : 7r, N 2 10 (or 30 or 50. depending on which sum we are calculating). DELTA7X = (XfMAX — XfMIN)/N, and RIGHT_ENDPOINT : X_MIN + DELTA_X. 2 Repeat steps 2a. 2b in sequence until RIGHT_ENDPOINT > X_MAX. 2a Add sin (RIGHTVENDPOINT) t0 SUM. 2b Add DELTA7X to RIGHT_ENDPOINT. SECTION 5.1 AREAS AND DISTANCES 437 At the end of this procedure, (DELTA_X) ~ (SUM) is equal to the answer we are looking for. We find that 7r 10 in 7r 3° . 1'71" 7r 50 . z'7r ___ -__ N :__ _ m _ , R =__ —— z1%%. R10 7 7; sm<10> ~ 1.9835. R30 30 1; sm ( 30> 1 9982 and 50 50 i; s1n 50 It appears that the exact area is 2. Shown below is program SUMRIGHT and its output from a TI—83 Plus calculator. To generalize the program. we have input (rather than assigned) values for Xmin. Xmax. and N. Also. the function. sin m. is assigned to Y1. enabling us to evaluate any right sum merely by changing Y1 and running the program. PRDGRHM:SUMRIEHT =B+S , :PPDNPL Hm1n :PPOMPt Xmax =PPOMPt H_ =(Hmax-HM1HJIH+D =Hm1n+D+R =FDP(I:1:N) :S+91(R)+S =R+D+R =End :D$S+E =D15P 2 PPSNSUHRIGHT Xm1n=?B Xmax=?n N=?1B 1.98352353? Done . We can use the algorithm from Exercise 7 with X_MIN : 1. X_MAX : 2. and 1 /(RIGHT_ENDPOINT)2 instead 1 1° 1 f ' RIGHT ENDPOINT in ste 2a. We find that R : — u «8 0.4640. 0 sm( _ ) p 10 101; (1+i/10)2 1 30 77 dB 1 5 1 04926 I h h R :— ——20.48 .an :— fix . . ta earstatteexactarea 3” 30.23 (Hi/30)2 5° 50 2; (I-tr/50)2 pp is . In Maple, we have to perform a number of steps before getting a numerical answer. After loading the student package [command: with (student) ;] we use the command left_sum: =1eftsum (x‘ (1/2) , x=1 . . 4 , 10 [or 30. or 50]); which gives us the expression in summation notation. To get a numerical approximation to the sum. we use evalf (1eft_sum) ; . Mathematica does not have a special command for these sums. so we must type them in manually. For example. the first left sum is given by (3/10) *Sum[Sqrt [1+3 (i—1)/10] , {i,l,lO}].and we use theNcommand on the resulting output to get a numerical approximation. In Derive. we use the LEFT_RIEMANN command to get the left sums. but must define the right sums ourselves. (We can define a new function using LEFT_RIEMANN with k ranging from 1 to 71 instead of from 0 to n ~ 1.) (a) With f(x) : 1 S at S 4. the left sums are of the form Ln = E Z V 1 + 3(2 ~ 1). Specifically. n 1':1 n L10 :1 4.5148. L30 % 4.6165. and L50 % 4.6366. The right sums are of the form Rn : g 2 1+ —. Specifically, R10 8 4.8148. R30 % 4.7165. and R50 % 4.6966. 438 C CHAPTERS INTEGRALS (b) In Maple. we use the leftbox and rightbox commands (with the same arguments as leftsum and right sum above) to generate the graphs. 2.1 2.1 1 1 4 0 0 left endpoints. n : 30 left endpoints. n : 50 2.l 2.1 1 1 l 4 0 0 right endpoints, n : 10 right endpoints. n = 30 right endpoints. n : 50 (c) We know that since fl is an increasing function on (1. 4). all of the left sums are smaller than the actual area. and all of the right sums are larger than the actual area. Since the left sum with n = 50 is about 4.637 > 4.6 and the right sum with n : 50 is about 4.697 < 4.7. we conclude that 4.6 < L50 < exact area < R50 < 4.7. so the exact area is between 4.6 and 4.7. 10. See the solution to Exercise 9 for the CAS commands for evaluating the sums. n ' i 1 (a) With f(a:) : sin(sin a3), 0 g a: S the left sums are of the form Ln 2 21 2 sin (sin In 711:1 particular. L10 R: 0.8251. L30 z 0.8710, and L50 % 0.8799. The right sums are of the form Rn : 21 2 sin (sin . In particular. R10 % 0.9573. R30 % 0.9150. and R50 z 0.9064. n [:1 TL (b) In Maple. we use the leftbox and rightbox commands (with the same arguments as leftsum and right sum above) to generate the graphs. N|=1 left endpoints, n : 10 left endpoints. n : 30 left endpoints. n : 50 11. 12. 13. 14. 15. 16. SECTION 5.1 AREAS AND DISTANCES 3 439 Ill Ill right endpoints. n 2 50 7'» II right endpoints. n 2 10 NH right endpoints. n 2 30 (c) We know that since sin(sin as) is an increasing function on (0. [this is true because its derivative. — cos(sin m)(— cos x). is positive on that interval]. all of the left sums are smaller than the actual area. and all of the right sums are larger than the actual area. Since the left sum with n 2 50 is about 0.8799 > 0.87 and the right sum with n 2 50 is about 0.9064 < 0.91, we conclude that 0.87 < L50 < exact area < R50 < 0.91. so the exact area is between 0.87 and 0.91. Since 11 is an increasing function, L6 will give us a lower estimate and R6 will give us an upper estimate. L6 2 (0 ft/s)(0.5 s) + (6.2)(0.5) + (10.8)(0.5) + (14.9)(0.5) + (18.1)(0.5) + (19.4)(05) 2 0.5(69.4) 2 34.7 ft R5 2 0.5(62 + 10.8 + 14.9 + 18.1 + 19.4 + 20.2) 2 0.5(896) 2 44.8 ft (a)dmL5 2 (30ft/s)(125)+28-12+25-12+22~ 12+24- 12 2 (30+28+25+22+24)-122 129-1221548ft (b)dzR5 2(28+25+22+24+27)-122126-122 1512ft (c) The estimates are neither lower nor upper estimates since u is neither an increasing nor a decreasing function of 25. Lower estimate for oil leakage: R5 2 (7.6 + 6.8 + 6.2 + 5.7 + 5.3)(2) 2 (31.6)(2) 2 63.2 L. Upper estimate for oil leakage: L5 2 (8.7 + 7.6 + 6.8 + 6.2 + 5.7)(2) 2 (35)(2) 2 70 L. We can find an upper estimate by using the final velocity for each time interval. Thus. the distance d traveled after 62 seconds can be approximated by 6 d 2 Z v(t.)At.,- 2 (185 ft/s)(10 s) + 319 ~ 5 + 447 ~ 5 + 742 ~ 12 + 1325 ~ 27 + 1445-3 2 54.694 ft 121 For a decreasing function, using left endpoints gives us an overestimate and using right endpoints results in an underestimate. We will use M6 to get an estimate. At 2 1, so .Me 2 1[v(0.5) + 21(15) + v(2.5) + v(3.5) + 22(45) + v(5.5)] z55+40+28+18+10+42155ft For a very rough check on the above calculation. we can draw a line from (O. 70) to (6,0) and calculate the area of the triangle: é(70)(6) 2 210. This is clearly an overestimate. so our midpoint estimate of 155 is reasonable. For an increasing function. using left endpoints gives us an underestimate and using right endpoints results in an 3020_ _ 5 _; 6 ‘55—3600h‘720h‘ overestimate. We will use M6 to get an estimate. At 2 M6 2 fiber» + v(7.5) + v(12.5) + 21(175) + 11(225) + 14275)] = @(3125 + 66 + 88 + 103.5 + 113.75 + 119.25) = 710621.75) m 0.725 km For a very rough check on the above calculation. we can draw a line from (0, 0) to (30, 120) and calculate the area 17. 18. 19. 20. 21. 23. 3 CHAPTERS INTEGRALS of the triangle: %(30)(120) : 1800. Divide by 3600 to get 0.5. which is clearly an underestimate. making our midpoint estimate of 0.725 seem reasonable. Of course. answers will vary due to different readings of the graph. f(m) : 1 g m S 16. Ax = (16 — 1)/n = 15/71. and x, : 1 +iAz‘ : 1 + 151/71. 11:11,“ Rn:1im Z (x.)A:c:1im : 4 1+%‘ —> . n 00 n—mozyil n—DOOI_1 15 n . l f(x)=£x§.3g:cg10. A$=(10—3)/n:7/nandsc;:3+1Am:3+7i/n. . _ n . 7‘ 1n(3+7i/n) 7 AZIIHIRn:llm xiA =1 —————-—. n—>oo n—>oo7§1 ( ) a: n f(x):xcosm039:§§ Ax:(§—0)/n=g/nandmi=0+iA:c:§i/n. A21 R7121” ” iAzl' "L7? 11.: .52.. .32.. (a) m 2n 2n . n 2 2‘ 1° . 11m Z Z (5 + can be interpreted as the area of the region lying under the graph ofy = (5 + x)10 on the i=1 2—0 2 21' interval [0.2], since fory = (5 + 26)“) on [0, 2] with Ax = — = —. an : 0 +1LAzc = —. and w? : am, the n n n . . . " . " 22‘ 1° 2 . . expresston for the area is A : 11111 E f(:1:i ) A3: 2 11m E 5 + — -. Note that the answer is not unique. "goo i=1 "#00 1:1 n n We could use y = $10 on [5, 7] or. in general. y = ((5 e n) + $)10 on [n,n + 2]. n 1 11111 E 1 tan E can be interpreted as the area of the region lying under the graph of y = tan a: on the interval naoo :1 4n 4n . . 4 — , ’ 1 [0,1].smce fory = tanm on [0, 3] With An: = W/ 0 : L, 1:,- : 0 + zAw : E. and 36,; : cm. the 4 4 n 471 4n TL n . m 7r . ex ression for the area is A : lim w? Ax = lim tan — —. Note that this answer is not um ue. p ngmzfl I) woo: 4n 4” q 1: 2:1 since the expression for the area is the same for the function y : tan(x — kw) on the interval [km k7r + g] . where k is any integer. . n n . 3 1—0 1 1 . 2 1 A:——:—-d.: 'A :—.A:1' Rn:l' .Azl — -—. (a) m n nan ac 0+2 m n 711—1130 m n , " 2'3 1 _ 1 ",3 . 1n(n+1)2 .(n+1)2 1. 1 2_1 (b)n1L”éo;ns‘n*JE‘;on4;Z11320714 2 2.113;. 4712 1.132.. 1+” *4 2—0 2 2i : : 5_A :——:— d ,‘ZO 2—. (a)y f(9c) a: m n nan :a +2 x n n 11252 “3252 .64". ’1 n2(n —l—1)2(2n2 —l— 277. — 1) b .5CéS ()72211 ———-———12 t 64 n2(n+1)2(2n2+2n—1) 64 . (n2+2n+1)(2n2+2n—1) (c)11m - — 11m n—wo n6 12 12 n—>oo n2 -n2 16 1 2 1 -16 -2 ‘11:“;(Hfi'n—2X2W‘m)'31”3 SECTION 5.2 THE DEFINITE INTEGRAL ‘ 2 n —2z'/n - n ~21/n _ 6‘2(62 _ d 24. From Example 3(a). we have A 2 11m E Z 6 . Usmg a CAS. 2 e — Egg/n 1 3" n—voo V [:1 7,:1 _ D441 lim — - x 2 872(62 — 1) z 0.8647. whereas the estimate from Example 3(b) using A110 n—wo n 82/" — 1 was 0.8632. b — 0 b , b4 25. y2f(;1:) 2cossc. Aw: — 2 — andmi 20+2Aav2 —Z. n n n TL—*OO A 2 lim Rn = "lingogflmflfix 2 nlgigcos<%) .gcés lim Ifb2§.thenA2sin§ 21. 26. (a) 0 The diagram shows one of the n congruent triangles. AAOB. with r central angle 27r/nt 0 is the center of the circle and AB is one of the sides of the polygon. Radius 00 is drawn so as to bisect ZAOB. It A I. 3 follows that 00 intersects AB at right angles and bisects AB. Thus. C AAOB is divided into two right triangles with legs of length %(AB) 2 rsin(7r/n) and rc0s(7r/n). AAOB has area 2 - %[T‘ sin(7r/n)][r cos(7r/n)] 2 r2 sin(7r/n) cos(7r/n) 2 $79 sin(27r/n). so An 2 n - area(AAOB) 2 $717"? sin(27r/n). sin0 (b) To use Equation 3.4.2. tling 2 1. we need to have the same expression in the denominator as we have in the argument of the sine function—in this case. 27r/n. . 2 . lim An 2 lim énfisinQn/n) 2 lim énrzm ~ E 2 lim WWT2. Lett9 2 n—roo n—mo n—>oo 27T/71 TL n—‘00 27r/n Tl Then as n ~> oo, 6 ——> 0, so lim WWW 2 lim smenrg 2 (1) 7l'T‘2 2 7l'7"2 n—voo 27r/n 9—0 5.2 The Definite Integral \ 4 1. R4 2 Z Am [33: 2 x; is a right endpoint and A26 2 0.5] i=1 = 0.5 [HO-5) + f(1) + f(1-5) + f(2)l : 0.5 [1.75 + 1 + (20.25) + (22)] : 0.5(05) = 0.25 W) = 2 — $21 The Riemann sum represents the sum of the areas of the two rectangles above the x—axis minus the sum of the areas of the two rectangles below the m-axis; that is. the net area of the rectangles with respect to the m—axis. ...
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This note was uploaded on 12/08/2009 for the course MATH 101 taught by Professor Dr.tahir during the Fall '08 term at King Fahd University of Petroleum & Minerals.

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5.1 - 5 Cl INTEGRALS 5.1 Areas and Distances 1. (a) Since f...

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