Integration CHAPTER 5 EXERCISE SET Endpoints 0, 1 n, 2 n,..., n 1, 1; using right endpoints, 2 n A n =

Transkript

1 CHAPTER 5 Itegratio EXERCISE SET 5.. Edpoits,,,...,, ; usig right edpoits, ] A = A Edpoits,,,...,, ; usig right edpoits, A = ] 5 5 A Edpoits, π, π )π,...,, π; usig right edpoits, A = siπ/) + siπ/) + + siπ )/) + si π] π 5 5 A Edpoits, π, π )π,...,, π ; usig right edpoits, A = cosπ/) + cosπ/) + + cos )π/) + cosπ/)] π 5 5 A Edpoits, +, +,...,, ; usig right edpoits, A = ] 5 5 A Edpoits π, π + π, π + π,..., π )π + A = cos π + π ) + cos π + π ) + + cos 5 5 A , π ; usig right edpoits, π ) )π π ) ] π + + cos 5

2 6 Chapter 5 7. Edpoits,,,...,, ; usig right edpoits, ) ) ) A = A Edpoits, +, + ),..., +, ; usig right edpoits, ) ) ) A = A Edpoits, +, +,...,, ; usig right edpoits, ] A = e + + e + + e e + e 5 5 A Edpoits, +, +,...,, ; usig right edpoits, A = l + ) + l + ) l ) ] + l 5 5 A Edpoits,,,...,, ; usig right edpoits, ) ) ) ] A = si + si si + si ) 5 5 A Edpoits,,,...,, ; usig right edpoits, ) ) ) ] A = ta + ta ta + ta ) 5 5 A ). 5 ) 5. + ) 6. ) 7. + ) ) 8. )

3 Eercise Set false; the area is π. false; cosider the left edpoit approimatio o, ].. true. true; a differetiable fuctio is cotiuous. A6) represets the area betwee = ad = 6; A) represets the area betwee = ad = ; their differece A6) A) represets the area betwee = ad = 6, ad A6) A) = 6 ) = 6.. A9) = 9 /, A ) = ) /, ad the area betwee = ad = 9 is give b A9) A ) = 9 ) )/ = B is also the area betwee the graph of f) = ad the iterval, ] o the ais, so A + B is the area of the square. 6. If the plae is rotated about the lie = the A becomes B ad vice versa. 7. The area which is uder the curve lies to the right of = or to the left of = ). Hece f) = A ) = ; = Aa) = a, so take a = or a = to measure the area to the left of = ). 8. f) = A ) =, = Aa) = a a, so take a = or a = ).. Ituitivel it is the area represeted b a set of tall thi rectagles, stretchig from = a to = b, each havig height C; i other words Cb a). Aalticall it is give b b a f) + C) f)] d = Cb a). EXERCISE SET 5.. a) d = + + C + b) + )e d = e + C. a) b) d si cos + C) = cos cos + si = si d ) d d + C + / = = ) / d + 5] = d d d ] = + ) so so + 5 d = C + ) d = + + C d )] cos ) si = d d si cos ] = si d so so cos ) d = si ) + C si d = si cos + C

4 8 Chapter 5 9. a) 9 /9 + C b) 7 /7 + C c) 9 9/ + C. a) 5 5/ + C b) C = C c) 8/8 + C. 5 + ] 5 d = 5 d + 5 d = 5 + ) C = C.. / 7/5 + ] d = 9 / + 8 ] d = / d d 7/5 d + 9 d = / 5 / C / d + 8 d = 5 5/ C. + ] / d = d / d + d = ) / / + ) / + C = / / C )d = / + 5 /5 + C + + )d = C / + )d = + )d = + + C + )d = / / + / ) + C t )dt = t t + C / / + 7/ )d = / 7 7/ + / + C... ] + e d = l + e + C t ] e t dt = l t e t + C si sec ] d = cos ta + C csc t sec t ta t] dt = cot t sec t + C sec + sec ta )d = ta + sec + C csc si + cot ) d = + csc cot ) d = csc + C

5 Eercise Set sec θ cos θ dθ = sec θ dθ = ta θ + C si d = cos + C 9. sec ta d = sec + C. φ + csc φ)dφ = φ / cot φ + C. + si θ)dθ = θ cos θ + C sec + ] d = ta + + C ] + d = si ta + C ] + si si d = + cos d = d = sec + + ) d = sec + + +ta +C si sec cos d = sec ta ) d = ta sec + C cos d = sec d = ta + C 7. true 8. true; both are atiderivatives ot the same C though) 9. false; ) =. true. 5. c/ c/ -5. a) ) = / d = / + C, ) = + C =, C = 5 ; ) = / + 5 π ) b) t) = si t + ) dt = cos t + t + C, = + π + C = /, C = π ; t) = cos t + t + π c) ) = / + / )d = / + / + C, ) = = 8 + C, C = 8, ) = / + / 8. a) ) ) = 8 d = 6 + C, ) = = 6 + C, C = 6 ; ) = 6 + 6

6 Chapter 5 b) t) = sec t si t) dt = ta t + cos t + C, π ) = = + + C, C = ; t) = ta t + cos t c) ) = 7/ d = 9 9/ + C, ) = = C, C = ; ) = 9 9/ 5. a) = e d = e + C, = ) = + C, C =, = e b) t) = t dt = l t + C, ) = C = 5, C = 5; t) = l t a) = b) t dt = si t + C, ) = = π + C, C = π, = si t π d d = +, = ] d = ta + C, + ) = π = π + C, C = π, = ta + π 7. st) = 6t + C; st) = 6t + 8. st) = si t + C; st) = si t + 9. st) = t / + C; st) = t / 5 5. st) = e t + C; st) = e t e 5. f ) = / + C ; f) = 5 5/ + C + C 5. f ) = / + si + C, use f ) = to get C = so f ) = / + si +, f) = /6 cos + + C, use f) = to get C = so f) = /6 cos d/d = +, = + )d = + + C; = whe = so ) + ) + C =, C = 6 thus = f ) = m = + ), so f) = + ) d = + ) + C; f ) = 8 = + ) + C = + C, = 8 + = 5, f) = + ) f ) = m = si so f) = si )d = cos + C; f) = = + C so C =, f) = cos d/d =, = d = / + C; = whe = so ) / + C =, C = 7/ thus = / + 7/ 57. d/d = 6d = + C. The slope of the taget lie is so d/d = whe =. Thus ) + C =, C = 6 so d/d = 6, = 6)d = 6 + C. If =, the = 5 ) = so ) 6) + C =, C = 7 thus =

7 Eercise Set a) f) = si 7 si + cos b) f) = a) 6 b) c) f) = / 6. a) 6 b) c) = e + )/ 6. This slope field is zero alog the -ais, ad so correspods to b) This slope field is idepedet of, is ear zero for large egative values of, ad is ver large for large positive. It must correspod to d).

8 Chapter This slope field has a egative value alog the -ais, ad thus correspods to c) This slope field appears to be costat approimatel ), ad thus correspods to differetial equatio a) Theorm 5.. sas that ever atiderivative is of the form F ) + C, for some C. I particular, the atiderivative of is C for some C, ot for ever C. Differet problems will have differet values of C). 66. The first equalit is icorrect because the right had side evaluates to / + C / + C ) = C C. 67. a) F ) = ) +, G ) = + + / = + = F ) b) F ) = π/; G) = ta ) = π/, ta + ta /) = π/ c) Draw a triagle with sides ad ad hpoteuse +. If α deotes the agle opposite the side of legth ad if β deotes its complemet, the ta α = ad ta β = /, ad siα + β) = si α cos β + si β cos α = + + =, ad cosα + β) = cos α cos β + si α si β = + =, so the cosie of α + β is zero ad the sie of α + β is ; + cosequetl α + β = π/, i.e. ta + ta /) = π/.

9 Eercise Set a) For, F ) = G ) =. But if I is a iterval cotaiig the either F or G has a derivative at, so either F or G is a atiderivative o I. b) Suppose G) = F ) + C for some C. The F ) = ad G) = + C, so C =, but F ) = ad G ) =, a cotradictio. c) No, because either F or G is a atiderivative o, + ). 69. sec )d = ta + C a) cos )d = si ) + C b) d 7. For >, d sec ] =, ad for <, d d sec ] = d d sec )] = ) = which ields formula ) i both cases. 7. v = 87 7 csc )d = cot + C + cos ) d = + si ) + C T / dt = 87 7 T / +C, v7) = 87 = 87+C so C =, v = 87 7 T / ft/s 7. dt/d = C, T = C + C ; T = 5 whe = so C = 5, T = C + 5. T = 85 whe = 5 so 5C + 5 = 85, C =., T =. + 5 EXERCISE SET 5.. a) u du = u / + C = + ) / + C b). a) b). a) b). a) b) 8 5. a) b) u du = u / + C = cos )/ + C si u du = cos u + C = cos + C u / du = u/ + C = C sec u du = ta u + C = ta + ) + C u / du = 6 u/ + C = 6 + ) / + C u / du = π π u/ + C = π si/ πθ) + C u /5 du = 5 9 u9/5 + C = ) 9/5 + C u du = u + C = cot + C u 9 du = u + C = + si t) + C

10 Chapter 5 6. a) b) 7. a) cos u du = si u + C = si + C sec u du = ta u + C = ta + C u ) u / du = u 5/ u / + u / )du = 7 u7/ 5 u5/ + u/ + C b) 8. a) = 7 + )7/ 5 + )5/ + + )/ + C csc u du = cot u + C = cotsi ) + C si u du = cos u + C = cos π) + C b) du u = u + C = C 9. a) du = l u + C = l l + C u b) e u du = 5 5 eu + C = 5 e 5 + C. a) u du = l u + C = l + cos θ + C du b) u = l u + C = l + e ) + C. a) u =, du + u = ta ) + C b) u = l, du = u si l ) + C. a) u =, u u du = sec ) + C b) u = du, + u = ta u + C = ta ) + C 5. u =, 6. u = 5 +, 7. u = 7, 7 8. u = /, u 9 du = u + C = ) + C u du = 6 u/ + C = ) / + C si u du = 7 cos u + C = cos 7 + C 7 cos u du = si u + C = si/) + C 9. u =, du = d; sec u ta u du = sec u + C = sec + C

11 Eercise Set u = 5, du = 5d; 5. u =, du = d; sec u du = 5 ta u + C = ta 5 + C 5 e u du = eu + C = e + C. u =, du = d;. u =,. u =, 5. u = 7t +, du = t dt; u du = l u + C = l + C u du = si ) + C + u du = ta ) + C 6. u = 5, du = d; u / du = u/ + C = 7t + ) / + C u / du = 5 u/ + C = C 7. u =, du = d, du = ) ) u u + C = 8. u = +, du = + ) d, u du = + + C ) + C 9. u = 5 +, du = d,. u =, du = d, du u du = u + C = 5 + ) + C si u du = cos u + C = cos ) + C. u = si, du = cos d; e u du = e u + C = e si + C. u =, du = d; e u du = eu + C = e + C. u =, du = 6, 6 e u du = 6 eu + C = 6 e + C. u = e e, du = e + e )d, 5. u = e, u du = l u + C = l e e + C + u du = ta e ) + C 6. u = t, u + du = ta t ) + C 7. u = 5/, du = 5/ )d; 5 si u du = 5 cos u + C = 5 cos5/) + C 8. u =, du = d; sec u du = ta u + C = ta + C

12 6 Chapter 5 9. u = cos t, du = si t dt, u du = 5 u5 + C = 5 cos5 t + C. u = si t, du = cos t dt; u 5 du = u6 + C = si6 t + C. u =, du = d; sec u du = ta u + C = ta ) + C. u = + si θ, du = 8 cos θ dθ; 8. u = si θ, du = cos θ dθ;. u = ta 5, du = 5 sec 5 d; 5 5. u = ta, 6. u = cos θ, u du = u du = si ta ) + C u + du = ta cos θ) + C 7. u = sec, du = sec ta d; u + C = + si θ) + C u / du = 6 u/ + C = 6 si θ)/ + C u du = u + C = ta 5 + C u du = 6 u + C = 6 sec + C 8. u = si θ, du = cos θ dθ; si u du = cos u + C = cossi θ) + C 9. e d; u =, du = d; e u du = e u + C = e + C 5. e / d; u = /, du = d/; e u du = e u + C = e / + C = e + C 5. u =, du = d;, e u du = e u + C = e + C 5. u = +, du = d; + e u du = e u + C = e + + C 5. u = +, du = d; u ) du = u 6 u/ u + C = 6 + )/ + + C 5. u =, du = d; u) u du = 8 u/ + 5 u5/ + C = 5 )5/ 8 )/ + C 55. si θ si θ dθ = cos θ) si θ dθ; u = cos θ, du = si θ dθ, u )du = u + 6 u + C = cos θ + 6 cos θ + C

13 Eercise Set sec θ = ta θ +, u = θ, du = dθ sec θ dθ = ta u + ) sec u du = 9 ta u + ta u + C = 9 ta θ + ta θ + C ) dt = t + l t + C t 58. e l = e l =, >, so e l d = d = + C 59. le ) + le ) = le e ) = l = so le ) + le )]d = C 6. cos d; u = si, du = cos d; si du = l u + C = l si + C u 6. a) si /) + C b) / 5) ta / 5) + C c) / π) sec / π) + C 6. a) u = e, + u du = ta e /) + C b) u =, du = 9 u si /) + C, c) u = 5, u u du = sec 5/ ) + C 6. u = a + b, du = b d, a + b) d = u du = b 6. u = a + b, du = b d, d = b du u / du = b a + b)+ b + ) b + ) u+)/ + C = + C b + ) a + b)+)/ + C 65. u = sia + b), du = b cosa + b)d u du = b b + ) u+ + C = b + ) si+ a + b) + C 67. a) with u = si, du = cos d; u du = u + C = si + C ; with u = cos, du = si d; u du = u + C = cos + C b) because the differ b a costat: ) si + C ) cos + C = si + cos ) + C C = / + C C 68. a) First method: 5 + )d = C ; secod method: 5 u du = 5 u + C = 5 5 ) + C

14 8 Chapter 5 b) 5 5 ) + C = ) + C = C ; the aswers differ b a costat = + d = )/ + C; = ) = 6 + C, 5 so C = 6 = , ad = )/ = + si ) d = cos + C ad π ) = = π C, C = π, = π + cos 7. = e t dt = et + C, 6 = ) = + C, = et + 7. = 5 + 9t dt = ) 5 ta 5 t + C, π = 5 ) = π 5 + C, C = π 6, = 5 ta 5 t ) + π 6 7. a) u = +, du = d; u du = u + C = + + C b) a) u = +, du = d; b) u du = l u + C = l + ) + C 75. f ) = m = +, f) = + ) / d = 9 + )/ + C f) = = 9 + C, C = 7 9, so f) = 9 + )/ pt) = +.t) / dt = +.t)5/ + C; = p) = 5/ + C, C = / 8.8 so that

15 Eercise Set 5. 9 p5) = + 5.))5/ + /.5 so that the populatio at the begiig of the ear is approimatel, u = a si θ, du = a cos θ dθ; du a u = θ + C = si u a + C 78. If u > the u = a sec θ, du = a sec θ ta θ dθ, du u u a = a θ = a sec u a + C EXERCISE SET 5.. a) = 6 b) = 55 c) = d) = 6 e) = f) =. a) + + = b) + + = c) π + π + + π = π d) = terms) e) f) =. k= k. k= k 5. k= k 6. 8 k ) 7. k= 6 ) k+ k ) 8. k= 5 ) k+ k k= 9. a). a). 5 k= k 5 ) k+ a k k= b) 5 ) k+ b k k= b) c) ) + ) = k= 5 k= k ) a k k k= k + k= d) 5 a 5 k b k k= = 7 )) + = 5, ))) = 87. k k= kk ) = k k) = k 6 k k= k= k= k= k= 6 k = 6)7) 6) 7) = k= k = k= k = + ) = + ) k= k = 87 = 856 k= k = ) ) )) =,65

16 Chapter k= k= k= k = k= k = k= 5 k ) = 5 k = 6 )) ) = ) ) 6 k = ) = ) k= k= k = 5 ) + ) =. true. false; the value of a fuctio at the midpoit of a iterval eed ot be the average of the values of the fuctio at the edpoits of the iterval. false; if a, b] cosists of positive reals, true; but false o, e.g., ].. false; e.g. si o, π] 5. a) b) + ), + 6 ), + 9 ) ),..., ) +, + ) Whe, 5] is subdivided ito equal itervals, the edpoits are, +, +, +,..., + ), + = 5, ad the right edpoit approimatio to the area uder the curve = is give b the summads above. + k ) gives the left edpoit approimatio. k= 6. is the umber of elemets of the partitio, k is a arbitrar poit i the k-th iterval, k =,,,...,,, ad is the width of a iterval i the partitio. I the usual defiitio of area, the parts above the curve are give a + sig, ad the parts below the curve are give a sig. These umbers are the replaced with their absolute values ad summed. I the defiitio of et siged area, the parts give above are summed without cosiderig absolute values. I this case there could be lots of cacellatio of positive areas with egative areas. 7. Edpoits,,, 5, 6; = ; a) Left edpoits: f k) = = 6 b) c) Midpoits: k= f k) = = 5 k= Right edpoits: f k) = = 58 k= 8. Edpoits,, 5, 7, 9, = ; a) Left edpoits: f k) = b) Midpoits: k= f k) = k= ) = ) = 5 8

17 Eercise Set 5. c) Right edpoits: f k) = k= ) = Edpoits:, π/, π/, π/, π; = π/ a) Left edpoits: f k) = + / + ) / π/) = π/ b) c) Midpoits: k= f k) = cosπ/8) + cosπ/8) + cos5π/8) + cos7π/8)] π/) k= Right edpoits: = cosπ/8) + cosπ/8) cosπ/8) cosπ/8)] π/) = ) f k) = / + / π/) = π/ k=. Edpoits,,,, ; = a) f k) = = b) c) k= f k) = = k= f k) = + + = k=. a).7877,.7588, b).66877,.6888, c).69856, , a).76969,.7775,.6875 b).58586,.6886, c) , , a).8877, , b) , , c) 5.779, , a).997, , b).88,.5767,.5 c).7688,.87558, =, k = + k; f k) = k = f k) = ] + k = + k= k= k= A = lim + + )] = + ) = = 5, k = + k 5 ; f k) = 5 k) = + ) k = ] + ) = + ] k k ) 5 = 5 5 k ] +

18 Chapter 5 f k) = k= A = k= lim k = 5 5 k= )] = 5 5 = 5 7. =, k = + k ) ; f k) = 9 9 k ) f k) = 9 9 k = 7 ) k k= k= k= ] A = lim 7 7 ) + k = 7 7 = 8 k= + ) = 5 5 = 7 7 k= ) + k 8. =, k = k f k) = f k) = k= A = k) ] = k= lim = 7 + k= k 9k ] = 7k + ) + ) 6 + ) + ) = 9 8 ) + )] = 9 )) = 9/ 8 9. =, k = + k f k) = k) = k= A = f k) = lim + = = + ] k = + 6 k= k= k + + ] k = k + 8 k= ) ) + ) + ) + + = + ) + )) + ) ] = k= k ] + 6 k + k + 8 ] k 6 + ) + ) + 8 ] + ) + ) + ) + + ) ]. =, k = + k ; f k ) = k) ] = + ) ] k f k) = k= = + ) + ) ) k ) ] + ) k 8 k ] ]

19 Eercise Set 5. A = lim = ) = + ) ) + ) + ) ]. =, k = + k ) f k) = k = + k ) ] = + k ) 9 ] f k) = ] + 9 k ) = + 9 ] ) k= k= k= A = lim )] = + 9 = 5 = + 9. = 5, k = 5 k ) f k) = 5 k) = 5 5 ] 5 k ) = 5 5 k ) f k) = 5 5 k ) = 5 5 k= k= k= A = lim 5 5 )] = = 5. =, k = + k ) ; k f ) k) = 9 9 f k) = k= A = k ) 9 9 k= lim = ] = 7 ) + + = 8 k= ] k ) ) = 7 7 k= k + 5 k= k 7. =, k = k ) f k) = f k) = k= A = lim + k) ] = k= 7 k= k + 7 9k ) k= ] = 7k k 7 k= = ) + ) + ) = 9 + ) + ) ) + + 7k )] + + = 9 )) = 9/ 8 5. Edpoits,, 8 ),...,, =, ad midpoits, 6,,..., 6,. Approimate the area with the sum ) k = 6 ] + ) 6 eact) as +. k=

20 Chapter 5 6. Edpoits, +, + 8 +, + 6, + ),..., k )) = k= ),..., +, + = 5, ad midpoits k= which is eact, because f is liear. 7. =, k = k f k) = f k) = k= k ) ) = k k + k= k k=, 5 6 k + 8 k + Usig Theorem 5.., A = lim f k) = = k= k= 8. =, k = + k f k) = + k ) = 8k 8k + f k) = 8 k 8 k + k= A = lim + k= k= k= f k) = =. Approimate the area with the sum ) = 6 + ) + 8 = 8 =, 9. =, k = + k f k) = + k ) = + k f k) = + k= A = lim + k= k= f k) = k = + + ) = + + The area below the -ais cacels the area above the -ais. 5. =, k = + k f k) = + k ) = + 9 k k= A = f k) = ) lim + k= f k) = =

21 Eercise Set 5. 5 The area below the -ais cacels the area above the -ais that lies to the left of the lie = ; the remaiig area is a trapezoid of width ad heights,, hece its area is + =. 5. =, k = k k ) f k) = ] = 8k f k) = 8 k= A = lim + k= k= k k= f k) = 6 6 = = 8 + ) + ) 6 5. =, k = + k f k) = + k ) = + k k k + 6 f k) = + + ) + ) + ) ) 6 k= A = lim + k= f k) = = 5. a) With k as the right edpoit, = b, k = b k ) f k) = k) = b k, A = b lim + k= + ) = b / f k) = b k = b k= + ) b) First Method tedious) = b a, k = a + b a k f k ) = k) = a + b a ] k b a = b a a + a b a) k + f k) = b a) k= A = lim + k= f k) ab a) k b a) + k a + a b a) ) + ) ab a) ] + + ) b a) = b a) a + a b a) + ab a) + ] b a) = b a ) ] Alterative method: Appl Part a) of the Eercise to the iterval, a] ad observe that the area uder the curve ad above that iterval is give b a. Appl Part a) agai, this time

22 6 Chapter 5 to the iterval, b] ad obtai b. Now subtract to obtai the correct area ad the formula A = b a ). 5. Let A be the area of the regio uder the curve ad above the iterval o the -ais, ad let B be the area of the regio betwee the curve ad the iterval o the -ais. Together A ad B form the square of side, so A + B =. But B ca also be cosidered as the area betwee the curve = ad the iterval o the -ais. B Eercise 7 above, B =, so A = =. 55. If = m the m + m ) = m k = k= mm + ) m+ if = m + the m + ) + m ) = k ) m+ m+ = k = k= k= m + )m + ) k= m + ) = m + ) = = kk+) = k + k = k= k= ) ) ) = ) + ) ) = ) + ) + + ) 9 = = ) + ) + + ) = k= = mm + ) = + ; 6 + = 8, a) b) 6. a) b) k= lim + k= lim + k )k + ) = + = kk + ) = = k= + = = = k= k ) + k ) k + ) + 5 ] = + + k + ) = + = + ) + 5 ) )] + ) + ) + + ) +

23 Eercise Set i ) = i = i= i= i= i= i= i= i but = i = so i ) = = i thus i= 6. S rs = ar k k= k= ar k+ = a + ar + ar + + ar ) ar + ar + ar + + ar + ) = a ar + = a r + ) so r)s = a r + ), hece S = a r + )/ r) 65. both are valid 66. d) is valid 67. a k b k ) = a b ) + a b ) + + a b ) = a + a + + a ) b + b + + b ) = a k k= k= k= b k 68. a) b) c) d) meas add to itself times, which gives the result. k= k= k= k= k = + ) = +, so k = + ) + ) 6 lim + k = k= = , so k = ) + ) = + +, so lim + lim + k = k= k = k= EXERCISE SET 5.5. a) /)) + 5/)) + )) = 7/6 b). a) /)π/) + )π/) + )π/) + /)π/) = )π/8 b) π/. a) 9/)) + )) + 6/6)) + 5)) = 7/6 b). a) 8)) + )) + )) + 8)) = b) d 6. )d 8. π/ d si d

24 8 Chapter 5 9. a) lim. a) lim ma k k= ma k k= k k ; a =, b = b) lim k k, a =, b = ma k k= k k + k; a =, b = b) lim ma k k= + cos k) k, a = π/, b = π/. Theorem 5.5.a) depeds o the fact that a costat ca move past a itegral sig, which b Defiitio 5.5. is possible because a costat ca move past a limit ad/or a summatio sig.. If f) for all i a, b] the we kow that positivit or oegativit) is preserved uder limits ad sums, hece also b Defiitio 5.5.) for itegrals.. a) A = )) = 9/ b) A = ) + ) = / A 5 A c 6 A c) A + A = + 8 = 5/ d) A + A = 5 A A A. a) A = )) = A 5 c) A = /)) = / 6 b) A = )/ + /) = A $ A d) A A = / = / A A A

25 Eercise Set a) A = 5) = b) ; A = A b smmetr A 5 A c 6 A c) A + A = 5)5/) + )/) 5 = / A A 6. a) A = 6)5) = A 5 6 d) b) π) ] = π/ A A + A = because A = A b smmetr A $ A c) A + A = )) + )) = 5/ d) π) = π 7. a) b) c) d) 8. a) A A f) d = + ) d Triagle of height ad width, above -ais, so aswer is. f) d = + ) d + ) d Two triagles of height ad base ; aswer is. 6 d = ) d + 6 ) d Triagle of height ad base together with a triagle of height ad base, so + 8 =. 6 f) d = + ) d + + ) d + A ) d + 6 ) d Triagle of height ad base, below ais, plus a triagle of height, base above ais, aother of height ad base above ais, ad a triagle of height ad base, above ais. Thus f) = =. d =area of a triagle with height ad base, so.

26 5 Chapter 5 b) c) d) d = d + d Two triagles of height ad base o opposite sides of the -ais, so the cacel to ield. d Rectagle of height ad base 9, area = 8 / d + 5 d Trapezoid of width / ad heights ad, together with a rectagle of height ad base, so / + + = / + 8 = 5/ 9. a).8 b).6 c).8 d).. a) b) 9 c) 8 d) 75. f)d + g)d = 5 + ) =. f)d g)d = ) =.. 5 f)d = 5 f)d f)d = ) = f)d = f)d = f)d + ] f)d = 6) = 5. d 5 d = 5 / + )/) = d d = ) )/ = 8 d + d = / + π/) = + π)/ d + 9 d = + π) )/ = 6 + 9π/ 9. false; e.g. f) = if >, f) = otherwise, the f is itegrable o, ] but ot cotiuous.. true; cos is strictl positive o, ], so the itegrad is positive there, so the itegral is positive.. false; e.g. f) = o, +]. true; Theorem a) b) >, < o, ] so the itegral is egative cos > for all ad for all ad > for all > so the itegral is positive

27 Eercise Set a) >, > o, ] so the itegral is positive b) 9 <, + > o, ] so the itegral is egative 5. If f is cotiuous o a, b] the f is itegrable o a, b], ad, cosiderig Defiitio 5.5., for ever partitio ad choice of f ) we have m k f k) k M k. This is equivalet to mb a) result. k= k= k= f k) k Mb a), ad, takig the limit over ma k we obtai the k= , so + d ) d = π5) / = 5π/ 8. + )d = 5/. 9 ) d = π) / = 9π/ d = π) / = π. a) The graph of the itegrad is the horizotal lie = C. At first, assume that C >. The the regio is a rectagle of height C whose base eteds from = a to = b. Thus b) b a C d = area of rectagle) = Cb a). If C the the rectagle lies below the ais ad its itegral is the egative area, i.e. C b a) = Cb a). Sice f) = C, the Riema sum becomes lim f ma k) k = lim C k ma k = lim Cb a) = Cb a). k ma k k= B Defiitio 5.5., b a k= f) d = Cb a).. For a partitio of, ] we have f ) = or f ) = ; accordigl, either we have f k) k = k =, or we have k= f k) k = k= k= k =. k= This is because f) = for all ecept possibl, which could be. Both possibilities ted to i the limit, ad thus f) d =.. Each subiterval of a partitio of a, b] cotais both ratioal ad irratioal umbers. If all k are chose to be ratioal the f k) k = ) k = k = b a so lim f k) k = b a. k= k= If all k are irratioal the lim k= ma k k= the precedig limits are ot equal. ma k k= f k) k =. Thus f is ot itegrable o a, b] because

28 5 Chapter 5. Choose a large positive iteger N ad a partitio of, a]. The choose i the first iterval so small that f ) > N. For eample choose < /N. The with this partitio ad choice of, f k) k > f ) > N. This shows that the sum is depedet o partitio k= ad/or poits, so Defiitio 5.5. is ot satisfied. 5. a) f is cotiuous o, ] so f is itegrable there b Theorem 5.5. b) c) d) f) so f is bouded o, ], ad f has oe poit of discotiuit, so b Part a) of Theorem f is itegrable o, ] f is ot bouded o -,] because lim f) = +, so f is ot itegrable o,] f) is discotiuous at the poit = because lim si does ot eist. f is cotiuous elsewhere. f) for i, ] so f is bouded there. B Part a), Theorem 5.5.8, f is itegrable o, ]. EXERCISE SET 5.6. a) b) c) )d = /) ] = / = ] d = = ) ) = + )d = / + ) ] = 9/ + / + ) = 6. a) c). a) 5 ] 5 d = / = 5/ b) + )d = / + ) = *, f*)) ] b) 9 = / + 6 / ) = / = f*) = ] 9 5d = 5 = 59) 5) = c) = + *, f*)). a) 5.5 = *, f*)) b) 5 = 5 f*) = 5 c) = + *, f*)) ] d = / = 8/ 6/ = 65/ 6. ] d = 5 /5 = /5 )/5 = /5

29 Eercise Set d = / ] = 6 = 8. 7 ] 7 / d = / = ) = 6 9. l e d = ] l e = ) =. 5 ] 5 d = l = l 5 l = l 5. a) b) ] d = / = = f ) ), so f ) =, = + ) d = + ] = 5, so f ) )) = 5, ) + =, = 6, 7 but ol 7 lies i the iterval. f 7) = 9 7 =, so the area is that of a rectagle wide ad high.. a) f ave = π. b) f ave = π π 6 + ) d = si d = ; si =, = π,, π d = ; ) =, = ] ] + = + 8 ) = 8. ] ) d = ) = 8 6 ) = ] d = = + = 6. ] 9 5 5/ = 8/ d = ] 5 5 = /6 d = ] = + = ] π/ ] π/ 9. cos θ =. ta θ = π/ ] π/. si =. sec ) ] π/ = π π/ 9. 5e ] l ] = 5e 5) = 5e. l )/ = l )/ / ] / 5. si = si / ) si = π/ ] 6. ta = ta ta ) = π/ π/) = π/ 7. sec ] = sec sec = π/ π/ = π/

30 5 Chapter 5 ] / 8. sec = sec / ) + sec ) = 5π/6 + π/ = π/ 9. t t /)] =. )] π/ cot = π /9 + π/6. a) b). a) π/ d = cos d + / π/ π/ d + ) d + / cos )d = si ] / ) d = ) ] π/ ] π/ si = / π/ ] + d = )/ + )/] + ] + ) = 5 / = ) + 8 ) = 8 + ) b). a) b) π/ cos /) d + e )d+ π/ π/ / cos ) d = si /) ] π/ ] π/ + / si ) π/ = / π/6) + π/ π/6 /) = π/ e )d = e ) d + d = l ] ] +e ) ] + l = e )+e = e+/e ] = l + l + l =. a) The fuctio f) = 5 is a eve fuctio ad chages sig at =, thus + b) f) d = f) d = = 8 ta ) + 6 ta ) / d = ) = si = π 6 + si ) / f) d + ] ) f) d 5. a) 7/6 b) F ) = / d + / ] d + = π + π +, + 6, >

31 Eercise Set a) d + b) F ) = d = ] / ] = 7/ /, < + 5, 7. false; cosider F ) = / if ad F ) = / if 8. true 9. true. true, = ; d = ] = /..5767; π/..6789;..9865; 5. A = 6. A = 7. A = 8. A = π/ + )d = ) d = 9. Area = ] π/ si d = cos = sec d = ta = ta.855 ] d = l = l )] + = si d = cos )] = = 6 ] π/ d = ] = 5/ ) d + = 9/ ) d = 5/6 + /6 = A A

32 56 Chapter 5 5. Area = π si d π π/ si d = + = A i 6 A 5. Area = e ) d + e ) d = /e + e A A 5. Area = )/ d + / )/ d = / + / = A A 5. a) A =.8 ].8 d = si = si.8) b) The calculator was i degree mode istead of radia mode; the correct aswer is a) the icrease i height i iches, durig the first te ears b) the chage i the radius i cetimeters, durig the time iterval t = to t = secods c) the chage i the speed of soud i ft/s, durig a icrease i temperature from t = F to t = F d) the displacemet of the particle i cm, durig the time iterval t = t to t = t hours 56. a) Let the areas i quadrats IV, III, I ad II be A, A, A ad A respectivel. The it appears that A > A, ad A > A. Sice the total area is give b A + A A A, the area is positive. b) Area = a) F ) = b) ) d = 59 6 t ) dt = t t) ] = +, ad d d + ) =

33 Eercise Set a) cos b) F ) = si t ] π/ = si, F ) = cos 59. a) si b) e 6. a) 6. cos 6. u + b) l 6. F ) = + 9, F ) = + 9 a) b) 5 c) 6. F ) = ta, F ) = + a) b) π/ c) / 65. a) F ) = = whe =, which is a relative miimum, ad hece the absolute + 7 miimum, b the first derivative test. b) icreasig o, + ), decreasig o, ] c) 66. F F ) = + 7) = ad o 7, + ) 7 ) + ) + 7) ; cocave up o, 7), cocave dow o, ) 5 t 67. a), + ) because f is cotiuous there ad is i, + ) b) at = because F ) = 68. a), ) because f is cotiuous there ad is i, ) b) at = because F ) = 69. a) amout of water = rate of flow)time) = t gal, total amout = ) = gal b) amout of water = c) amout of water = 6 + t/)dt = gal + t)dt = gal 7. a) The maimum value of R occurs at : P.M. whe t =. b) 6.t )dt = 58 cars

34 58 Chapter k= Thus ) πk = lim π sec + k= k= π sec + k = + k / so + k = for. Thus f k) where f) = sec, k = πk ad = π for π. ) πk π/ ] π/ = lim f k) = sec d = ta = + lim k= + k= k= + k = k= f k) where f) = +, k = k, ad = lim + k= f k) = + d = π. 7. Let f be cotiuous o a closed iterval a, b] ad let F be a atiderivative of f o a, b]. B F b) F a) Theorem 5.7., = F ) for some i a, b). B Theorem 5.6., b a b a f) d = F b) F a), i.e. b a f) d = F )b a) = f )b a). EXERCISE SET 5.7. a) displ = s) s) = b) c) dist = dt = displ = s) s) = dist = vt) dt = displ = s) s) = dist = vt)dt = dt = dt = t)dt + vt) dt = t t /) ] t )dt = t t /) + t / t) ] ] t / t) ] ] + t / t) = / = /; d) displ = s) s) = dist = = t / vt)dt = tdt + ] + t ] dt + tdt + 5/ + 5t t ) dt + 5 t)dt + ] 5/ 5 t)dt = t / + t 5t) 5/ ] t 5)dt 5/ = ] ] ] + t + 5t t ) = /;. v 8 t

35 Eercise Set a) vt) = + t au)du; add areas of the small blocks to get v) = 5. m/s b) v6) = v) + 6 au)du = 5. m/s. a) egative, because v is decreasig b) speedig up whe av >, so < t < 5; slowig dow whe < t < c) egative, because the area betwee the graph of vt) ad the t-ais appears to be greater where v < compared to where v > 5. a) st) = t t + C; = s) = C, so st) = t t + b) vt) = cos t + C ; = v) = + C, C =, so vt) = cos t +. The st) = si t + t + C ; = s) = C, so st) = si t + t + 6. a) st) = t cos t + C ; = s) = + C, C =, so st) = t cos t b) vt) = t t + t + C ; = v) = C, so C =, vt) = t t + t. The st) = t t + t + C ; = s) = +C, so C =, st) = t t + t 7. a) st) = t + t + C; = s) = C, C = ad st) = t + t b) vt) = t + C, = v) = + C, C = ad vt) = t + so st) = l t + t + C, = s) = + C, C = ad st) = l t + t + 8. a) st) = t / dt = 5 t5/ + C, s8) = = 5 + C, C = 96 5, st) = 5 t5/ 96 5 tdt b) vt) = = t/ + C, v) = = 8 + C, C =, vt) = t/, st) = t/ ) dt = 5 t5/ t + C, s) = 5 = 5 + C = 5 + C, C = 9 5, st) = 5 t5/ t a) displacemet = sπ/) s) = distace = π/ si t dt = m b) displacemet = sπ) sπ/) = distace = π π/ cos t dt = π/ π π/ π/ π/ ] π/ si tdt = cos t = m ] π cos tdt = si t cos tdt + π π/ π/ = m cos tdt = m

36 6 Chapter 5. a) displacemet = distace = b) displacemet = distace = t ) dt = t dt = / t dt = 5 m t dt = 5 m. a) vt) = t t + t = tt )t ) displacemet = distace = b) displacemet = distace =. a) displacemet = distace = b) displacemet =. v = t distace = displacemet = distace = t ) dt + t t + t)dt = 9/ m vt) dt = vt)dt + t )dt = 6 m vt) dt = t t) dt = 8 m t t dt = m t + dt = m t + dt = m t ) dt = m t dt = m / vt)dt + vt)dt = 6 m t ) dt = + 8 = m vt)dt = / m. vt) = t t displacemet = distace = 5 5 ) t t dt = / m t t dt = ) t t dt + 5 ) t t dt = 7/ m 5. v = / t + dt = t + + C; v) = / so C = /, v = t + + / 5 ) displacemet = t + + dt = 96 7 m 5 ) distace = t + + dt = 96 7 m

37 Eercise Set vt) = cos t + displacemet = distace = π/ π/ π/ cos t + )dt = π + )/ m cos t + dt = π/ π/ π/ cos t + )dt = π + )/ m 7. a) s = si πt dt = π cos πt + C b) s = whe t = which gives C = π so s = π cos πt + π. a = dv dt = π cos πt. Whe t = : s = /π, v =, v =, a =. v = t dt = t + C, v = whe t = which gives C = so v = t s = t dt = t + C, s = whe t = which gives C = so s = t a) s = Whe t = : s = /, v = /, v = /, a =. cos πt dt = πt si π + C s = whe t = which gives C = π so s = πt si π π a = dv dt = π πt si ; whe t = : s = π π, speed = v =, a = π b) v = e t dt = e t + C, v = e whe t =, hece C = so v = et. The st) = vt) dt = e t t + C, s = e whe t =, so C = ad st) = e t t. Whe t =, s) =, v) =, speed =, a) =. 9. B ispectio the velocit is positive for t >, ad durig the first secod the particle is at most 5/ cm from the startig positio. For T > the displacemet of the particle durig the time iterval, T ] is give b T vt) dt = 5/ + T 6 t /t) dt = 5/ + t / l t) ad the displacemet equals cm if T / l T = /, T.7 s ] T = / + T / l T,. The displacemet of the particle durig the time iterval, T ] is give b T vt)dt = ta T.5T. The particle is cm from its startig positio whe ta T.5T = or whe ta T.5T = ; solve for T to get T =.9,.5, ad.95 s.. st) = t t + ) dt = t 55t + t + C. But s = whe t =, so C = ad s = t 55t + t. Moreover, at) = d vt) = t. dt

38 6 Chapter st) vt) at) 6. at) = t, vt) = t t +, st) = t 5t + t 5; st) vt) at). true; if at) = a the vt) = a t + v. true 5. false; cosider vt) = t o, ] 6. true 7. a) positive o,.7) ad.97, 5), egative o.75,.97) b) For < T < 5 the displacemet is disp = T/ sit ) + T cost ).5 8. a) the displacemet is positive o, ).5 b) For < T < the displacemet is disp = π + T π cos πt T si πt π

39 Eercise Set a) the displacemet is positive o, 5).5 b) For < T < 5 the displacemet is disp = T + T + )e T 5. a) the displacemet is egative o, ). b) {. a) at) = For < T < the displacemet is disp = T ) lt + ) + ) T l T + T +, t <, t > { b) vt) =. 5, t < 65 t, t > 5 a t v t 6 8 c) t) = { 5t, t < 65t 5t 8, t >, so 8) =, ) = d) 6.5) =.5. Take t = whe deceleratio begis, the a = so v = t + C, but v = 88 whe t = which gives C = 88 thus v = t + 88, t a) b) v = 5 mi/h = 66 ft/s, 66 = t + 88, t = s v = the car is stopped) whe t = 8 s s = v dt = t + 88)dt = t + 88t + C, ad takig s = whe t =, C = so s = t + 88t. At t = 8, s = 5. The car travels 5 ft before comig to a stop.. a = a ft/s, v = a t + v = a t + ft/s, s = a t / + t + s = a t / + t ft; s = ft whe v = 88 ft/s. Solve 88 = a t + ad = a t / + t to get a = whe t = 5,

40 6 Chapter 5 so s =.t + t, v = 5 t +. a) a = 5 ft/s b) v = 55 mi/h = c) v = whe t = 6 s ft/s whe t = 7 s. dv/dt = 5, v = 5t + C, but v = v whe t = so C = v, v = 5t + v. From ds/dt = v = 5t + v we get s = 5t / + v t + C ad, with s = whe t =, C = so s = 5t / + v t. s = 6 whe t = thus 6 = 5) / + v ), v = 5 m/s 5. The truck s velocit is v T = 5 ad its positio is s T = 5t + 5. The car s acceleratio is a C = ft/s, so v C = t, s C = t iitial positio ad iitial velocit of the car are both zero). s T = s C whe 5t + 5 = t, t 5t 5 = t + 5)t 5) =, t = 5 s ad s C = s T = t = 5 ft. 6. Let t = correspod to the time whe the leader is m from the fiish lie; let s = correspod to the fiish lie. The v C =, s C = t 5; a L =.5 for t >, v L =.5t + 8, s L =.5t + 8t. s C = at t = 5/ 9.58 s, ad s L = at t = , so the challeger wis. 7. s = ad v = whe t = so vt) = t +, st) = 6t + t a) v) = 6 ft/s, v5) = 8 ft/s b) v = whe the projectile is at its maimum height so t + =, t = 7/ s, s7/) = 67/) + 7/) = 96 ft. c) s = whe it reaches the groud so 6t + t =, 6tt 7) =, t =, 7 of which t = 7 is whe it is at groud level o its wa dow. v7) =, v = ft/s. 8. s = whe t = so st) = 6t + v t +. But s = whe t = thus 6) + v ) + =, v = ft/s. 9. a) st) = whe it hits the groud, st) = 6t + 6t = 6tt ) = whe t = s. b) The projectile moves upward util it gets to its highest poit where vt) =, vt) = t + 6 = whe t = / s. 8. a) st) = s gt = 8 6t ft, st) = whe t = 6 = 5 b) vt) = t ad v5 ) = ft/s = 5.8 mi/h. st) = s + v t gt = 6t.9t m ad vt) = v gt = 6 9.8t m/s a) b) c) d) vt) = whe t = 6/ s s6/9.8) 8.67 m aother 6. s; solve for t i st) = to get this result, or use the smmetr of the parabola s = 6t.9t about the lie t = 6. i the t-s plae also 6 m/s, as see from the smmetr of the parabola or compute v6.)). a) the are the same b) st) = v t gt ad vt) = v gt; st) = whe t =, v /g; v) = v ad vv /g) = v gv /g) = v so the speed is the same at lauch t = ) ad at retur t = v /g).. st) =.9t + 9t + 5 ad vt) = 9.8t + 9

41 Eercise Set a) b) the projectile reaches its maimum height whe vt) =, 9.8t + 9 =, t = 5 s s5) =.95) + 95) + 5 = 7.5 m c) the projectile reaches its startig poit whe st) = 5,.9t + 9t + 5 = 5,.9tt ) =, t = s d) e) f) v) = 9.8) + 9 = 9 m/s st) = whe the projectile hits the groud,.9t + 9t + 5 = whe use the quadratic formula) t.6 s v.6) = 9.8.6) , the speed at impact is about 7. m/s. take s = at the water level ad let h be the height of the bridge, the s = h ad v = whe t = so st) = 6t + h a) s = whe t = thus 6) + h =, h = 56 ft b) First, fid how log it takes for the stoe to hit the water fid t for s = ) : 6t + h =, t = h/. Net, fid how log it takes the soud to travel to the bridge: this time is h/8 because the speed is costat at 8 ft/s. Fiall, use the fact that the total of these two times must be s: h 8 + h =, h + 7 h =, h + 7 h =, ad b the quadratic formula h = 7 ± 7) + ), reject the egative value to get h 5.5, h 9.5 ft. 5. If g = ft/s, s = 7 ad v is ukow, the st) = 7 + v t 6t ad vt) = v t; s = s ma whe v =, or t = v /; ad s ma = 8 ields 8 = sv /) = 7 + v v /) 6v /) = 7 + v /6, so v = 8. ft/s. 6. s = + v t )t = + v t 6t ; s = whe t = 5, so v = )/5 = 8 ft/s, ad v5) = v t = ft/s EXERCISE SET 5.8. a) f ave = c) 8 d = b) =, =. a) f ave = d = / b) ) = /, = ±/, but ol / is i, ]

42 66 Chapter 5 c). f ave =. fave = d = ] = 6 8 / d = 8 ) 9 π / ] 8 = 5 5. fave = π si d = π cos ] π = π 6. fave = π π/ sec ta d = ] π/ π sec = π 7. fave = e e d = l e l ) = e e 8. fave = 9. fave = + l 5 l 5 e d = + l 5 5 e ) ] d + = ta = ] d. fave = = / si = π /. f ave =. fave = π ) π e d = 8 e ] = 8 e 8 ) π/ π/ sec d = π ta ] π/ π/ = π π π ) = π. a) 5 f.) + f.8) + f.) + f.6) + f.)] = ] = 5.8 b).) +.) ) +.) ] = 86 =.5 c) f ave = d = ] = d) Parts a) ad b) ca be iterpreted as beig two Riema sums = 5, = ) for the average, usig right edpoits. Sice f is icreasig, these sums overestimate the itegral.. a) b)

43 Eercise Set c) f ave = + ) ] d = + l ) = + l.6978 d) Parts a) ad b) ca be iterpreted as beig two Riema sums = 5, = ) for the average, usig right edpoits. Sice f is decreasig, these sums uderestimate the itegral. 5. a) b) vt) dt = vt) dt = t) dt + t dt + dt + t ) dt =, so v ave = 6 t + 5) dt = + + =, so v ave = 5 6. Fid v = ft) such that ft) dt =, ft), f 5) = f ) =. Let ft) = ct5 t); the 5 5 ct5 t) dt = ct ] ] 5 5 ct = c 5 ) = 5c =, c =, so v = ft) = t5 t) satisfies all the coditios. ) a + b 7. Liear meas fα + β ) = αf ) + βf ), so f = fa) + fa) + fb) fb) =. 8. Suppose at) represets acceleratio, ad that at) = a for a t b. The the velocit is give b vt) = a t + v, ad the average velocit = b a t + v ) dt = a b a a b + a) + v, ad the ) a + b a + b velocit at the midpoit is v = a + v which proves the result. 9. false; f) =, g) = o, ]. true; Theorem 5.5.a). true; Theorem 5.5.b). false;f) = g) = o, ], fave = gave = /, fg)ave = /. a) v ave = b) v ave = s) s). a) a ave = 5 b) a ave = 5 vπ/) v) π/ t + )dt = = 7 t + )dt = 7/ 789 = 6 = / = = )/π π/ 5. time to fill tak = volume of tak)/rate of fillig) = π) 5]/) = 5π, weight of water i tak at time t = 6.) rate of fillig)time) = 6.t, weight ave = 5π 5π 6.t dt = π =.8 lb 6. a) If is the distace from the cooler ed, the the temperature is T ) = 5 +.5) C, ad T ave = 5 +.5)d =.5 C

44 68 Chapter 5 b) B the Mea-Value Theorem for Itegrals there eists i, ] such that f ) = 5 +.5)d =.5, =.5, = 5 7..t )dt = 9 cars, so a average of 9 = 97 cars/mi. 8. Vave = From the chart we read t, t dv dt = ft) =, t t +, t 5 ] e.7t dt = e.7t = $,.96 It follows that costats of itegratio are chose to esure that V ) = ad that V t) is cotiuous) t, t V t) = t, t t + t, t 5 Now the average rate of chage of the volume of juice i the glass durig these 5 secods refers to the quatit 5 V 5) V )) = 5 = 8, ad the average value of the flow rate is f ave = 5 ft) dt = 5 ] t dt + dt + t + ) dt 5 = ] = 8. 5 π. a) J ) = cossi t) dt b) π c) d) J ) = if =.86.5 = J ) Solve for k : k ] k d = 6k, so / = k / = 6k, k = ) = 7 5. w t) = kt, wt) = kt / + w ; w ave = kt / + w ) dt = k t Solve 66k/ + w = kt / + w for t, t = 66/, so t 9.76 ] w = 66 k + w EXERCISE SET 5.9. a) 5 u du b) 5 9 u du c) π π/ π/ cos u du d) u + )u 5 du

45 Eercise Set a) 7 u 8 du b) 5/ / u du c) u du d) u )u / du. a) e u du b) u du. a) π/ π/ u du b) / du u 5. u = +, 6. u =, 7. u =, 8. u =, 6 u du = ] 8 u = or ] 8 + ) = u du = 6 u ] 6 = 8, or ] ) = 8 6 u du =, because u is odd o, ]. u 8 du = ] 7 u9 = 9, or ] )9 = u = +, 9 u )u / du = 9 u / u / )du = 5 u5/ u/ ] 9 = 9/5, or 5 + )5/ + )/ ] 8 = 9/5. u =, u) u du = )/ )5/ 5 π/. u = /, 8 ] u/ ] 5 u5/ = 6/5 or = 6/5 ] π/ si u du = 8 cos u = 8 ] π/, or 8 cos/) = 8. u =, π/ cos u du = ] π/ si u = /, or ] π/6 si = /. u = +,. u =, π/ 6 u du = ] u = /8, or 6 π/ + ) ] = /8 ] π/ sec u du = ta u = 8, or ta π/ )] +π = 8 π

46 7 Chapter 5 5. u = e +, du = e d, u = e l + = + = whe = l 7 ] 7 u = e l + = + = 7 whe = l, du = l u = l7) l/) = l/), or / u / ] l le + ) = l 7 l/) = l/) l 6. u = e, du = e d, u = whe =, u = 7 whe = l 5 7 u du = ] 7 8 u = 6, or ] l 5 8 e ) = 6 7. u =, ta ] = π/6 / 8. u = e, / 9.. ] l/ ) si e l ] u + du = ta u ] / du = u si u = π + π 6 = π/6 = ta ta ) = π/ π/) = π/6 or / 5 5 u du = ] 5 π5) = 5 6 π. u du = = si + si = π + π 6 = π 6 or u du = π) ] = π/8 m 6 u du = ] π) = π.. 9 u du = π) / = 9 π si πd = ] π cos π = ) = /π m π. A = A = 7. A = π/8 cos d = ] π/8 si = / ] ] 9 d = 9 + ) = 9 + ) = 6 /6 8. = si, A = ] d + ) = = + ) d = / 9 du = ] / u si u = π/8 π/ ] π/ si d = cos =

47 Eercise Set u =, 9 du = ] 9 u =. u ] 5 )/ = 8/5 5. ] + 9) / = ). u = cos +, 6 u 5 du =. u = + + 7, 8 ] 8 u / du = u / = 8 = 7 ). ) d = ] = / 5. si ] π/ = u =, π π ] π si u du = cos u = π ] π/ ta )/ = / 7. ] π 5 si ) = 9. u = θ, π/ π/ sec u du = ] π/ ta u = )/ π/. u = cos θ, / ] u du = l u / = l. u =, = u), d = du 6 8u + u du = 7 u / 7 = 7 6u / 8u / + u / )du u / 6 u/ + 5 u5/ ] = 6/5. u = 5 +, 9 u 5 u du = 9 u / 5u / )du = u/ u / ] 9 = 8/. ] e l + e) = l le) l e) =. ] e = e e )/ 5. u =, ] du = u u si = π ) = π 6 6. u = ], du = u u si = π/ π/6) = π/6 7. u =, ] + u du = ta u = π = π 9

48 7 Chapter 5 8. u =, + u du = u ] ta = π π/ π/6) = 9. b) 5. b) π/6 π/ π/ si si ) cos d = ta sec ) d = ta 5 si5 ) 7 si7 π/ π/ π/ π/ = + ta + ) π/ π/ π/6 sec ) d = = 8 = + π = + π 5. a) u = +, c) u =, / 5. u =, 5. si = cosπ/ ), π/ si d = 5. u =, fu)du = 5/ b) u =, fu)du = / m ) d = = 55. t) = 8.7) π/ π/ fu)du = / u) m u du = 9 u u) m du = cos π/ )d = cos u du u = π/ ) π/ cos u du = u)u du = = π/ cos d b replacig u b ) u)u du = + ) + ) fu)du = 5/ ) m d u u + )du = + + e.58t dt = 5.959e.58t + C; ) = 75 = C, C = 5., t) = 5.959e.58t + 5., ) 8,,5, 56. st) = 5 + e.5t )dt = 5t e.5t + C a) b) 57. a) b) s) s) = 5 e.5 ) = 5 / e 8.69 ft es; without it the distace would have bee 5 ft ] = si. +.8) d = a) V rms = /f = fv p /f Vp si πft)dt = fv p t πf siπft) ] ] /f /f cosπft)]dt = V p, so Vrms = V p /

49 Eercise Set b) V p / =, V p = 69.7 V 59. k e d =, e ] k =, ek ) =, e k = 7, k = l 7 6. The area is give b k = / + k )d = / k) ta k) =.6; solve for k to get 6. a) si πd = /π 6. a) Usig Part b) of Theorem with g) = it follows that I b) = u, d = u du, I = is impossible because + /u /u )du = d =. u du = I so I = which + is positive o, ]. The substitutio u = / is ot valid + because u is ot cotiuous for all i, ]. 6. a) Let u = the b) a a f)d = a a f u)du = a a so, replacig u b i the latter itegral, a a f)d = a a f)d, a a a f u)du = fu)du a f)d =, a The graph of f is smmetric about the origi so thus a a a a so a a f)d = f)d = f)d = a f)d = a a f)d + a a f) d + a f u)du = f)d + a f)d = a f)d = a f)d, let u = i a a f u)du = f)d = a a f)d is the egative of a fu)du = f)d f)d to get a f)d a f)d The graph of f) is smmetric about the -ais so there is as much siged area to the left of the -ais as there is to the right. 6. Let u = t, the du = d ad t ft )g)d = t fu)gt u)du = t fu)gt u)du; the result follows b replacig u b i the last itegral. 65. a) I = = a a fa u) fa u) + fu) du = du a a fa u) + fu) fu) du fa u) + fu) fu) du, I = a I so I = a, I = a/ fa u) + fu)

50 7 Chapter 5 b) / c) π/ 66. a) B Eercise 6a), b) cos ) d = u = π/, du = d, siu + π/) = si u, cosu + π/) = si u π π/ π/ si 8 cos 5 d = si 8 u si 5 u) du = si u du = b Eercise 6a). π/ π/ EXERCISE SET 5.. a) b) c) t.5 t e t. t ] ac ] /c. a) l t = lac) = l a + l c = 7 b) l t = l/c) = 5 c) ] a/c ] a l t = la/c) = 5 = d) l t = l a = l a = 6 ] a. a) l t c) l t ] /a = l a / = ] a l a = 9/ b) l t = l + 9 ] a = l 9 d) l t = 9 l 5. l ; l 5 =.6979; magitude of error is <.6 6. l.9865; l =.98689; magitude of error is <. 7. a), > b), c), < < + d), < < + e), > f) l +, > g), < < + h) e, > 8. a) fl ) = e l = e l/9) = /9

51 Eercise Set b) fl ) = e l + e l = + e l/) = + / = 7/ 9. a) π = e π l b) = e l. a) π = e l π b) l = e. a) =, lim + ) = lim + ) ] / = lim + ) ] / = e / b) =, lim + ) / = lim + ) /] = e. a) = /, lim + ) ] = + b) lim + lim + ) / = lim + ) /] / = e / + ) ] = e. g ) =. g ) = cos 5. a) ) = 6. a) + b) 7. F ) = si +, F ) = + ) cos si + ) b) e l = ) ) si a) b) c) 8. F ) = +, F ) = + a) b) c) 6/ 9. true. true. false; itegral does ot eist. true. a) b). a) b) d d d d d d t + tdt = + ) = + t + tdt = + ) / ) 5/ 5 a a g) ft)dt = d d ft)dt = d d ft)dt = f) a g) a ft)dt = fg))g )

52 76 Chapter 5 5. a) cos b) ta + ta sec = ta 6. a) + ) b) cos ) ) 8. If f is cotiuous o a ope iterval I ad g), h), ad a are i I the g) h) so d d ft)dt = g) h) a h) ft)dt + g) a ft)dt = h) ft)dt = fh))h ) + fg))g ) a ft)dt + 9. a) si ) ) si )) = si ) si ) b) + ) ) = g) a ft)dt = cos /). F ) = 5 5) ) = so F ) is costat o, + ). F ) = l 5 so F ) = l 5 for all >.. from geometr, ft)dt =, = 5 7 ft)dt = 6, 7 t 7)/dt = a) F ) =, F ) =, F 5) = 6, F 7) = 6, F ) = 5 ft)dt = ; ad 7 ft)dt b) F is icreasig where F = f is positive, so o /, 6] ad 7/, ], decreasig o, /] ad 6, 7/] c) critical poits whe F ) = f) =, so = /, 6, 7/; maimum 5/ at = 6, miimum 9/ at = / d) F) F is icreasig resp. decreasig) where f is icreasig resp. decreasig), amel o, ) ad 7, ) resp. 5, 7)). The ol edpoit commo to two of these itervals is = 7, ad that is the ol poit of iflectio of F.. < : F ) = : F ) = = t)dt t] = ), t)dt + t dt = + ; F ) = { )/, < + )/,

53 Eercise Set : F ) = > : F ) = 5. ) = + 6. ) = t dt + t dt =, dt = + ) = ; F ) = t ] + dt = + t + l t) = + l + t t / + t / )dt = / + / = / + / 8 7. ) = + sec t si t)dt = ta + cos / π/ ] 8. ) = + d = + l l t = + l l e l e 9. P ) = P + T rt)dt idividuals. st ) = s + vt)dt { /,, >. II has a miimum at =, ad I has a zero there, so I could be the derivative of II; o the other had I has a miimum ear = /, but II is ot zero there, so II could ot be the derivative of I, so I is the graph of f) ad II is the graph of ft) dt.. b) lim k k k ) = d dt t] = l t=. a) where ft) = ; b the First Derivative Test, at t = b) where ft) = ; b the First Derivative Test, at t =, 5 c) at t =, or 5; from the graph it is evidet that it is at t = 5 d) at t =, or 5; from the graph it is evidet that it is at t = e) F is cocave up whe F = f is positive, i.e. where f is icreasig, so o, /) ad, ); it is cocave dow o /, ) ad, 5) f).5.5 F) 5. a) erf)

54 78 Chapter 5 c) e) g) erf ) > for all, so there are o relative etrema erf ) = e / π chages sig ol at = so that is the ol poit of iflectio lim erf) = +, lim erf) = + 5. C ) = cosπ /), C ) = π siπ /) a) cos t goes from egative to positive at kπ π/, ad from positive to egative at t = kπ + π/, so C) has relative miima whe π / = kπ π/, = ± k, k =,,..., ad C) has relative maima whe π / = k + )π/, = ± k +, k =,,.... b) si t chages sig at t = kπ, so C) has iflectio poits at π / = kπ, = ± k, k =,,...; the case k = is distict due to the factor of i C ), but chages sig at = ad siπ /) does ot, so there is also a poit of iflectio at = 6. Let F ) = lim h h +h l tdt, F F + h) F ) ) = lim = lim h h h h l tdt = l +h l tdt; but F ) = l so 7. Differetiate: f) = e, so + provided e a =, a = l )/. ft)dt = + a a e t dt = + e t ] a = + e e a = e 8. a) The area uder /t for t + is less tha the area of the rectagle with altitude / ad base, but greater tha the area of the rectagle with altitude / + ) ad base. b) + dt = l t t ] + = l + ) l = l + /), so / + ) < l + /) < / for >. c) from Part b), e /+) < e l+/) < e /, e /+) < + / < e /, e /+) < + /) < e; b the Squeezig Theorem, lim + + /) = e. d) Use the iequalit e /+) < + /) to get e < + /) + so + /) < e < + /) From Eercise 8d) e + 5) 5 < 5), ad from the graph 5) < F ) = f), thus F ) has a value at each i I because f is cotiuous o I so F is cotiuous o I because a fuctio that is differetiable at a poit is also cotiuous at that poit

55 Review Eercises, Chapter 5 79 REVIEW EXERCISES, CHAPTER / + C. u / u + 7u + C. cos + si + C. sec ta + )d = sec + + C 5. / 5e + C 6. l ta + C 7. ta + si + C 8. sec + + C 9. a) ) = / + C; ) =, so C =, ) = / b) ) = si 5e + C, ) = = 5 + C, C = 5, ) = si 5e + 5 c) ) = + d) ) = t / dt = + ] t/ = 5 + / te t dt = e. The directio field is clearl a odd fuctio, which meas that the solutio is eve ad its derivative is odd. Sice si is periodic ad the directio field is ot, that elimiates all but, the solutio of which is the famil = / + C.. a) If u = sec, du = sec ta d, sec ta d = udu = u / + C = sec )/ + C ; if u = ta, du = sec d, sec ta d = udu = u / + C = ta )/ + C. b) The are equal ol if sec ad ta differ b a costat, which is true.. ] π/ sec = ] π/ ) = / ad ta = ) = /. u =, du = d, du u u = sec u + C = sec ) + C. + / d = / / + d; u = / +, du = / d u / du = u / + C = / + ) / + C 5. u = 5 + si, du = 6 cos d; 6 u du = u/ + C = 5 + si + C 6. u = +, du = d; 7. u = a + b, du = a d; 8. u = a, du = ad; a udu = u/ + C = + ) / + C au du = au + C = a + ab + C sec udu = a ta u + C = a taa ) + C

56 8 Chapter 5 9. a) k= k + )k + ) b) 9 k=5 k )k ). a) k b) k ) = k. lim k= + k=. lim 6 = lim + + k= k= k ) ] k + ) 5k ) k= = = + ) = lim ) + ) 6 k k ) k= ] ] 5k ) 5 = = lim + 6 ] =..5577,.5596, , , a) + = c) 5 ) = 5 b) = d) 8. a) e) ot eough iformatio f) ot eough iformatio + = 5 b) ot eough iformatio c) ot eough iformatio d) ) = 9. a) d + d = ) + π) / = + π/ b) + ) / ] c) u =, du = d; π) / = / ) 9π/ u du = π) / = π/8. a) k= lim + k = k= k= k = f k) where f) =, k = k/, ad = / for. Thus / d =

57 Review Eercises, Chapter 5 8 b) c) k= lim + k= lim ) k = e k/ + k= k= f k) where f) =, k = k/, ad = / for. Thus k= ) k = d = 5 = f k) where f) = e, k = k/, ad = / for. Thus k= e k/ = lim + k= f k) = e d = e )] + 7 = 8. d = ] = /. )] sec = / sec) 6. + )] 5 5 = 8/ 5/ + )] 8 = 79/ 6 t t/ + )] t = 55/ / π/6 9 )d + / / si ) d + ] 9 d = / ] / ] )d = ) + ) = 9/ + / = 5/ / π/ π/6 si /) d = / + cos ) ] π/6 ] π/ cos + /) π/6 = π/ + /) π/ + / + π/) = π/ = 7 ) = 5/. /5 d = 5 ] /5 = 5 /5 ). ] ] e e e d = e = e e. d = l = l =. A = + )d = + )] = /6. The ol positive zero of f is b = + 5, ad the area is give b A = 5. A = A + A = 6. A = A + A = )d + = ) ] + )/ )d = / + / = / ] + d + + ] d ) ] + + )/ = + + b f) d = =

58 8 Chapter 5 7. a) + b) F ) = )] t + t = + 5 ; F ) = + 8. a) F ) = b) F ) = ] t = ; F ) = 9. e 5. cos cos 5. cos + si 5. l ) 56. a) F ) = + 7 ; icreasig o, ],, + ), decreasig o, ] b) F ) = ; cocave dow o, ), cocave up o, + ) + 7) c) lim F ) =, so F has o absolute etrema. ± d).5 F) a) F ) = /) / ) = so F is costat o, + ). b) F ) = + t dt + + t dt = ta = π/, so F ) = ta + ta /) = π/. 58., ) because f is cotiuous there ad is i, ) 59. a) The domai is, + ); F ) is if =, positive if >, ad egative if <, because the itegrad is positive, so the sig of the itegral depeds o the orietatio forwards or backwards). b) 6. F ) = The domai is, ]; F ) is if =, positive if <, ad egative if < ; same reasos as i Part a). t dt, F ) =, so F is icreasig o, ]; F ma = F ) t + ad F mi = F ) a) f ave = / d = /; = /, = b) fave = e e d = e l ] e = e ; = e, = e

59 Review Eercises, Chapter Mar to Ju 7 is weeks, so wt) = t s t ds =, so the weight o Jue 7 will be gm For < < the area betwee the curve ad the -ais cosists of two triagles of equal area but of opposite sigs, hece. For < < 5 the area is a rectagle of width ad height. For 5 < < 7 the area cosists of two triagles of equal area but opposite sig, hece ; ad for 7 < < the curve is give b = t 7)/ ad average is ) =.. 6. fave = 7 t 7)/ dt =. Thus the desired l e + e ) d = l e + e ) d = l l/) l/) l l l 65. If the acceleratio a = cost, the vt) = at + v, st) = at + v t + s. 66. a) o, sice the velocit curve is ot a straight lie b) 5, ] c) s = d) e) vt) dt ) = 5.8 = 5 ft )/8 =.8/8 =.85 ft/s o sice the velocit is positive ad the acceleratio is ever egative f) eed the positio at a oe give time e.g. s ) 67. st) = t t + )dt = t t + t + C, s) = ) ) + + C =, C =, st) = t t + t vt) = cos t dt = si t + C, v) = si + C =, C =, vt) = si t, st) = si t )dt = cos t t + C, s) = cos + C =, C =, st) = cos t t 69. st) = t )dt = t t + C, s) = ) ) + C = 5, C = 7, st) = t t vt) = st) = cos t t) dt = si t t + v ; but v = so vt) = si t t vt)dt = cos t t / + C : s) = = + C, C =, st) = cos t t / + 7. displacemet = s6) s) = distace = 7. displacemet = distace = t dt = t dt = 6 t )dt = t t) t dt = / m t)dt + t )dt ] 6 = m t )dt = t t ) t )dt = / m ] ] 6 + t t) = m