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Unformatted text preview: 1. 2. 3. Math 101 Solutions April 1994 (a) First, partial fractions will be found. £51,; = 65%;?) = g +
£335,561, which gives (A + 3)::2 + Cm + 2A 2 2: + 1. Equating powers
gives A = %, B = —%, and C = 1. The integral is now + 3113 — Whiz = %lnz + 7‘;tan"(%) — §1nx2 +2 +C. ‘ ' ._ 2 __ _ 1d:
(b) Integration by parts With U — a; , du — 2a: dz, dv — m,
and v = —(9 — 2:2); yields fx3(9 — $2)‘§da: = 2:2\/9:c2 + 2x\/9 —x2da: =12VQ—z2 — Z(9—;1,2)% +0,
3 (c) First, integration by parts with u = tan‘1 3, du = ﬁfty, dv = 9: dz, and v = #22 is applied. In: tan'1 Ida: = %:c2 tan‘1 a: — fi—gfn The second integral can be solved with the substitution :1: = tan t, which
7 gives 1 + x2 = sec2t and d1: = sec2 tdt. Now I a—‘Ey = ftanztdt = f sec2 tdt — f dt = tant — t = a: — tan‘1 2:. Thus the indeﬁnite integral is #12 tan‘1 a: + §tan“l a:  #1: + C, which when evaluated over the interval 0 5 :1: S 1 gives £312. (d) Using the substitution 6% = sec t, which implies that y/e' — =
tan2t and dz = 2tantdt, the integral is transformed as follows.
f\/e‘ — Ida: = 2ftan2tdt = 2f(sec2t — 1)dt = 2tant — 2t+ C = 2V6": — — 25ec“(e§) + C. A small segment of length is d3 = ‘/ iii—f? + G?)2 = VQts + t10 dt =
t3x/t4 + 9dt. Thus the length is L = f: ds = f: t3\/t4 + 9dt = %(t4 + 9%: 45‘! (a) The volume is V = flﬁnR(z)2da: = «flﬁ(ﬁ:—’T'—f)2dz =
«flﬁu + Elﬁdz = 7r(:c — 3311“? = 7r 2%. (b) The volume V can be found most easily by shells. Thus
V = flﬁ27rzf($)dx = ZNfIﬁﬂg‘T—ldz : 27rf1ﬁx/1+z2dz, which can nogv be transformed,r via a: = tant and dz = secztdt
to V = 21r sec3 tdt = 21r faﬂsect + sect tan2 t)dt. The ﬁrst of
4 these integrals can be integrated immediately to ln(sec t + tan t), while
for the second integral use integration by parts with u = tant and
dv = secttantdt, which give du = seczt and v = sect. Note
here that f vdu is the very integral we are trying to compute, and
so fsect tang tdt = %tant sect. Thus V = 27r(ln(sect + tan t) + %tant sec 0: = 21r(ln(2 + J5) — ln(1 + J?) + \f —— 715). By symmetry §A = moi r109)2 d0 + [if r2(0)2d0 + ff; r3(0)2d9),
12 12 where 11(0) = 113(0) = 23in 29 and r2(0) = 1. As can also be seen from the graph, the areas over which r1(9) and r3(9) are integrated are equal. Thus, A = 8 for"! sin2 20 d0+2 [:13 d0 = 2m: +2(20—sin 20 cos 26m,” =
§1r _ y/S. 5. Since ﬁ; = 1—z+x2—2:3+x4...,1—+1—,g =1—t2+t4—t6+t8—....
Thus, 7231;?) = t‘% —t% +13% —tlé" +t11é —.... Integrating this power series gives femi 4% +t% —t‘2L +tli§ —...)dt = (2t% —’g’t5 +§t3 —
13 17 1
T25” + 12—,” —...)§ = 2y/Z(1—§t?+§t4 — £13“ + 11—7t8 — ...)5 = 1 — $141 + ﬁ; — 7331; + ")4 — . . .. Since this is an alternating series,
and 73—33; < 10“, then H — 13 < 10‘4, where [3 denotes the power
series truncated after three terms. Thus, I3 2: 1 — $7 + 5%; N 0.9879
is a suitable approximation to the integral. After differentiating this equation, we find that f’ (2:) = 3:1:2 f (1:).
Manipulating this gives 1,153 = 3:32, and integrating both sides shows
In f(:c) = $3+C, or f(1:) = eruc = ke’s, where k is some constant. To
determine k, substitute the trial function f(:1:) 2 Ice”3 into the original
equation, i.e. Ice$3 = 2 + 3f01t2 ke‘3 dt = 2 + ke‘alﬁ = (2 — k) + ke’s =
ke‘a. This is satisﬁed when k = 2, and so f(:l:) = 26:3. (3) If f(z) = ln(sin:c), then f’(2:) = cots: and f”(:c) = —csc2x =
;i—n1r;. Since sin2 ml 5 1, f”(a:) is at a maximum when sinzx is at
a minimum on g g a: S sin2 a: is increasing on this interval, so the
smallest value occurs at a: = 1,}, where If”(§) = 4 = M. (b) Here a = g, b = a and M = 4. Therefore, IETI g J—TLMIZT 3 = BIL}. For ET 5 104, n2 2 T51: 2g: 3 3827.94, or n 2 61.87. Thus
n = 62 will give I z Tn accurate to within 10“. 8. (a) As can be calculated, the vertex of the parabola occurs at (0, ——%).
At points of tangency, :1 115(3), where y1(a:) = %(:1:2 —— 1) and 10. y2(a:) = 1 2t V2 — :52. Differentiation yields yi(:1:) = a: and y§(x) =
iﬁ. Setting these two expressions equal, and keeping in mind that the square root function is positive, a: = ﬁ, means that 1:2 = 1, or
x = :lzl. Therefore the points of tangency are at (:l:1,0). (b) The area is A = f_I1(1 —— V2 — 2:2 — are? — 1)) d2: = f01(3 — $2 —
2V2 — 2:2)da: = (31:  §z3 — :cVZ  2:2 — 2sin'l(:‘ﬁ)l},=1°g3". 2 (a) By symmetry, 5: = 17, where i: = ﬁ/{zdan Here A = inaz = ﬁe since A is a quarter circular disk. Thus, in = #5 f0" :m/a2  2:2 d1: = —13;: (a2 was = (b) As in (a), symmetry dictates that :E = g. The area A = 4 ——
g : 164‘", and fxdz = [012:(2 — v1—x2)d:c+ fl22rcdx = HIE, — A
fol :cx/l —:1:2d2: = 4+ %(1 —z2)%(1,= Therefore :3 = 17 = 48:43". (a) This inﬁnite series should look familiar, since it consists of n
subintervals each of which has a given height, i.e. an integral (in the
limit). The integral begins at (Inn — lnn) = 0 = ml and ends at (ln 2n —~ 1n n) = ln 2. Thus the sum can be written as lnxdm, which
can be integrated by parts with u 2 him, du 2 df, do = dz, and v = 2:.
Now, fl2lnxda: = :51an — flzda; = (:clna: —:1:)'f = 2ln2 — 1. (b) It helps to write out this series to see what is going on. 3—2 =
3323.33~4_;111111111 8+§r+§r+?—+...— (8+§§+§§+'811+§5+§5+§1+'81+§7+'81+
...)=3((§+§7+§%+§;+...)+(—§+§§
(€1+...)+...)=3(%(1+(%)
(§)2+(,§)3+(%)4+...)+gs 1
3(1+(§)+(%)2+(§—)3+ Remembering that 1: G)“ + . Thus the cap above is really the produc 3(—1—) 4— — 1), with a: = ' 8 L_B
sumis3 7 7—49. 11. 12. (a) If it can be shown that the integral is smaller than another in
tegral that is ﬁnite, then it follows that the integral is ﬁnite, i.e.
converges. Given a function g(z) such that g(x) 3 NO + 1:), then 1 1 _ ﬁ _ g
7—” I > 7—: (1+3). For a: > 1, let g(:r) — x8 and for a: g 1, let g(a:) —. a: .
T 00 d: l d: 00 dz < 1 d: °° dz _ hen "[0 :1: (1+:) f" 2: (1+2) + II 1 (1+1) " ° 2—! + II 3 gm; I}, — 3x‘ii ‘,’0 = g, which is ﬁnite. Therefore the integral converges. (b) If it can be shown that the integral is larger than another integral
that is nonﬁnite, then it follows that the integral is non—ﬁnite, i.e. does not converge. Given a function 9(1) such that 9(3) 2 2:5(1 +93%),
1 1 ~ _
then m _<_ For 1' Z 1, let 9(3) — 22:, and for a: S 1, _ 00 d __ l d 00 4
let 9(3) — 22'}. Then f0 m _ f0 W+fl I (1:: ) Z 124: 002d:
0 j+fl T Therefore the integral does not converge. 12:; l+2lna: °° = Z+21noo = 00, which is inﬁnite.
3 0 1 3 (8) Given that % = 2 —— 2t (the 2 litres of fresh water entering the
tank minus the 2t litres of mixed solution leaving the tank per unit
time) and V(O) = 3, it is easy to integrate and ﬁnd V(t) = 2t — t2 + 3. (b) If Q(t) is the amount of salt in the tank, then 98% is the amount of
salt in the tank per unit volume. Since 2t litres of mixed solution leave the tank per minute, then d—Q  Q ‘ 2t. From (a), V(t) = 2t —t2 +3, dt — _ V(t
and so if? = ﬁéﬁg. This can be integrated using partial fractions, 1 i.e. 33%;“: ﬁg+t+l,soan(t)= glnt—3+%lnt+1+C,or Q(t) = C(t — 31ml: + 1)%. Since (2(0) = 1,0 = 7157. The amount of
water in the tank is at a maximum when “7‘; = 2 — 2t = 0, or at t = 1.
Therefore Q(1) = 742; kg of salt are in solution when the amount of water in the tank is at a maximum. ...
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This note was uploaded on 01/30/2011 for the course MATH 101 taught by Professor Broughton during the Spring '08 term at UBC.
 Spring '08
 Broughton
 Math

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