Solucao impares Calculo 6ed SEC.5.5

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61. Sincep0({) = ({), p =U₀4({) g{ =U₀49 + 2{g{ =k9{ +4₃{3@2l4 0= 36 +

32

3 0 = 1403 = 4623 kg. 63. Letv be the position of the car. We know from Equation 2 that v(100) v(0) =U₀100y(w) gw. We use the Midpoint Rule for

0 w 100 with q = 5. Note that the length of each of the ve time intervals is 20 seconds = 20

3600hour = 1801 hour. So the distance traveled is

U100

0 y(w) gw 1801 [y(10) + y(30) + y(50) + y(70) + y(90)] = 1801 (38 + 58 + 51 + 53 + 47) =247180 1=4 miles. 65. From the Net Change Theorem, the increase in cost if the production level is raised

from 2000 yards to 4000 yards isF(4000) F(2000) =U₂₀₀₀4000F0({) g{. U4000 2000 F0({) g{ = U4000 2000 3 0=01{ + 0=000006{2_{g{ =}_{3{ 0=005{}2_{+ 0=000002{}34000 2000= 60,0002,000 = $58,000 67. (a) We can nd the area between the Lorenz curve and the line| = { by subtracting the area under | = O({) from the area

under| = {. Thus,

coefcient of inequality= area between Lorenz curve and line| = { area under line| = { =

U1 0 [{ O({)] g{_U 1 0 { g{ = U1 0 [{ O({)] g{ [{2_@2]1 0 = U1 0 [{ O({)] g{ 1@2 = 2 U₁ 0 [{ O({)] g{

(b)O({) =₁₂5{2+₁₂7{ O(50%) = O₂1=₄₈5 +₂₄7 = 19₄₈= 0=39583, so the bottom 50% of the households receive at most about40% of the income. Using the result in part (a),

coefcient of inequality= 2U1₀[{ O({)] g{ = 2U1₀{ ₁₂5{2₁₂7{g{ = 2U1₀₁₂5{ ₁₂5{2g{ = 2U1₀ 5 12({ {2) g{ = 56 ₁ 2{213{3 1 0=56 ₁ 213 =5 6 ₁ 6 = 5 36

5.5 The Substitution Rule

1. Letx = {. Then gx = g{> so g{ = gx. Thus,Uh{g{ =Uhx(gx) = hx+ F = h{+ F. Don’t forget that it is often very easy to check an indenite integration by differentiating your answer. In this case,

g

g{(h{+ F) = [h{(1)] = h{, the desired result. 3. Letx = {3+ 1. Then gx = 3{2g{ and {2g{ =1₃gx, so

] {2s_{3_{+ 1 g{ =}] _x1 3gx = 1₃x_3@23@2+ F = 1₃· 2₃x3@2_{+ F =} 2 9({3+ 1)3@2+ F. 5. Letx = cos . Then gx = sin g and sin g = gx, so

]

cos3_{sin g =}] _x3_{(gx) = x}4

4 + F = 14cos4 + F.

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9.Letx = 3{ 2. Then gx = 3 g{ and g{ = 1₃gx, soU(3{ 2)20g{ =Ux201₃gx=1₃·₂₁1x21+ F = ₆₃1(3{ 2)21+ F.

11.Letx = 2{ + {2. Thengx = (2 + 2{) g{ = 2(1 + {) g{ and ({ + 1) g{ = 1₂gx, so ] ({ + 1)s2{ + {2_{g{ =}] _x1 2gx = 1₂ x_3@23@2+ F = 1 3 2{ + {23@2_{+ F.}

Or: Let x =2{ + {2_{. Then}_x2 _{= 2{ + {}2 _{2x gx = (2 + 2{) g{ x gx = (1 + {) g{, so} U

({ + 1)2{ + {2_{g{ =}U_{x · x gx =}U_x2_{gx =}1

3x3+ F = 13(2{ + {2)3@2+ F. 13.Letx = 5 3{. Then gx = 3 g{ and g{ = 1₃gx, so

] g{ 5 3{ = ] 1 x 1 3gx = 1 3ln |x| + F = 13ln |5 3{| + F.

15.Letx = w. Then gx = gw and gw =1gx, soUsin w gw =Usin x1gx= 1( cos x) + F = 1cos w + F. 17.Letx = 3d{ + e{3. Then gx = (3d + 3e{2) g{ = 3(d + e{2) g{, so

] _{d + e{}2 3d{ + e{3g{ = ] 1 3gx x1@2 = 1₃ ] x1@2_{gx =} 1 3· 2x2+ F = 23 s 3d{ + e{3_{+ F.} 19.Letx = ln {. Then gx = g{ {, so ] _{(ln {)}2 { g{ = U x2_{gx =}1 3x3+ F = 13(ln {)3+ F. 21.Letx =w. Then gx = gw 2wand 1 wgw = 2 gx, so ] cosw w gw = U

cos x (2 gx) = 2 sin x + F = 2 sinw + F. 23.Letx = sin . Then gx = cos g, soUcos sin6 g =Ux6gx = 1₇x7+ F = 1₇sin7 + F.

25.Letx = 1 + h{. Thengx = h{g{, soUh{1 + h{g{ =U x gx =2₃x3@2+ F = 2₃(1 + h{)3@2+ F.

Or: Let x =1 + h{_{. Then}_x2 _{= 1 + h}{_and_{2x gx = h}{_{g{, so} U

h{_{1 + h}{_{g{ =}U_{x · 2x gx =} 2

3x3+ F = 23(1 + h{)3@2+ F. 27.Letx = 1 + }3. Thengx = 3}2g} and }2g} = 1₃gx, so

] }2 3 1 + }3g} = ] x1@31 3gx =1 3 ·32x2@3+ F = 12(1 + }3)2@3+ F.

29.Letx = tan {. Then gx = sec2{ g{, soUhtan {sec2{ g{ =Uhxgx = hx+ F = htan {+ F.

31.Letx = sin {. Then gx = cos { g{, so

] _{cos {} sin2_{g{ = ] ₁ x2gx = ] x2_{gx = x}1 1 + F = 1x+ F = 1sin {+ F [or csc { + F ].

33.Letx = cot {. Then gx = csc2{ g{ and csc2{ g{ = gx, so ]

cot { csc2_{{ g{ =}] _{x (gx) = x}3@2

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35.

] _{sin 2{}

1 + cos2_{g{ = 2

] _{sin { cos {}

1 + cos2_{g{ = 2L. Let x = cos {. Then gx = sin { g{, so 2L = 2] _{1 + x}x gx₂ = 2 ·1

2ln(1 + x2) + F = ln(1 + x2) + F = ln(1 + cos2{) + F.

Or: Let x = 1 + cos2_{. 37.

]

cot { g{ =] cos {_{sin {}g{. Let x = sin {. Then gx = cos { g{, so] cot { g{ =] _x1gx = ln |x| + F = ln |sin {| + F. 39. Letx = sec {. Then gx = sec { tan { g{, so

U

sec3_{{ tan { g{ =}U_sec2_{{ (sec { tan {) g{ =}U_x2_{gx =}1

3x3+ F = 13sec3{ + F. 41. Letx = sin1{. Then gx = 1

1 {2 g{, so ] _g{ 1 {2_sin1_{= ] ₁ xgx = ln |x| + F = lnsin1{ +F. 43. Letx = 1 + {2. Thengx = 2{ g{, so ] 1 + { 1 + {2g{ = ] 1 1 + {2g{ + ] { 1 + {2 g{ = tan1{ + ] 1 2gx x = tan1{ +12ln|x| + F = tan1_{{ +}1 2ln1 +{2 +F = tan1{ +12ln 1 + {2_{+ F [since 1 + {}2_{A 0].} 45. Letx = { + 2. Then gx = g{, so ] { 4 { + 2g{ = ] x 2 4 x gx = ] (x3@4_2x1@4_{) gx =} 4 7x7@4 2 ·43x3@4+ F = 4 7({ + 2)7@483({ + 2)3@4+ F

In Exercises 47–50, leti({)denote the integrand andI ({)its antiderivative (withF = 0).

47. i({) = {({2₁₎3_. _{x = {}2_{1 gx = 2{ g{> so} U {({2₁₎3_{g{ =}U_x31 2gx =1 8x4+ F =18({2 1)4+ F Wherei is positive (negative), I is increasing (decreasing). Where i changes from negative to positive (positive to negative),I has a local minimum (maximum).

49. i({) = sin3_{{ cos {. x = sin { gx = cos { g{, so} U

sin3{ cos { g{ =Ux3gx = 1

4x4+ F = 14sin4{ + F Note that at{ = ₂,i changes from positive to negative and I has a local maximum. Also, bothi and I are periodic with period , so at { = 0 and at{ = , i changes from negative to positive and I has local minima.

51. Letx = { 1, so gx = g{. When { = 0, x = 1; when { = 2, x = 1. Thus,U₀2({ 1)25g{ =U₁1 x25gx = 0 by Theorem 7(b), sincei(x) = x25is an odd function.

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53.Letx = 1 + 2{3, sogx = 6{2g{. When { = 0, x = 1; when { = 1, x = 3. Thus, U1 0{2 1 + 2{35_{g{ =}U3 1x5 ₁ 6gx =1 6 ₁ 6x6 3 1= 361(36 16) = 361(729 1) = 72836 =1829 . 55.Letx = w@4, so gx = 1₄gw. When w = 0, x = 0; when w = , x = @4. Thus,

U 0 sec2(w@4) gw = U@4 0 sec2x (4 gx) = 4 tan x@4₀ = 4tan 4 tan 0 = 4(1 0) = 4. 57.U_@6@6 tan3_{g = 0 by Theorem 7(b), since i() = tan}3_{is an odd function.}

59.Letx = 1@{, so gx = 1@{2g{. When { = 1, x = 1; when { = 2, x = 1₂. Thus, ] 2 1 h1@{ {2 g{ = ] 1@2 1 h x_{(gx) =}_hx1@2 1 = (h1@2 h) = h h. 61.Letx = 1 + 2{, so gx = 2 g{. When { = 0, x = 1; when { = 13, x = 27. Thus,

] 13 0 g{ 3 t (1 + 2{)2 = ] 27 1 x 2@31 2gx =k1 2· 3x1@3 l27 1 = 3 2(3 1) = 3.

63.Letx = {2+ d2, sogx = 2{ g{ and { g{ =1₂gx. When { = 0, x = d2; when{ = d, x = 2d2. Thus, ] d 0 { s {2_{+ d}2_{g{ =}]2d 2 d2 x 1@21 2gx = 1 2 k 2 3x3@2 l2d2 d2 = k 1 3x3@2 l2d2 d2 = 1 3 k (2d2₎3@2_(d2₎3@2l₌ 1 3 22 1d3 65.Letx = { 1, so x + 1 = { and gx = g{. When { = 1, x = 0; when { = 2, x = 1. Thus,

] 2 1 { { 1 g{ =] 1 0 (x + 1) x gx =] 1 0 (x 3@2_{+ x}1@2_{) gx =}k2 5x5@2+23x3@2 l1 0= 2 5+23 =1615. 67.Letx = ln {, so gx = g{

{ . When{ = h, x = 1; when { = h4;x = 4. Thus, ] h4 h g{ {ln {= ] 4 1 x 1@2_{gx = 2}k_x1@2l4 1= 2(2 1) = 2.

69.Letx = h}+ }, so gx = (h}+ 1) g}. When } = 0, x = 1; when } = 1, x = h + 1. Thus, ] 1 0 h}_{+ 1} h}_{+ }}g} = ] h+1 1 1 xgx = k ln |x|lh+1 1 = ln |h + 1| ln |1| = ln(h + 1). 71.From the graph, it appears that the area under the curve is about

1 +a little more than1₂· 1 · 0=7, or about1=4. The exact area is given by D =U1

0

2{ + 1 g{. Let x = 2{ + 1, so gx = 2 g{. The limits change to 2 · 0 + 1 = 1 and 2 · 1 + 1 = 3, and D =U₁3x1 2gx = 1 2 k 2 3x3@2 l3 1= 1 3 33 1=3 1 3 1=399.

73.First write the integral as a sum of two integrals: L =U₂2 ({ + 3)4 {2_{g{ = L1}_{+ L2}₌U2 2{ 4 {2_{g{ +}U2 23 4 {2_{g{. L1}_{= 0 by Theorem 7(b), since}

i({) = {4 {2_{is an odd function and we are integrating from}_{{ = 2 to { = 2. We interpret L2}_{as three times the area of} a semicircle with radius2, so L = 0 + 3 ·1₂ · 22= 6.

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75. First Figure Letx ={, so { = x2andg{ = 2x gx. When { = 0, x = 0; when { = 1, x = 1. Thus, D1=U₀1h{_{g{ =}U1 0hx(2x gx) = 2 U1 0 xhxgx. Second Figure D2=U₀12{h{g{ = 2U₀1xhxgx.

Third Figure Letx = sin {, so gx = cos { g{. When { = 0, x = 0; when { = ₂,x = 1. Thus, D3=U₀@2hsin {_{sin 2{ g{ =}U@2

0 hsin {(2 sin { cos {) g{ = U1

0 hx(2x gx) = 2 U1

0 xhxgx. SinceD1= D2= D3, all three areas are equal.

77. The rate is measured in liters per minute. Integrating fromw = 0 minutes to w = 60 minutes will give us the total amount of oil that leaks out (in liters) during the rst hour.

U60 0 u(w) gw = U60 0 100h0=01wgw [x = 0=01w, gx = 0=01gw] = 100U0=6₀ hx_{(100 gx) = 10,000}_hx0=6 0 = 10,000(h0=6 1) r 4511=9 r 4512 liters 79. The volume of inhaled air in the lungs at timew is

Y (w) =U₀wi(x) gx =U₀w1 2sin ₂ 5 x gx =U₀2w@51 2sin y ₅ 2gy substitutey =2₅ x, gy =2₅ gx = 5 4 cos y2w@5₀ = 5 4 cos2 5 w + 1= 5 4 1 cos2 5 w liters 81. Letx = 2{. Then gx = 2 g{, soU₀2i(2{) g{ =U₀4i(x)₂1gx= 1₂U₀4i(x) gx =1₂(10) = 5. 83. Letx = {. Then gx = g{, so Ue di({) g{ = Ue di(x)(gx) = Ud e i(x) gx = Ud e i({) g{ From the diagram, we see that the equality follows from the fact that we are reecting the graph ofi, and the limits of integration, about the |-axis. 85. Letx = 1 {. Then { = 1 x and g{ = gx, so

U₁ 0{d(1 {)eg{ = U₀ 1(1 x)dxe(gx) = U₁ 0xe(1 x)dgx = U₁ 0{e(1 {)dg{. 87. _{1 + cos}{ sin {₂_{ = { ·_{2 sin}sin {₂_{ = { i(sin {), where i(w) = _{2 w}w ₂. By Exercise 86,

] 0 { sin { 1 + cos2_{g{ = ] 0 { i(sin {) g{ = 2 ] 0 i(sin {) g{ = 2 ] 0 sin { 1 + cos2_{g{ Letx = cos {. Then gx = sin { g{. When { = , x = 1 and when { = 0, x = 1. So

2 ] 0 sin { 1 + cos2_{g{ = ₂ ] 1 1 gx 1 + x2 = ₂ ] 1 1 gx 1 + x2 = ₂ tan1_x1 1 = ₂[tan1_{1 tan}1_{(1)] =} 2k4 ₄l= ₄2