Unformatted text preview: CHAPTER 5 REVIEW 485 52. On the interval [1, 7;] a: is increasing and sin a: is decreasing so sizm is decreasing. Therefore. the largest value
sinzc 7t 71' . sin(7r/4) ﬁ/2: 2_\/_—2_ 2\/§
—— ——=—~ BProert 8withM:——we et
0f :5 on [4 2l 1s 7r/4 7T/4 7r y p y 7r g
"/2 Slnac 2\/§ 7r 7r ﬂ
71/4 an 7r 2 4 2 53. cosa: g 1 => 605 cosx g 6” => fol excosmdcc S fol em dzc =[ez](1): e —.1
54. F0r0 S x S 1.0 5 sin‘1 :0 5 %.so folwsin"1:vdw S fol $05) da: 2 [£332]; 2 g.
55. Let me) : W on [0,1]. The Midpoint Rule with n : 5 gives fol x/lt—Jie'dx x [f (0.1) + f(0.3) + f(0.5) + f(0.7) + f(0.9)]
:[m1 W 1 1 W] m1110 56. (a) displacement = f05(t2 ~ t) dt : [ UIIH U‘HIH t3 — 1162]: : 1—35 — 3255 : % : 29.10 meters bah—I (b) distance traveled = I; lt2 , ti dt: f: t( (ti 1)dt= [01(15 7152) dt+ f15t2(— t)dt
= [$9 * $53]; + lets * ithi —% i 01035 2—25) (% ) 7 ¥ — 29.5 meters NIH 57. Note that r(t)( : b’(t) where b(t) : the number of barrels of oil consumed up to time t. So. by the Net Change Theorem f 0 t)—dt — b(3 )— b(0) represents the number of barrels of oil consumed from Jan. 1, 2000. through
Jan. 1. 2003. 58. Distance covered : f05'0 v(t) dt 22 M5 : 505 0 [0(05) + 21(1.5)—l— v(2.5) + 143.5) + 0015)]
:1(4.67 + 8.86 + 10.22 + 10.67 + 10.81): 45.23 m 59. We use the Midpoint Rule with n 2 6 and At = 241; 0 = 4. The increase in the bee population was [024 110011 m M6 2 40(2) + r(6) + 1(10) 1 1(14) + r(18) + 7122)]
m 4[50 + 1000 + 7000 + 8550 + 1350 + 150]
=4(18,100) = 72,400 60. A1 : ébh : %(2)(2) : 2. A2 = ébh— — (—1)(1)—— and since
— iv 1 — m2 for 0 S m g 1 represents a quartercir:le with radius 1.
A3 : l7r'r‘2 : l7r(1)2 : g. SO
f_3f 1})dl‘ — A1 A2 A3 : 2 y=~x—1 y % 4 ﬂﬁ 71'). 61. By the Fundamental Theorem of Calculus, we know that F (ac) : f: t2 sin(t2) dt is an antiderivative of
f(a:) : m2 sin($2). This integral cannot be expressed in any simpler form. Since f: f dt : 0 for any a, we can take a : 1. and then F(1) : 0. as required. So F(m) = f: t2 sin(t2) dt is the desired function. ...
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 Calculus

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