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Chapter 5 44

Chapter 5 44 - 476 CHAPTER 5 INTEGRALS 83...

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Unformatted text preview: 476 CHAPTER 5 INTEGRALS 83. Letu=1—\$.Thenm=1—uanddw= —du so folma(1—m)bdm:f10(1vu)aub(—du)= fOu 1117(liu)adu:f0mb (l—w) doc. 84. Letu:7r~w.Thendu=—d:c.Whenw=7r.u:0andwhenx=0,u=7r. So f0"a1:f(sina:)d1::—f0(7r—u)f(sin(7r—u))du—j07r (7r—u)f(sinu)du =7rf07rf(sir1u)du—f0 uf(sinu)du=nf0"f(sina:)da:—f07rccf(sin:1c)da: => 2f;r mf(sinm)d;1: :7rf07r f(sinx)dm =¢ f0" mf(sinz)d ~ 2 “"ffo (sinx) xsinw sinw t 85. ———— : -— : 1+c082\$ a: 2—sin2x ”(51%) Where f(t ): 2—75 " cesium 7r 7r 7‘ 7r 7' sinm /0 1+c052:c :c /0 mf(s1na:) a: 2/02 f(sma:)d:c 2/0 1+c0s2mdm Letuzcosm. Thenduz—sinxdzv. Whenx:7r.u:71 andwhencc=0,u=1. So 7r/7r sinm dm n/Tl du 7r 1 du ”1‘53 _1 ]1 _ —— :—— : — = _ n 2 o 1+cosga: 2 1 1+u2 2 _11+u2 2 “—1 : ghanil 1 i tan—1(—1)]—g[E—(—E)] : 71; 5.6 The Logarithm Defined as an Integral 1. (a) x : 1.5. The area of the rectangle BCDE is % - g : 3 1 5 observe that 5 < In 1.5 < E‘ (b) With f(t) : 1/t. n : 10. and A3: : 0.05, we have In 1.5: I: 5(1/t)d ~(0.05)[f(1.025) + f(1.075) + - - ~ + f(1.475)] :'(0 05)[1.1025 1.375 + ‘ ' ' + 1.1175] x 04054 2 — 1 . . l . 2. (a) y : %. y' = —ti2. The slope of AD is 1; 7 1 : ~é. Let cbe the t-coordmate of the po1nt on y : T w1th ;. c2 — 2 c 7 \/§ since c > 0. Therefore. the tangent line is given by — gt + ﬂ. tangent line. that is. above the line segment BC. Now [AB| = and ICD| : —1 + ﬂ. So the area of the trapezoid ABCD is %[(*i + V?) + (—1 + ﬁn] : —g + ﬁx 0.6642. So ln2 > area of trapezoid ABCD > 0.66. We interpret In 1.5 as the area under the curve y : 1/ac from an : 1 to l. The area of the trapezoid ABCD is % ~ %(1 + g) = % Thus. by comparing areas, we Since the graph of y : 1/t is concave upward, the graph lies above the ’é‘i’ﬁ ...
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