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Test 1 Practice Test

# Test 1 Practice Test - é M Practice Problems for Test 1 1...

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Unformatted text preview: é/ M} Practice Problems for Test 1 1. Evaluate each of the following integrals. _ W4 2 4 3 _ 2 ——i—2—\$—dm,/ msec2(m)dm,/ —4a:2e”3dg:7/ __3_\$_____d\$ (x2+1)(x—~1) 0 0 1 x2+5m+6 1 3 3 71-2/4 COS(\/E) so 16 — 9:2 25dm,/ mln x dx,/ Lainv/ —-————dm. /0 ( ) 1 ( ) 0 V25 — 902 «2/9 x/E 2. Differentiate each of the following with respect to an: i t x2 73-3 5m 2 1 3 (t) /x sm(e)dt,[1 t2+3dtﬁlcﬂ2 ln(1+t )dt,/0 t e 5 dt. 3. Find an integer n so that the midpoint approximation to f13/2 cos(a:2)da: is within 1/ 100 of the value of this integral. Write the Riemann sum corresponding to this midpoint approximation for this value of n. 4. Find an integer n so that the midpoint approximation to fol xgemdw is within 1/100 of the value of this integral. Write the Riemann sum (midpoint approximation) corresponding to this value of n. 5. For each of the following, determine whether the integral converges or diverges. oo 1 oo ‘ oo oo 1 2 3 _. m — m /2 xa/zdm,/q; cos(2:c)d:c,/_Ooe dm,/1 \$2+In(m)dm,/_me dx. 6. Find the area bounded by the graphs of y : m3 and y : 5c. 7. Find the area bounded by the graphs of y = cos(m) and y = sin(a:) for 0 S x 3 7r. 8. Find the area bounded by the graphs of y = 1/2 and y : cos(m) for O S a: g 271'. 9. Write an integral for the length of the graph of y : tan(r) for 7r / 4 S m 3 7r/ 3. 10. Find the length of the graph of y z 223/2 for 0 g a: g 4. » 11. Find the length of the graph of y = %(m2 + 2)3/2 for 0 g x S 3. 12. Find the volume of the solid formed by rotating the area between the graph of y : x/sin(x) and thepzr— axis around the :c— axis, for 0 S x g 7r/2. 13. A region is bounded by the graph of y = :1: — 1:2 and the 33- axis. Find the volume of the solid formed by revolving this region about the x~— axis. 14. Let B be the region bounded by the graph of y = sin(a:) and the x— axis for 0 S :2: g 77. B is the base of a solid having the following property: the cross section of the solid at a: is a square. Find the volume of this solid. 15. Let B be the same region as in (14), but now the cross section of the solid at m is an equilateral triangle. Find the volume of the solid. 16. A rectangular trough is 4 feet wide by 12 feet high by 20 feet long. It is ﬁlled to a height of 3 feet with water (weighing 62.5 pounds per cubic foot). Find the work required to empty the trough (that is, lift the water to the top of the trough). 17. A hemispherical bowl has radius 15 feet. It is ﬁlled to a depth of 6 feet with water. Find the work required to empty the bowl (that is, to lift the water to the top of the bowl). Solutions 1. For fm‘EE—fﬁﬁdm, write 4—230 A Bm—vC (m2+1)(m—1) ac—l \$2—-1’ For this to be true, we must have A(m2 + 1) + (B30 + C)(a: — 1) = 4 ~ 2:13. Letmzltoget 2A:2,soA=1. Letzr:0togetA—C:4. SinceAzl, then C = —3. Finally, let x = —1 to get 2A — 2(C' — B) 2 6. But then 2—2(—3—B)=6 4 — 2113 _ 1 m _ 3 (\$2+1)(\$—1)— 96—1 m2+1 932+1 Then 4 — 2 1 /ie:3xffn“””mx‘”‘§[email protected]”+U—3M“wWW+0 For OW/4xsec2(x)da:, integrate by parts, with u : m and d7; : seC2(\$)d\$. Then du 2 ale: and v 2 tan(9:), so 7r/4 7r/4 / xse02(x)dm = [as tan(:c)]g/4 — / tan(a:)da: o 0 7r ‘ M4 —sin a: Z) +/0 cos(§:))d\$ + [1n |cos(m)ng/4 7rta =—— n 4 #4:) A z E + ln(cos(7T/4)) ~—1n(cos(0>) =g+m(%)=§—§ma For f0? —41L‘26\$3d\$, let u = 933. Then du : 3x2daz, so xzdac : édu. When 110:0, uzO. Whenx:2, uz8. Then 2 I 8 / w4m2em5dm = —:1/ eudu : E (1 — es). 0 3 0 3 3 sac—2 - For f1 mains, ﬁrst write 318-2 3:5—2 A B w2+5m+6: (m+3)(m+2) :x+3+az+2' Then A(\$+2)+B(m+3) =3ac—2. Let a: = —2 to get B : —8. Let a: = ~3 to get A = 11. Then 3 ' 3 3 3m—2 11 8 Ax2+5x+6dm‘/1x+3dx_/1m+2dm : [111n(\$ + 3)]? »— [81n(ac + 2)}? = 11[ln(6) —1n(4)] — 8[ln(5) ~* 111(3)]. This can also be written 1n<<2>“)—1n(<2>8>7 OI‘ For fol m(16 — m2)25dm, let u = 16 — :62. Then du : ——2mda:, so mda: = ~é—du. When 90 = O, u = 16. When a: = 1, u : 15. Then 1 225d _ 15 125 11 2615 0 22(16—10) m— 16 ——§u duz—52—6[u L6. _ 1 26 26 _ 52 (16 15 ) . This is the way the answer should be left. However, if you carry out the arith— metic, the value of the integral is 16 494 734 359 545 3%; 094 693 100 895 391 o For x1n(\$)daz, integrate by parts. Let u :- ln(:c) and d1; = mdcc. Then (in : ids: and v : éxz. Then /30cln(:c)dxw 1932ln(z) 3—/1x23d 1 _ 2 1 2 a: “3 9 1 3 9 1 3 : 51n(3) — — /1 acdac = ~2—1n(3) w ~[m2]1 2 4 9 1 9 For f: «ngdx, let u = 25 - 302. Then du = —2mdm, so Jada: = wédu. When a: = 0, u = 25. When cc 2 3, u = 16. Then 3 16 16 / _____x_dm:/ _l%du: [_u1/2] o V25—m2 25 211/ 25 = —x/I€ + V55 2 1. For fig ﬂ\/la.:@211:lc,1et‘, u = Then (in = %%dx, so ﬁdw : 2du. When a: = «2/9, 11 = 7r/3. When m = 7T2/4, u 2 71/2. Then . “2/4 ”/2 / cosh/'11?) dx 2 / 2 cos(u)du = [2 sin(u)]:/§ 7r'2/9 ﬂ ‘ "/3 = 2sin(7r/2) — 251n(7r/3) = 2 <1 — . 2 2. d 1 a; sin(et)dt = - sin(e\$) cl “52 t~— 3 x2 - 3 a; 4 t2 +3dt : (223) (x4 +3) d 5:17 (1—3: 1n(1 + t2)dt 2 51n(1 + 253:2) — 2m1n(1 + 934) m2 1 i t3e°°s(t)dt = 0. dm 0 (Since fol t3eC°S(t)dt is a deﬁnite integral, it is a number, and so has zero deriv- ative). 3. For 13/2 cos(m2)dm we have f(m) : cos(nc2), so f’(:c) : —2x 3111079) and f”(:c) = —251n(a:2) — 4x12 COS(\$2). For 1 S x S 3/2, |f”(m)| 1—2 sin(x2) — 4m? cos(m2)| g l~251n(x2)| -l— [—4902 cos(:c2| S 2 lsin(a:2)| + 4:1:2 lcos(a:2)l 2 gz+4<§> =11. (Any larger number is also a bound on f” Use K = 11 in the error bound. Now choose n so that K(b— a)3 m 11(3/2~ 1)3 < 1 24712 _ 2477.2 — 100' Then we need 2- > 11(100) - 8(24) ’ 10¢ﬁ m 10x/ﬁ «8(24) _ 482(3) AREA/1.1. _8 3—4 3' We need an integer n to satisfy this inequality. Just to get started, try n = 2. Then we would need TL SO Q 511 2 > — . " 4 3 For this to be true, we would need 8 2 5w/11/3. Squaring both sides would then give 64 Z 25 or 192 Z 275, which is false, so n = 2 will not work. Try n = 3. Now we would need 511 3, Z 4 3 ‘ F For this to be true, we would have to have 12 2 5 11/3, or 144 2 25%. This is true if 3(144) 2 25(11), or 432 Z 275, which is true. Choose n : 3. Subdivide [1,3/ 2] into three subintervals of length 1/6. These are [1,7/6], [7/6, 8/6], [8/6,9/6]. The midpoints are, respectively, 13/12, 15/12 and 17/12. The midpoint approximation is the sum 3 _ l ZfWﬂg i=1 2 é(cos((13/12)2) + cos((15/12)2) + cos((17/12)2)). If you compute this sum, you get approximately —0.0045551. Here we got by with n = 3. Certainly any larger value of n will also work. A better approximation is obtained by taking n = 30, yielding the approximate value— .0053318. 4. For 01 xzemdm, f(ac) = 9326”, so f’(:c) = 232:8QC + 50263 and f”(m) 2 2e“ + 4906'” + \$26”. Since 61 = e < 3, then for 0 g :E g 1, MW)! 3 2(3) + 4(3) + 1(3) = 21. Any number larger than 21 could also be used as a bound for | f” although larger choices could lead to more terms, and hence more work, in the midpoint approximation. To make the arithmetic as simple as possible without costing much work in the sum, we will choose K : 24. Now we must choose n so that K(b — a)3 _ 24(1 — 0)3 _ 24n2 _ 24m2 _ n 100' This requires that n2 2 100, or n 2 10, so choose 71 = 10. Subdivide [0,1] into 10 subintervals of length 1/10. These subintervals are: [0, 1/10], [1/10, 2/10], [2/10, 3/10}, [3/104/10], [4/10, 5/10], [5/10, 6/10] [6/10, 7/10], [7/10, 8/10], [8/10, 9/10], [9/10, 10/10]. The midpoints are, respectively, 1 3 5_7_9 1113151719 20’20’20’20’20’20’20’20’20’20' The midpoint approximation is 1 1 2 1 3 2 5 2 7 2 __ __ /20 _ 3/20 __ 5/20 _ 7/20 10 ((20) e +<20) e +(20/ e +(20) e 1 9 2 9 11 2 13 2 15 2 __ __ /20 _ 11/20 _ 13/20 __ 15/20 +10 ((20) e +<20> e +<20> e +<20) e 1 17 2 17/20 19 2 19/20 +1.00%) 8 +96) 6 _ 2. 8286 x 10—2 + 0.284 2 + 0.402 40 If you compute this sum, you get-approximately 0.71489. If you use n = 30 in the midpoint approximation, you get the approximate value 0.7179 for the integral. 5. For E” 9331/2 dx, compute /2T #dx : [—22'1/2]; = 2—3 wm/iasraoo. y; This integral converges to For cos(22)dx, compute / cos(2:r)d23 : % sin(2r). This has no limit as 'r —) 00., so this integral diverges. For foo emdx consider f0 emdm + foo emdx Now ——oo ’ —oo 0 ‘ 0 /emdm=1—er~+1asr—>—oo, r so fix) ezdx converges (to 1). However, 00 7" / ewdm = lim exdzz; : lim (eT — 1) oo, 0 T—NDO 0 ’I‘—->OO so [00° exdcc diverges. Therefore [3°00 exdx diverges. Ffr floo Waist, we have no hope of integrating directly. However, for a: Z 7 1 1 <—————<—— 0'“ w2+ln(a:) “ m2 1 00 00 and we have seen that f1 Elydzr converges, so f1 m comparison test. For L200 chains, compute dx converges by the 2 . . 1 hm e3zdzr = hm — 'r—r—oo r 7‘-—->~OO 3 (66 __ 637‘) 2 £86, so the integral converges to §e6. 6. These graphs intersect at (—1,-1), (0,0) and at 1,1). For —1 S x the graph of y = 9:3 is above the graph of y = 3:. They reverse roles on [ Thus S 0, 0,1]. 0 1 area 2/ (m3 - ac)da: +/ (x — \$3)d:c = ~1 0 2 One could also argue by symmetry that the total area is twice that to the right of the m— axis, so ‘2 1 41 1 __2 303—235 91: Mi area /0(x—a:)1:— [—2 ———4 0—2 7. On [0,7T], the graphs intersect at (7r/4,1 On [Om/4], the cosine graph is the upper boundary, and on [7r/4, 7r], the sine curve is the upper bound— ary. Then ' 7r/4 7r area = / (c0s(:c) ~ sin(x))dac + / (sin(m) — cos(a:))da: 0 7r/4 : {sin(93) + cos(ac)]g/4 + [— cos(ac) ~ sin(a:)]:/4 =[1/\/§+ 1/\/§ ~— 1] + [1 + 1/\/§+ 1N5] = 2ﬁ. Mummw—MM. r hr” W y A“ MW is m ,3, m l l j 8. For 0 S m 3 7r / 3, the line is the lower boundary, and the cosine curve the upper boundary. These reverse from 7r / 3 to 57r / 3, then reverse again from 57? / 3 to 27r. The area area = /07T/3 (005(92) — dm + [:73 — cos(m)> doc + <COS(:E) — da: =<éeéwl+<§w+ﬁ>+<ée§w> 1 9. The length is 7T/3 length = / x/l + sec4(x)dx. 7r/4 This integral cannot be evaluated exactly. If you use the midpoint rule with n = 20, you get the approximate value 0.77895. 10. The length is 4 3 2 4 / 9 lengthz/ 1+ (—ml/Z) dmz/ 1+—a:dz. 0 2 0 4 Unlike the preceding problem, this integral can be evaluated exactly. Let u = 1 + gar/4. Then clu : gdn When :16 = 0, u = 1. When an : 4, u =10. Then 10 4' 8 0‘ _-—- —_ d —_ _.._ . length 1 9/11 u 27 (10 1) 11. The length is 3 1 2 length 2 / 1 + + 2)1/2(2m)) dx 0 3 3 2/ 1+m2(\$2+2)dm:/ \/1+\$4+2:c2dm 0 0 3 3 2/ x/(m2LI—1)2dx:/ (x2+1)dm:12. 0 0 12. For 0 S m g 77/ 2, the slice at :1: through the solid and perpendicular to the 1;— axis is a disk of radius a /sin(:r), and therefore has area A(a:) = 7r = 7rsin(:1:). l l l l l l The volume of the solid is 77/2 volume = / 7rsin(a:)dm = [—7r cos(m)]3/2 = 7r. 0 13. The graph intersects the ac— axis at (0, 0) and at (1, 0). The slice through the solid and perpendicular to the 2:- axis is a disk of radius :r — 333 and area A(m) = 7r (:3 — m2)2. The volume is 1 volume 2 / 7r(:c — x2)2da: : 0 1 =7r [1363+lm5—lm4] = ~71. 0 30 3 5 2 14. At any :1: in [0, 7r], the slice at x perpendicular to the z— axis is a square having side length sin(ac) and area A(ac) : (sin(m))2. The volume of this solid is 7r 7r _ 2 . volume : / sin2(a:)dac : / 1—wdm 0 0 - 1 7T 2 - 21— sin(2m)]0 : in. 15. The cross section of the solid at an is an equilateral triangle having side length sin(:c). The area of an equilateral triangle of side length a is Agag. Therefore the area of this triangle is A(m) : 43 sin2(a:). Then volume : i3 sin2(:1c)dx : é / (1 — cos(2x))d:z: 0 4 8 0 «5 1 7‘ V3 : 73— [m - 5 sin(2:r)]0 = \$71 16. Put an 3:- axis along the side of a vertical end of the trough, with a: = O at the bottom of the trough.' Subdivide [0,3], where the water is, into subintervals. The ith subinterval {\$141,331} determines a 20 by 4 by xi — x¢_1 slab through the water. The force acting on this slab of water is 62.5(20)(4)(x¢ — cot-4), which is volume of this slab multiplied by the weight per cubic foot of the water. The slab must be lifted a distance of approximately 12 — mi feet. The sum of the work done in lifting these slabs is Z 62.5(20)(4)(12 — 50¢)(xi — xH). This sum approximates the work done in lifting the water to the top of the tank. In the limit as n —> 00, this sum, which is also a Riemann sum, approaches an integral for the work. The work needed to empty the tank is ‘ 3 3 work = /0 (62.5)(20)(4)(12 .- m)dm = 5000 /O (12 — x)dm 3 z 5000 [1295 — 3x2] = 5000 (Q) 2 0 2 = 157500 foot—pounds. 17. Imagine the bowl sitting on a table, and put an m— axis vertically down through the center of the bowl, with at = 0 at the bottom. The water lies in the interval [0, 6}. Partition this interval into n subintervals and look at the layer of water between xi_1 and 93¢. The force acting on this slab is (62.5)(volume of the slab). Let n be the radius of the bottom disk of this slab. Then 7»? + (15 — 3602 = 15?, SO 2 7: . r? = 30901 ~ 93 The volume of this slab is approximately 7F7‘1-2(£Ei — acid). The force on this slab of water is approximately 62.57r(30:c¢ — 9:2)(334 — \$14), so the work done in lifting this slab a distance of approximately 15 — 3:1- feet to the top of the bowl is approximately 62.57r(30\$1~ — \$§)(15 — 330(931‘ — acid). Sum these to approximate the work done in emptying the bowl. In the limit as n —> 00, this Riemann sum approaches 6 work 2/ 62.57r(30m —- 3:2)(15 — a:)dac 0 6 = 62.57r/ (30m — 3:2)(15 ~— :13)da: 0 : 62.57r(5184) : 3240000 foot—pounds. 10 (18:13. 85.31. 910; 95 1 ,. as?a,233,5X55is?‘5335:;n.22,K.2VéééEﬁﬁgtévtgrzxpxxweE»imagz 0.4 0. i6 :12 gZXiszx 143,32 . qauozﬁﬂﬁé // , is? H x. _ , (Mb; wwzmwam MWW w / - _. ...
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Test 1 Practice Test - é M Practice Problems for Test 1 1...

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