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Unformatted text preview: 476 C CHAPTERS INTEGRALS 83. Letu = 1 7:3. Then]: =1—uanddw : —du,s0
fol 950(1 7 m)” dx : f1" (1 — u)” ub(~du) = f0] ub(1 — u)“ du : f01 xb(17 my dm.
84. Letu: 7r—w. Thendu: Adm. Whena: :7r.u =0andwhenm:0,u:7r. So
f0" scf(sin 11:) dsc = — ff(7r A u)f(sin(7r — 11)) du = f0"(7r — u)f(sin u) du
:7rf07r f(sin u)du — f0” uf(sinu)du : ﬁfe” f(sinm)d$ — f0" xf(sin ac)d:I:
:> 2ft)7r xf(sin:c)d:c = 7rf07r f(sin:c)dac => f0" mf(sinm)d:c = g f0" f(sinac)d:c. x sin w sin :c t 85. _— Z ._ Z . : ,
1+cos2x x 2_sin2:c $f(51n$)~Whefe f(t) 2_t2. By Exerc1se 84.
7r xsinm " 7r 7‘ 7r 7' sins:
— d : . d : _ I : —
f0 1+Coszm :6 f0 If(sm:c) as 2/0 f(smm)dx 2/0 —1+m82mdﬂc
Letu : COSm. Then du = —sinxd;z;_ Whenx : 7r‘u : ,1 and whsnx : O‘u :1. SO 71' /" sins: dw 71' f4 du 71' 1 du aha _1 ]1
_ —_——— :—— : — I _ n
2 0 1+cos2m 2 1 1+u2 2 _11+u2 2 “‘1 =g[tan7117 tan—1(—1)] : — (—EH : 7r— 5.6 The Logarithm Defined as an Integral We interpret In 1.5 as the area under the curve y : 1 /:c from x : 1 to w : 1.5. The area of the rectangle BCDE is i  g : The area of the 1. (a) trapezoid ABCD is % ~ %(1 + = Thus. by comparing areas. we 1 5
observe that g < In 1.5 < E, (b) With f(t) : 1/25, n : 10. and A3: : 0105, we have
In 1.5 = [1"5(1/t)dt z (0.05)[f(1.025) + f(1.075) +   ~ + f(1.475)] : (0'05)[14325 + 11575 + ’ ' ' + 1.4175] x 04054 2 — . i 1 .
2. (a) : ’ = —l. The 510 e of AD is 1/ 1 : ~l. Let cbe the t—coordmate of the mm on y : — w1th
y t y t2 p 2 i 1 2 p t
1 1 1 2 . . . .
slope 2. Then CZ — i ~> c — 2 => 0 : x/i smce c > 0. Therefore. the tangent line IS given by
y $5: %(t —> y étlx/i Since the graph of y : l/t is concave upward. the graph lies above the
tangent line. that is. above the line segment BC. Now AB = *% + x/i
and ICD : —1 + So the area of the trapezoid ABCD is Hp; +\/§) + (—1+\/§)1] : _g +\/§z0.6642. So
ln2 > area of trapezoid ABC'D > 0.66. SECTION 5.6 THE LOGARITHM DEFINED AS AN INTEGRAL E 477 1 1 1 1 "1
' _ — — — —dt=l .
3. TheareaofR,1sZ,+1ands02+3+ ln</1 t nn 1 1 "1
TheareaofSiisl,andsol+—+~~+———>/ —dt:lnn.
2 2 71—1 1 t 1 1 1 1 1
Thus.§+§+~~~+T—l<lnn<1+§+~~+nir 4. (a) From the diagram, we see that the area under the graph of y = 1/3: between a? : 1 and m : 2 is less than the area of the square. which'is 1. So In 2 = f12(1/:c) dx < 1. To show the other side of the inequality. we must ﬁnd an area larger than 1 which lies under the graph of y = 1/50 between ac : 1 and :v : 3. One way to do this is to partition the interval [1, 3] into 8 intervals of equal length and calculate the resulting Riemann sum. using the right endpoints: >1 4 + + + ‘27,720 1i+i+11111+1728271
5/4 3/2 7/4 ' 2 9/4 5/2 11/4 3 and therefore 1 < f13(1/:c)d:c = ln 3‘ A slightly easier method uses the fact that since y : l/m is concave upward. it lies above all its tangent lines.
Drawing two such tangent lines at the points (g, and (g. we see that the area under the curve from m : 1
to m : 3 is more than the sum of the areas of the two trapezoids. that is. g + g = %2. Thus,
1< 4% < ff(1/x)dm : ln3. (b) By part (a). 1n 2 < 1 < ln 3. But 6 is deﬁned such that lne : 1. and because the natural logarithm function is
increasing. we have 1112 < lne < ln3 <=> 2 < e < 3. 5. Iff(x) = ln(mT). then f’($) : (l/mT)('r:cr_1) : r/x. But ifg(.z') = rlnx. then g’(m) = r/z‘. Sofandg must
differ by aconstant: ln($7i) = rlnac + C. Put x : 1: ln(1") : rln 1 + C :> C : 0. so ln(wr) = rlnm. 6. Using the second law of logarithms and Equation 10, we have ln(ez/ey) = ln ex — 1n 6" 2 ac — y : ln(em’y).
Since In is a one—to—one function. it follows that ex/ey : eky. 7. Using the third law of logarithms and Equation 10, we have In em : m: = 7" In em : ln(e“C )T. Since In is a oneto—one function. it follows that e” : (elf. 8. Using Deﬁnition 13 and the second law of exponents for ex. we have ezln a a: mlnaiylna: *—
eylna ay.
arc—y : C(miyﬂna : e 9. Using Deﬁnition 13. the ﬁrst law of logarithms. and the ﬁrst law of exponents for ex. we have (ab)x : exln(ab) : em(lna+lnb) : exlna+wlnb : ecclnaexlnb 2 axbx. 478 CHAPTER 5 INTEGRALS 10. Let Ioga :c : r and loga y = 3. Then aT : m and as : y. (a) my : aras : ah” :> loga(xy) = 7" + s = loga z + loga y
:3 ar . as b — 2 — 2 rib :> 1 — : i : — ( ) y as a oga y r s logaw logay (c) my : (ar)y : ary => loga(a:y) : Ty = yloga a: 5 Review
CONCEPT CHECK 1. (a) 2le f (an?) Act is an expression for a Riemann sum of a function f. at: is a point in the ith subinterval [31.561] and A2: is the length of the subintervals. (b) See Figure 1 in Section 5.2. (c) In Section 5.2. see Figure 3 and the paragraph beside it. 2. (a) See Deﬁnition 5.2.2.
(b) See Figure 2 in Section 5.2. (c) In Section 5.2. see Figure 4 and the paragraph above it. 3. See the Fundamental Theorem of Calculus after Example 9 in Section 5.3.
4. (a) See the Net Change Theorem after Example 5 in Section 5.4. (b) fit: 7’(t) (it represents the change in the amount of water in the reservoir between time t1 and time t2. 5. (a) [61020 v(t) dt represents the change in position of the particle from t = 60 to t = 120 seconds.
(b) £31020 v(t) dt represents the total distance traveled by the particle from t = 60 to 120 seconds. (C) [61020 a(t) dt represents the change in the velocity of the particle from t = 60 to t = 120 seconds. 6. (a) f f(;z) dz is the family of functions {F  F’ = f}. Any two such functions differ by a constant.
(b) The connection is given by the Evaluation Theorem: f (3:) dsc : [f f (x) dx]: if f is continuous. 7. The precise version of this statement is given by the Fundamental Theorem 0f Calculus. See the statement of this
theorem and the paragraph that follows it at the end of Section 5.3. 8. See the Substitution Rule (5.5.4). This says that it is permissible to operate with the dm after an integral sign as if it were a differential. TRUE—FALSE OUIZ 1. True by Property 2 of the Integral in Section 5.2.
2. False. Try a : 0. b : 2. f(:c) 2 9(93) : 1 as a counterexample. 3. True by Property 3 of the Integral in Section 5.2. ...
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 Fall '08
 Dr.Tahir
 Math

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