2 ) 5. (a) (1)(3) + (1)(2) = 5 (b) {area of shaded region in Fig. 24b} < 5

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1 Odd Aswers: Chapter Four Cotemporary Calculus PROBLEM ANSWERS Chapter Four Sectio 4.. (a) ()() + (8)(4) = 5 (b) ()() ()(8) = 76. bh + b(h h) = bh + bh bh = b ( h + H ) 5. (a) ()() + ()() = 5 (b) {area of shaded regio i Fig. 4b} < 5 7. A() =, A() =.5, A() = 4.5, A(4) = 6, A(5) = 7 9. C() =.5, C() = 4, C() = 7.5, ad C(x) = rect. + triagle areas = x + x. x = x + x. Distace = "area" = ()() + ()() = = 75 feet.. (a) A: secods to stop. B: 4 secods to stop. (b) A: ()(8) = 8 feet to stop. B: (4)(4) = 8 feet to stop. 5. miles, dollars, cubic feet, kilowatt. hours, people, square meals Sectio = 9. (+) + (+) + (+) = 9 5. cos() + cos(π) + cos(π) + cos(π) + cos(4π) + cos(5π) = + ( ) + + ( ) + + ( ) = k 9. k k= k= 7. k. k k=. (a) (+) + (+) + (+) = = (b) (++) + (++) = 5. (a) = = (b) 5. ( + + ) = 5. 6 = 7. (a) + + = = 4 (b) ( + + ) = 6 = 6 9. f() + f() + f() + f() = = 4.. f() +. f() +. f() +. f() = = 8. g() + g() + g() = = 8 5. g () + g () + g () = = 6 7. h() + h() + h(4) = = 6

Odd Aswers: Chapter Four Cotemporary Calculus 9 f()h() + f()h() + f()h() = ()() + (4)() + (9)( / ) =. ( ) + ( ) + ( ) + (4 ) +... + (7 6 ) = 7 = 49. ( ) + ( ) + ( 4 ) + ( 4 5 ) + ( 5 6 ) = 6 = 5 6 5.

(i) [,.8], [.8, 4.5], [4.5, 5.], [5., 7] (ii).8,.7,.7,.8 (iii) mesh =.8 (iv).8+.7+.7+.8 = 4 4. x + x + x +... + x = (x x ) + (x x ) + (x x ) +... + (x x ) = x x 45. (a) f()() + f()(.5) + f()(.

2 Odd Aswers: Chapter Four Cotemporary Calculus 9 f()h() + f()h() + f()h() = ()() + (4)() + (9)( / ) =. ( ) + ( ) + ( ) + (4 ) (7 6 ) = 7 = 49. ( ) + ( ) + ( 4 ) + ( 4 5 ) + ( 5 6 ) = 6 = ( ) + ( ) + ( ) ( 9 8 ) = 9 = 7. (i) [, ], [, 4.5], [4.5, 6], [6, 7] (ii),.5,.5, (iii) mesh =.5 (iv) = 5 9. (i) [, ], [, ], [,.5], [.5, ] (ii),,.5,.5 (iii) mesh = (iv) = 5 4. (i) [,.8], [.8, 4.5], [4.5, 5.], [5., 7] (ii).8,.7,.7,.8 (iii) mesh =.8 (iv) = 4 4. x + x + x x = (x x ) + (x x ) + (x x ) (x x ) = x x 45. (a) f()() + f()(.5) + f()(.5) = (4)() + ()(.5) + ()(.5) = 5.5 (b) f()() + f(.5)(.5) + f(.5)(.5) = ()() + (.75)(.5) + (.75)(.5) = a. (i) ad (ii) See the graph. (iii) f() =, f(π/4).77, f(π/) = (iv) ( π/4 )() + ( π/4 )(.77) + ( π/ )(). 49. (a) ()() + (5)() + (7)() RS (5)() + (7)() + (6)() so 9 RS 65. (b) ()() + (5)() + ()() + (7)() RS (5)() + ()() + (7)() + (6)() so 4 RS 58. (c) ()(.5) + (.5)(.5) + (5)() + ()() + (7)() RS (.5)(.5) + (5)(.5) + ()() + (7)() + (6)() so 4.65 RS (a) ()( π/ ) + ()( π/ ) RS ()( π/ ) + ()( π/ ) so RS π (b) ()( π/4 ) + (.77)( π/4 ) + ()( π/ ) RS (.77)(π/4) + ()(π/4) + ()(π/) so.56 RS.9 (c) ()(π/4) + (.77)(π/ ) + (.77)(π/4) + ()( π/4 ) RS (.77)(π/4) + ()(π/4) + ()(π/4)+ (.77)(π/4) so. RS (a) =.4 (b) = error (base)(height) = 4 5 (65 9) = 5 (56) = 5 = (a) () = 55 (b) = () =. () = = ( ) ( ) = () 9() = 45 = 65

3 Odd Aswers: Chapter Four Cotemporary Calculus 6. (k + k) = k k= k= () + k = ( ) + () = (55) + 55 = 8 k= Sectio x dx. + x dx. x dx 5. x dx = 8 7. x si(x) dx 9. l(x) dx x dx = / 5. x dx = 8 7. (a) (b) (c) 6 (d) 8 (e) 7 9. (a) see the graph (b) 4 feet. (c) 4 feet from the startig poit.. meters. feet = cubic feet 5. gram. meters 7. feet/secod = feet per secod 9. x = =. m i = (i ) ad M i = i so f(m i ) = { (i ) } ad f(m i ) = { i }. (a) LS = i= f(m i ) x = { (i ) i= } = 8 { i= i i= i + i= i } i= (b) = 6 4 { ( ) ( + + ) + ( + ) } = 6 4 { US = i= } = f(m i ) x = { i i= } = = 8 { i }= 6 i= 4 { } Sectio = = = 7..

Odd Aswers: Chapter Four Cotemporary Calculus 4 5. 7. 9.. (a) 8. 6 = 48 (b) 4. (a) (b) 8 = 64 5. 8 7..5 9. 4. 7 4. (a) graph y = A(x) = x / (b) graph y = A '(x) = x 45.

(a) f is ot cotiuous o [,4] (ot cot. at x = ad ) (b) f is ot differetiable o [,4] (ot diff. at x = ad ) (c) f is itegrable o [,4]. 4 5.

4 Odd Aswers: Chapter Four Cotemporary Calculus (a) 8. 6 = 48 (b) 4. (a) (b) 8 = (a) graph y = A(x) = x / (b) graph y = A '(x) = x 45. (a) graph of y = A(x) i Figure 45a (b) graph of y = A '(x) i Figure 45b 47. (a) f is cotiuous o [,4] (b) f is ot differetiable o [,4] (ot diff. at x. ad.) (c) f is itegrable o [,4] 49. (a) f is ot cotiuous o [,4] (ot cot. at x = ad ) (b) f is ot differetiable o [,4] (ot diff. at x = ad ) (c) f is itegrable o [,4] miles velocity dt = velocity dt + 4 velocity dt = = 85 Sectio 4.4. (a) see Figure (b) A() =, A() =.5, A() = 4, A(4) = 6.5 (c) A '() =, A '() =, A '() =, A '(4) =. (a) see Figure (b) A() =, A() =.5, A() =, A(4) = (c) A '() =, A '() =, A '() =, A '(4) =

(a) distace = t dt = t = feet. T (b) Fid T so 5 = t dt = t T = T. T = 5 7.7 secods.. (a) distace = T (b) 5 = 4t dt = t 4 =, feet. 4t dt = T 4. T = 4 5 8.4 secods.

5 Odd Aswers: Chapter Four Cotemporary Calculus 5 5. (a) see Figure (b) A() =, A() =, A() = 4, A(4) = 6 (c) A '() =, A '() =, A '() =, A '(4) = 7. (a) see Figure (b) A() =, A() = 4.5, A() = 8, A(4) =.5 (c) A '() = 5, A '() = 4, A '() =, A '(4) = 9. x = 9, x = 8, x =. x = 5, x = 4, x = 54. x 4 = 8, x4 = 8, x4 = 5. x = 54, x = 7, x = 7 7. x = 8, x = 6 9. (a) distace = t dt = t = feet. T (b) Fid T so 5 = t dt = t T = T. T = secods.. (a) distace = T (b) 5 = 4t dt = t 4 =, feet. 4t dt = T 4. T = secods.. (a) velocity = 75 t = whe t = 5 secods. (b) distace = 5 75 t dt = 75t t 5 = 5 feet. T (c) 5 = 75 t dt =75t t T = 75T T so T 75T + 5 = (solve usig Newto's method or by examiig the graph of y = x 75x + 5) ad T.74 secods.

6 Odd Aswers: Chapter Four Cotemporary Calculus 6 5. The total area is x dx = x = 9. (a) Fid T so. 9 = 9 = (b) Fid T so. 9 = = T x dx = T x dx = x T = T. T = x T = T. T = 7/ The fid T so. 9 = 6 = T x dx = x T = T. T = 8.6. Sectio 4.5. (a) A(x) = x. The A '(x) = x, ad A '() =, A '() =, ad A '() = 7. x t dt ) = x. A '() =, A '() =, ad A '() = 7. (b) A '(x) = D(. A '(x) = x so A '() =, A '() = 4, A '() = A '(x) = x so A '() =, A '() = 4, A '() = A '(x) = x so A '() =, A '() = 4, A '() = A '(x) = x so A '() =, A '() =, A '() = A '(x) = si(x) so A '().84, A '().9, A '() A '(x) = x so A '() =, A '() =, A '() =. 9. A '(x) = f(x) so A '() =, A '() =, A '() =.. A '(x) = f(x) so A '() =, A '() =, A '() =.. A '(x) = f(x) so A '() =, A '() =, A '() =.. A '(x) = f(x) so A '() =, A '() =, A '() =.. F() F() = 6 5 = 5. F() F() = 9 = 6 7. F(5) F().6 =.6 9. F() F(/). (.69) =.79. F(π/) F() = =. F() F().67 = F(7) F() = (7) / F(9) F() = = 9. F() F( ).9.4 = F(π/4) F() = =. F() F() = () /.4 5. F(x) = x. F() F( ) = 8 ( ) =. 7. F(x) = l(x). F(e) F() = =. 9. F(x) = x/. F() F(5) = 5 = F(x) = /x. F() F() =. ( ) =.9 4. F(x) = e x. F() F() = e.78

7 Odd Aswers: Chapter Four Cotemporary Calculus F(x) = ta(x). F(π/4) F(π/6).577 = The itegral goes from to so eve without kowig a atiderivative, si(x). l(x) dx =. 49. area = π π si(x) dx = cos(x) 5. area = (x ) dx = = ( ) ( ) = 5. area = x 4x + 4 dx = x x + 4x.5 INT(x) dx ) = () = 4.5. = =. 55. D( A(x) ) =. ta( x ), D( A(x ) ) = x. ta( x ), D( A( si(x) ) ) = cos(x). ta( si(x) ) x. ( 5 ) si(x). cos(x) 6. { ( x) + }. ( ) 6. cos(x) 65. ta( x ). x ta( x ) l(x). cos(. l(x) ). x

8 Odd Aswers: Chapter Four Cotemporary Calculus 8 Sectio 4.6. Left side = 4 x4 = 5 4. Right side = { x = 7 }. { x = } = 7 left side.. Left side = 4. Right side = ( ). ( ) = 6 left side. 5. si( x ) + C 7. cos( + ex ) + C 9. ta( si(x) ) + C. 5 l + x + C. cos( + x ) + C 5. 4 si( 4x ) + C ( 5 + x4 ) + C 9. l + x + C. ( l(x) ) + C. 4 ( + x ) 8 + C 5. sec( e x ) + C 7. π/ si( x ) = 9. cos( + e x ) = cos( ) cos( + e ) ( + x ) 6 = 9. 5 l + x = 5 l( 7 ) 5. ( x ) / = 7. 9 ( + x ) / = = x si( x ) + C 4. 4 si( x ) + C 4. x 4 si( x ) π = π x7 + 5 x5 + x + x + C 47. ex + e x + x + C x6 + 4 x4 + 5 x + 5x + C 5. ex + 4 e4x + C 5. 7 x7/ x5/ 4 x/ + C 55. x. l x + + C 57. x x + C 59. (divide first) x x + 7. l x + C 6. (divide first) x +. l x + C 6. x/ + 8 x / + C 65. (area of semicirle with radius ) = π() = π 67. (area of semicirle with radius ) = π() = 9 π 69. (area of rectagle) + (area of semicircle of radius ) = ()() + ( π() ) = 4 + π

9 Odd Aswers: Chapter Four Cotemporary Calculus 9 Sectio 4.7. betwee (usig left edpoits of itervals) ad 6 (usig right edpoits). betwee 4 (usig left edpoits of itervals) ad 6 (usig right edpoits) 5. Usig left edpoit widths: ()(4)+(7)(4)+(55)(4)+(9)(4)+()(4)+(5)(4) = 8,4 ft. Right edpoit widths (7, 55,...) ad average widths (7/, 5/,... ) give the same result, 8,4 ft. All of these are reasoable methods for estimatig the area of the islad e. π + 4 π 5. e 7. π 4 9. Estimate usig midpoits of uit itervals: 4 {f()()+f()()+f()()+f(4)()} = 9 4. About Estimate usig midpoits of uit itervals: {f()()+f()()} = 5. About 5.. average 5. average 5 7. average = 5 9. average =. average = π. (a) C = : average = (b) C = 9: average = (c) C = 8: average = 6 (d) C = : average = I geeral, average = C. 5. (a) Graphically, average. calls hour = 6 calls mi 5, calls mi. (b) About 58, calls mi. 7. (a) Similar to Example 5: work =,95 foot pouds (b) work =,.5 foot pouds 9. (a) work =, foot pouds (b) work = 6 foot pouds (c) work = 4 foot pouds 4. work =,75 foot pouds Sectio 4.8. Table #5, a = : arcta( x ) + C. Table #5, a = 5: x + 5 arcta( x 5 ) +C 5. Table #7, a = : ( ) l x + x +C 7. Table #5, a = : arcta( x ) + C 9. Table #5, a = : e x + 7 arcta( x ) + C

10 Odd Aswers: Chapter Four Cotemporary Calculus. Table #4, a = 5 :. arcsi( x 5 ) + C. Table #5, a = 5 : arcta( 5 x ) + C 5. First substitute u = x, du = dx. The use Table #4 with a = : 5. arcsi( u ) + C = 5. arcsi( x ) + C 7. Table #4 ad substitutio u = x: l x + + 9x + C 9. Table #8 ad substitutio u = x+: (x+). l x+ ( x+ ) + C or (x+). l x+ x + C. Table #8 ad substitutio u = 5x +7: { ( 5x +7 ). l 5x +7 ( 5x +7 ) } + C. Table #8 ad substitutio u = si(x): si(x). l si(x) si(x) + C 5. Table #44, a = : x x (4). l x + x C 7. Table #44, a = 4: x x (6). l x + x C 9. Table #5, a = 5: x + 5 arcta( x 5 ) = { arcta( 5 ) acrta( 5 ) }. Table #5, a = : acrta( x ) = { arcta( ) arcta( ) }. Table #4, a = 5 : arcsi( x 5 ) =. { arcsi( 5 arcsi( 5 } 5. Table #4, a = /: 5 arcsi( x / ). = 5 arcsi(. ) l(7) l(). l() l( ) 4. Table #9a: si (x). cos(x) cos(x) + C 45. Table #: cos4 (x). si(x) Table #9: x. si(x) + x. cos(x). si(x) + C 49. Average of si(x) o [,π] is π. Average of si (x) o [,π] is. π >. { aswer to umber 44 }

11 Odd Aswers: Chapter Four Cotemporary Calculus 5. Usig results from #5: (a) (c) e (b) 9 {. l() 9 } {. l() 99 } 6.5/99.65 (d) 5. (c) is largest. 55. approximately (a).57 (b).94 (c).4 (d).7 (e).9 99 {. l() 99 } 86.66/99 = 4. Sectio 4.9. h = (6 ) 4 =. T 4 = {. + (.8) + (.) + (.9) +. } = 5.5 S 4 = {. + 4(.8) + (.) + 4(.9) +. } = 5.5. h = ( ) 8 =.5. T 8 =.5 S 8 =.5 { 4. + (.8) + (.7) + (.5) + (.4) + (4.) + (.5) + (.) } = 7.5 { (.8) + (.7) + 4(.5) + (.4) + 4(4.) + (.5) + 4(.) } = / = T 4 = 4 7. T 4 = T 4 = S 4 = 4 S 4 = / =.666 S 4 = exact = 4 exact = / exact =. T 6 = T 6 = S 6 = S 6 = T 6 = S 6 = (a) M = so the error boud = (the trapezoidal approximatio is exact) (b) M 4 = so the error boud = (the Simpso's rule approximatio is exact) (c) = (d) = (must be a eve iteger) 9. M = { max. of 6x o [,] } = 6. f ''''(x) = so M 4 =. (a) error ()(4 (6) =.5 ) (b) M 4 = so the error boud = (the Simpso's rule approximatio is exact) (c) Set. = ()( ) (6) ad solve for 48 = ()(.) = 4, so = 6.5. Take = 64 (to be certai that we have eough subitervals, always roud UP). (d) = (must be a eve iteger)

12 Odd Aswers: Chapter Four Cotemporary Calculus. M = { max. of si(x) o [,π] } =. M 4 = { max. of si(x) o [,π] } =. π (a) error ()(4 ) () = (b) error π 5 (8)(4 4 () =.6645 ) (c) Set. = (d) Set. = π ()( ) () so = π 5 (8)( 4 ) () so 4 = π ()(.) π 5 (8)(.) =,58.9 ad = 5.8. = 5. =,7. ad = 6.4. = 8.. h =. S 6 = { 5 + 4(6) + (9) + 4(86) + (74) + 4(5) + 4 } =,4 square feet. ( T 6 =,7 square feet.) 5. h = 5. S 6 = 5 { + 4(5) + (7) + 4(6) + () + 4(7) + } = 7,66.7 square feet area. volume = (area)(depth) = ( 7,66.7 ft )( ft ) = 87,667 ft. T 6 = 5 { + (5) + (7) + (6) + () + (7) + } = 6,75 square feet area. volume = (area)(depth) = ( 6,75 ft )( ft ) = 88,5 ft. 7. Distace traveled by jogger is area uder the graph of velocity ( f(x) ) vs. time (x). area T = { } = 4, ft. 9.. o your ow. (a) L 4 =.5 (b) R 4 = 4.5 (c) M 4 = 4 (d) exact = 4 5. (a) L 4 =.75 (b) R 4 =.75 (c) M 4 =.65 (d) exact (a) L 4 =.896 (b) R 4 =.896 (c) M 4 =.5 (d) exact =

Math 21B-B - Homework Set 2

Math 21B-B - Homework Set 2 Math B-B - Homework Set Sectio 5.:. a) lim P k= c k c k ) x k, where P is a partitio of [, 5. x x ) dx b) lim P k= 4 ck x k, where P is a partitio of [,. 4 x dx c) lim P k= ta c k ) x k, where P is a partitio