NOT FOR SALE 6 APPLICATIONS OF INTEGRATION. 6.1 Areas Between Curves. Cengage Learning. All Rights Reserved. ( ) = 4 ln 1 = 45.

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1 6 APPLICATIONS OF INTEGRATION 6. Areas Between Curves ( ) ln ( ln 8) 4 ln 45 4 ln 8 ( ) ( ) ( ) ( ) ( 4) ( +6) + ( 8 + 7) 9 5. ( ) + ( +) ( + ) ( sin ) +cos The curves intersect when ( ) ( )( 4) or 4. 4 [ ( ) ] ( +5 4) 9 c 6 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

2 CHAPTER 6 APPLICATIONS OF INTEGRATION 8. The curves intesect when 4 6 ( 6) or 6. 6 [ ( 4)] 6 (6 ) 6 (6) (6) ( ) ln + ln + (ln + ) ln 9. By observation, sinand intersect at ( ) and ( ) for. sin cos ( ) 4 4. The curves intersect when ±. ( ) ( ) ( ) ( ) ( +6)( ) 6 or,so 6 or and (8 8 8) 64 c 6 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

3 SECTION 6. AREAS BETWEEN CURVES ±,so ( ) ( 6) 8 [by symmetry] 8 [(54 8) ] (6) ( ) or,so (4 ) (4 ) The curves intersect when 8cos sec 8cos cos 8 cos for.bysymmetry, (8 cos sec ) 8sin tan [( cos ) cos ] ( cos) sin (4 ) ±,so (4 + ) (4 ) [by symmetry] c 6 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

4 4 CHAPTER 6 APPLICATIONS OF INTEGRATION 8. The curves intersect when + + ( )( ) or. ( ) ( ) ( ) ( ) 6 9. By inspection, the curves intersect at ±. [cos (4 )] (cos 4 +) [by symmetry] sin ,sothecurves intersect when ( +)( ) [since ]. [( ) 4 )] The curves intersect when tan sin (on [ ]) sin sin cos sin cos sin sin ( cos ) sin or cos or ±. ( sin tan ) [by symmetry] cos ln sec [( ln ) ( )] ( ln ) ln. The curves intersect when ( ) ( +)( ) or ±. ( ) [by symmetry] c 6 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

5 SECTION 6. AREAS BETWEEN CURVES 5. The curves intersect when 8 ( ) ( 5 ) or 5 or 4,sofor 6, ( ) The curves intersect when cos cos (on []) cos cos. [cos ( cos )] + ( cos ) + ( cos) sin + sin +( ) + [( cos ) cos ] 5. By inspection, we see that the curves intersect at ± and that the area of the region enclosed by the curves is twice the area enclosed in the first quadrant. 5 [( ) 4 ] c 6 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

6 6 CHAPTER 6 APPLICATIONS OF INTEGRATION 6. sinh ( ) ln ln (or ln ). ln ( sinh ) + cosh ln ln (sinh ) + cosh + ln ( ) + (cosh + ) + ++cosh+,or ± and ±,sofor, ln ln 8 ln ( +6)( ) 6 or and + + ( +)( ) or,sofor, ( +) (a) Totalarea+79. (b) () () for and () () for 5,so 5 [() ()] [() ()] + 5 [() ()] [() ()] + 5 [() ()] () or ( ) ( 5+ 5) ( + ) 5 4 c 6 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

7 ( ) or. + + ln( + ) ln( + ) ln ln ln ( ) 6 ln (ln ) ln (ln) (ln) ln ln(ln ) ln or or [ or ] ln (ln ) (ln ) (ln ) ( ) 6 SECTION 6. AREAS BETWEEN CURVES 7. An equation of the line through ( ) and ( ) is ; through ( ) and ( ) is ; through ( ) and ( ) is An equation of the line through ( ) and ( ) is +; through ( ) and ( ) is + ; through ( ) and ( ) is +. ( +) + + ( +) The curves intersect when sin cos (on []) sin sin sin +sin ( sin )(sin +) sin. 6 6 sin cos (cos sin ) + 6 (sin cos ) sin +cos 6 + cos sin ( + ) + ( ) 4 c 6 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

8 8 CHAPTER 6 APPLICATIONS OF INTEGRATION 6. ln ln ln ln 4+ ln + ( ) + ln ln ln ln ln ( ) + ln ln ln ln 4 ln ln 7. From the graph, we see that the curves intersect at and 896,with sin( ) 4 on (). So the area of the region bounded by the curves is sin( ) 4 cos( ) 5 5 cos( ) From the graph, we see that the curves intersect (with )at and,where 5, with ( +) 5 on (). The area of the region bounded by the curves is ( +) (5 ) From the graph, we see that the curves intersect at 5 and 86 with +4 on ( ) and +4 on ( ). So the area of the region bounded by the curves is ( +4) ( ) + ( ) ( +4) ( +4) + ( + + 4) From the graph, we see that the curves intersect at 9 and 68. is the upper curve, so the area of the region bounded by the curves is 4 ln 5 c 6 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

9 SECTION 6. AREAS BETWEEN CURVES 9 4. Graph Y /(+xˆ4) and Y xˆ. WeseethatY Y on ( ),sothe area is given by + 4. Evaluate the integral with a command such as fnint(y -Y,x,-,) to get 8 to five decimal places. Another method: Graph () Y /(+xˆ4)-xˆ and from the graph evaluate () from to. 4. The curves intersect at ±. ( 4 ) The curves intersect at and tan The curves intersect at 997, 676,and ( +sin 4 ) cos + cos ( +sin 4 ) As the figure illustrates, the curves and enclose a four-part region symmetric about the origin (since and are odd functions of ). The curves intersect at values of where ;thatis,where ( 4 6 +). That happens at and where 6 ± 6 ± 6;thatis,at + 6, 6,, 6,and + 6.Theexactareais + 6 ( ) ( ) + ( 5 +6 ) CAS c 6 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

10 CHAPTER 6 APPLICATIONS OF INTEGRATION 46. The inequality describes the region that lies on, or to the right of, the parabola. The inequality describes the region if that lies on, or to the left of, the curve + if. So the given region is the shaded region that lies between the curves. The graphs of and intersect when + ( )( +) [for ]. By symmetry, ( ) second 6 hour, so s 6 h. With the given data, we can take 5to use the Midpoint Rule ,so distance Kelly distance Chris ( ) 5 [( )() + ( )() + ( )(5) 8 +( )(7) + ( )(9)] [( ) + (5 46) + (7 6) + (86 75) + (98 86)] 8 8 (+6+9++) 8 (4) 45 mile, or 7 feet 48. If distance from left end of pool and () width at, then the Midpoint Rule with 4and 8 4gives Area 6 4( ) 4(8) 9 m Let () denote the height of the wing at cm from the left end. 5 5 [() + (6) + () + (4) + (8)] 4( ) 4(58) 4 cm 5. For, () (), so the area between the curves is given by [() ()] ( ) , 4 7, 75, 8 9 This area A represents the increase in population over a -year period. 7, people 5. (a) From Example 5(a), the infectiousness concentration is cellsml. () 9() 9( )( )( +). Using a calculator to solve the last equation for givesustwosolutionswiththe lesser being 6 days, or the th day. (b) From Example 5(b), the slope of the line through and is. From part (a), ( ). Anequationofthe line through that is parallel to is ( ),or + +. Using a calculator, we c 6 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

11 SECTION 6. AREAS BETWEEN CURVES find that this line intersects at 4 78, or the 8th day. So in the patient with some immunity, the infection lasts about days less than in the patient without immunity. (c) The level of infectiousness for this patient is the area between the graph of and the line in part (b). This area is 4 () ( + + ) 78 6 ( ) From the figure, () () for. The area between the curves is given by [() ()] [(7 5 +) (7 + +6)] ( ) Thus, about.87 more inches of rain fell at the second location than at the first during the first two hours of the storm. 5. We know that the area under curve between and is () (),where () is the velocity of car A and A is its displacement. Similarly, the area under curve between and is B() B(). (a)afteroneminute,theareaundercurve is greater than the area under curve. So car A is ahead after one minute. (b) The area of the shaded region has numerical value A () B (), which is the distance by which A is ahead of B after minute. (c) After two minutes, car B is traveling faster than car A and has gained some ground, but the area under curve from to is still greater than the corresponding area for curve, so car A is still ahead. (d) From the graph, it appears that the area between curves and for (when car A is going faster), which corresponds to the distance by which car A is ahead, seems to be about squares. Therefore, the cars will be side by side at the time where the area between the curves for (when car B is going faster) is the same as the area for. From the graph, it appears that this time is. So the cars are side by side when minutes. 54. Theareaunder () from 5to representsthechangeinrevenue,andtheareaunder () from 5 to represents the change in cost. The shaded region represents the difference between these two values; that is, the increase in profit as the production level increases from 5 units to units. We use the Midpoint Rule with 5 and : 5 {[ (55) (55)] + [ (65) (65)] + [ (75) (75)] + [ (85) (85)] + [ (95) (95)]} ( ) (55) 55 thousand dollars Using would give us 5( ) 5 thousand dollars. c 6 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

12 CHAPTER 6 APPLICATIONS OF INTEGRATION 55. To graph this function, we must first express it as a combination of explicit functions of ;namely, ± +. We can see from the graph that the loop extends from to, and that by symmetry, the area we seek is just twice the area under the top half of the curve on this interval, the equation of the top half being +. So the area is +. We substitute +,so and the limits change to and,andweget [( ) ] ( ) We start by finding the equation of the tangent line to at the point ( ):, so the slope of the tangent is (), and its equation is ( ),or.wewouldneedtwointegralstointegratewith respect to, but only one to integrate with respect to. ( +) By the symmetry of the problem, we consider only the first quadrant, where. We are looking for a number such that (a) We want to choose so that (b) The area under the curve from to 4is 4 [take 4inthefirstintegralinpart(a)].Nowtheline must intersect the curve and not the line 4,sincetheareaundertheline 4 from to 4is only, which is less than half of. We want to choose so that the upper area in the diagram is half of the total 6 4 area under the curve from to 4. This implies that Letting,weget Thus, 6 ± ± 6 4.But c 6 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

13 SECTION 6. AREAS BETWEEN CURVES 59. We first assume that, since can be replaced by in both equations without changing the graphs, and if the curves do not enclose a region. We see from the graph that the enclosed area lies between and,andby symmetry, it is equal to four times the area in the first quadrant. The enclosed area is 4 ( ) So Note that 6 is another solution, since the graphs are the same. 6. It appears from the diagram that the curves cosand cos( ) intersect halfway between and, namely, when. We can verify that this is indeed true by noting that cos( ) cos( ) cos(). The point where cos( ) crosses the -axis is +. So we require that [cos cos( )] cos( ) + [the negative sign on the RHS is needed since the second area is beneath the -axis] [sin sin ( )] [sin ( )] + [sin() sin( )] [ sin( )] sin( )+sin + sin() sin sin +. [Herewehaveusedtheoddnessofthesinefunction,andthefactthatsin( ) sin]. So sin() sin() The curve and the line will determine a region when they intersect at two or more points. So we solve the equation ( +) ( + ) ( + ) ( + ) or + or or ±. Notethatif, this has only the solution, and no region is determined. But if, then there are two solutions. [Another way of seeing this is to observe that the slope of the tangent to ( +)at the origin is () and therefore we must have.] Note that we cannot just integrate between the positive and negative roots, since the curve and the line cross at the origin. Since and ( +)are both odd functions, the total area is twice the area between the curves on the interval. So the total area enclosed is + ln( +) [ln( +) ( )] (ln ) ln() + ln c 6 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

14 4 CHAPTER 6 APPLICATIONS OF INTEGRATION APPLIED PROJECT The Gini Index. (a) area between and area under [ ()] [ ()] (b) For a perfectly egalitarian society, (), so [ ]. For a perfectly totalitarian society, () if if so ( ).. (a) The richest % of the population in received (8) 498 5,or5%, of the total US income. (b) A quadratic model has the form () + +. Rounding to six decimal places, we get 5 57, 7 57,and The quadratic model appears to be a reasonable fit, but note that () 6 and is both decreasing and increasing. (c) [ ()] () + + Year Gini The Gini index has risen steadily from 97 to. The trend is toward a less egalitarian society. 4. Using Maple s PowerFit or TI s PwrReg command and omitting the point ( ) gives us () and a Gini index. [ ()] Note that the power function is nearly quadratic. _. _ c 6 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

15 SECTION 6. VOLUMES 5 6. Volumes. A cross-section is a disk with radius +, so its area is () ( +) ( + +). () ( + +) A cross-section is a disk with radius,so its area is (). 4 () A cross-section is a disk with radius, so its area is () ( ). 5 () 5 ( ) A cross-section is a disk with radius,so its area is () ( ) () ( ) c 6 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

16 6 CHAPTER 6 APPLICATIONS OF INTEGRATION 5. A cross-section is a disk with radius,soits area is (). 9 () (8) A cross-section is a disk with radius,soits area is () () (45 ) A cross-section is a washer (annulus) with inner radius and outer radius, so its area is () () ( ) ( 6 ). () ( 6 ) A cross-section is a washer (annulus) with inner radius and outer radius 6, so its area is () [(6 ) ]( 4 +). () ( 4 +) A cross-section is a washer with inner radius and outer radius, so its area is () () ( ) (4 4 ). () (4 4 ) c 6 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

17 . A cross-section is a washer with inner radius 4 and outer radius, so its area is () ( ) ( 4 ) ( ). () ( ) A cross-section is a washer with inner radius and outer radius, so its area is () ( ) ( ) ( + 4 ) ( + ) 4 +. () (4 + ) SECTION 6. VOLUMES 7. A cross-section is a washer with inner radius ( ) 4 and outer radius ( ) +, so its area is () ( +) (4) ( ). () ( ) A cross-section is a washer with inner radius ( + sec ) sec and outer radius, so its area is () (sec ) (4 sec ). () (4 sec ) 4 tan 4 (4 sec ) [by symmetry] 4 c 6 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

18 8 CHAPTER 6 APPLICATIONS OF INTEGRATION 4. A cross-section is a washer with inner radius sin ( ) and outer radius cos ( ), so its area is () (cos +) (sin +) (cos +cos sin sin) (cos +cos sin). 4 () 4 (cos +cos sin) sin +sin +cos (++) 5. A cross-section is a washer with inner radius and outer radius, so its area is () ( ) ( ) () ( 4 + ) For, a cross-section is a washer with inner radius ( ) and outer radius ( ), so its area is () ( )5. For, a cross-section is a washer with inner radius ( ) and outer radius ( ), so its area is () ( +) () ( + + 4) ln 5 + ( + ) +ln 5 + +ln (+ln) ( + ln ) c 6 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

19 7. From the symmetry of the curves, we see they intersect at and so ± washer with inner radius ( ) and outer radius, so its area is () ( ) ( + ) ( ) ( ) (5 ). () 5( ) [by symmetry] 6 SECTION 6. VOLUMES 9. A cross-section is a 8. For, a cross-section is an annulus with inner radius and outer radius 4, the area of which is () (4 ) ( ).For 4, a cross-section is an annulus with inner radius and outer radius 4, the area of which is () (4 ) ( ). 4 () (4 ) ( ) + 4 (4 ) ( ) (8 + ) R about OA (the line ): () (). R about OC (the line ): (). R about AB (the line ): (). R about BC (the line ): () ( ) ( ) [( ) ( ) ] ( + ) + ( +) + + [ ( + )] c 6 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

20 CHAPTER 6 APPLICATIONS OF INTEGRATION. R about OA (the line ): () 4 ( ) 4. R about OC (the line ): () [( 4 ) ] R about AB (the line ): () [ ( 4 ) ] ( 4 8 ) [ ( )] R about BC (the line ): () ( 4 ) ( 4 + ) R about OA (the line ): () 4 ( ) Note: Let R R R R.IfwerotateR about any of the segments,,, or, we obtain a right circular cylinder of height and radius. Its volume is (). As a check for Exercises 9,, and 7, we can add the answers, and that sum must equal. Thus, R about OC (the line ): () [ ( 4 ) ] ( 8 ) Note: See the note in Exercise 7. For Exercises, 4, and 8, we have R about AB (the line ): () [( 4 ) ( ) ] [( ) ( + )] ( 8 4 +) Note: See the note in Exercise 7. For Exercises, 5, and 9, we have c 6 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

21 . R about BC (the line ): () [( ) 4 ] ( + 4 ) [( + ) ( 4 + )] SECTION 6. VOLUMES Note: See the note in Exercise 7. For Exercises, 6, and, we have (a) About the -axis: ( ) 7585 [by symmetry] (b) About : [ ( )] [ ( )] 4. (a) About the -axis: [( +) ] ( + ) (cos ) cos 4 [by symmetry] 7 (b) About : [( ) ( cos ) ] [ ( cos +cos 4 )] ( cos cos 4 ) (a) About : ± c 6 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

22 CHAPTER 6 APPLICATIONS OF INTEGRATION (b) About : ± [Notice that this is the same approximation as in part (a). This can be explained by Pappus s Theorem in Section 8..] 4. (a) About the -axis: and ± ± ±786. ( ) ( 4 ) (b) About the -axis: ( ) + + ( ) ln( 6 +)and intersect at 49, 467,and 9. ln( 6 +) + ln( 6 +) and arctan intersect at 57 and 9. + (arctan ) 69 c 6 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

23 7. CAS 8 sin ( ) [ ( )] SECTION 6. VOLUMES 8. CAS ( ) ( ) sin sin describes the volume of solid obtained by rotating the region R ( ), sin of the -plane about the -axis. 4. ( ) describes the volume of the solid obtained by rotating the region R {( ), } of the -plane about the -axis. 4. (4 8 ) ( ) ( 4 ) describes the volume of the solid obtained by rotating the region R ( ) 4 of the -plane about the -axis [ ( ) describes the volume of the solid obtained by rotating the region R {( ) 4 } of the -plane about the -axis. 4. There are subintervals over the 5-cm length, so we ll use 5for the Midpoint Rule. 5 () [(5) + (45) + (75) + (5) + (5)] ( ) 7 cm 44. () 5 5 [() + () + (5) + (7) + (9)] ( ) (9) 58 m 45. (a) [()] 4 [()] +[(5)] +[(7)] +[(9)] (5) +() +(8) +() 96 units (b) 4 (outer radius) (inner radius) 4 4 (99) () + (97) () + (9) (56) + (87) (65) 88 units c 6 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

24 4 CHAPTER 6 APPLICATIONS OF INTEGRATION 46. (a) CAS ( +5 ) 5 (b) ( ) is graphed in the figure. Substitute 6, 4,,and 54 in the answer for part (a) to get CAS We ll form a right circular cone with height and base radius by revolving the line about the -axis. Another solution: Revolve + about the -axis Or use substitution with and to get. 48. ( ) ( ) + + ( ) + ( ) +( + ) ( + + ) Another solution: by similar triangles. Therefore, ( ) c 6 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

25 .Now ( ) [by Exercise 47] ( ) ( )( ) + + where and are the areas of the bases of the frustum. (See Exercise 5 for a related result.) SECTION 6. VOLUMES ( ) ( ) ( ) + ( ) + ( ) ( ), or, equivalently, ( ) 5. An equation of the line is +(-intercept) + +. () ( ) () + + ( ) + ( ) + ( ) + + ( ) + ( ) [Note that this can be written as + +,asinexercise48.] If, we get a rectangular solid with volume.if, we get a square pyramid with volume. c 6 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

26 6 CHAPTER 6 APPLICATIONS OF INTEGRATION 5. For a cross-section at height, we see from similar triangles that,so Similarly, for cross-sections having as their base and replacing,.so () [ where is the area of the base, as with any pyramid.]. 5. Consider the triangle consisting of two vertices of the base and the center of the base. This triangle is similar to the corresponding triangle at a height,so. Also by similar triangles, ( ) ( ). These two equations imply that ( ),and since the cross-section is an equilateral triangle, it has area () ( ),so 4 () 4 4 ( ) 5. A cross-section at height is a triangle similar to the base, so we ll multiply the legs of the base triangle,and4,bya proportionality factor of (5 )5.Thus,thetriangleatheight has area () ,so () cm ( 5 ) 5, A cross-section is shaded in the diagram. () (),so () 4( ) c 6 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

27 55. If is a leg of the isosceles right triangle and is the hypotenuse, then + () 4. () () ()() (4 ) 9 4 (6 9 ) SECTION 6. VOLUMES The cross-section of the base corresponding to the coordinate has length The corresponding equilateral triangle with side has area () ( ) Therefore, 4 4 () ( ) 4 ( + ) Or: ( ) ( ) [ ] The cross-section of the base corresponding to the coordinate has length. The corresponding square with side has area () ( ) +. Therefore, () ( + ) + + Or: ( ) ( ) [ ] 58. The cross-section of the base corresponding to the coordinate has length. ± The corresponding square with side has area () 4( ). Therefore, () 4( ) The cross-section of the base corresponding to the coordinate has length. The height also has length, so the corresponding isosceles triangle has area () ( ). Therefore, () ( ) ( + 4 ) [by symmetry] c 6 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

28 8 CHAPTER 6 APPLICATIONS OF INTEGRATION 6. The cross-section of the base corresponding to the coordinate has length. [ ± ] The corresponding cross-section of the solid is a quarter-circle with radius and area () 4 ( ) ( ). Therefore, () ( ) (4 ) 6. The cross-section of at coordinate,,isacircle centered at the point ( ) with radius ( ). Theareaofthecross-sectionis () ( ) 4 ( + 4 ) The volume of is () ( ) (a) () () (b) Observe that the integral represents one quarter of the area of a circle of radius,so (a) The torus is obtainedbyrotatingthecircle( ) + about the -axis. Solving for, we see that the right half of the circle is given by + () and the left half by (). So [()] [()] (b) Observe that the integral represents a quarter of the area of a circle with radius, so The cross-sections perpendicular to the -axis in Figure 7 are rectangles. The rectangle corresponding to the coordinate has abaseoflength 6 in the -plane and a height of () 6 and,since and. Thus, 4 () [Put 6,so ] (64) c 6 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

29 SECTION 6. VOLUMES (a) Volume( ) () Volume() since the cross-sectional area () at height is the same for both solids. (b) By Cavalieri s Principle, the volume of the cylinder in the figure is the same as that of a right circular cylinder with radius and height,thatis,. 66. Each cross-section of the solid in a plane perpendicular to the -axis is a square (since the edges of the cut lie on the cylinders, which are perpendicular). One-quarter of this square and one-eighth of are shown. The area of this quarter-square is. Therefore, () 4( ) and the volume of is () 4 ( ) 8( ) The volume is obtained by rotating the area common to two circles of radius,as shown. The volume of the right half is right So by symmetry, the total volume is twice this, or 5. Another solution: We observe that the volume is the twice the volume of a cap of a sphere, so we can use the formula from Exercise 49 with : ( ) We consider two cases: one in which the ball is not completely submerged and the other in which it is. Case : The ball will not be completely submerged, and so a cross-section of the water parallel to the surface will be the shaded area shown in the first diagram. We can find the area of the cross-section at height above the bottom of the bowl by using the Pythagorean Theorem: 5 (5 ) and 5 ( 5),so(). The volume of water when it has depth is then () () cm,. Case : 5 In this case we can find the volume by simply subtracting the volume displaced by the ball from the total volume inside the bowl underneath the surface of the water. The total volume underneath the surface is just the volume of a cap of the bowl, so we use the formula from Exercise 49: cap() (45 ). The volume of the small sphere is ball 4 (5) 5, so the total volume is cap ball (45 5) cm. c 6 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

30 CHAPTER 6 APPLICATIONS OF INTEGRATION 69. Take the -axis to be the axis of the cylindrical hole of radius. A quarter of the cross-section through perpendicular to the -axis, is the rectangle shown. Using the Pythagorean Theorem twice, we see that the dimensions of this rectangle are and,so () 4,and () The line intersects the semicircle when ±. Rotating the shaded region about the -axis gives us ( ) [by symmetry] 4 Our answer makes sense in limiting cases. As, 4, which is the volume of the full sphere. As,, which makes sense because the hole s radius is approaching that of the sphere. 7. (a) The radius of the barrel is the same at each end by symmetry, since the function is even. Since the barrel is obtained by rotating the graph of the function about the -axis, this radius is equal to the value of at,whichis. (b) The barrel is symmetric about the -axis, so its volume is twice the volume of that part of the barrel for. Also,the barrel is a volume of rotation, so [continued] c 6 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

31 SECTION 6. VOLUMES BY CYLINDRICAL SHELLS Trying to make this look more like the expression we want, we rewrite it as But ( ) Substituting this back into,weseethat + 5, as required. 7. It suffices to consider the case where R is bounded by the curves () and () for,where() () for all in [ ], since other regions can be decomposed into subregions of this type. We are concerned with the volume obtained when R is rotated about the line,whichisequalto [()+] [()+] [()] [()] + [() ()] + 6. Volumes by Cylindrical Shells. If we were to use the washer method, we would first have to locate the local maximum point ( ) of ( ) using the methods of Chapter 4. Then we would have to solve the equation ( ) for in terms of to obtain the functions () and () shown in the first figure. This step would be difficult because it involves the cubic formula. Finally we would find the volume using [ ()] [ ()]. Using shells, we find that a typical approximating shell has radius, so its circumference is. Its height is,thatis, ( ). So the total volume is ( ) A typical cylindrical shell has circumference and height sin( ). sin( ). Let.Then,so sin [ cos ] [ ( )]. For slicing, we would first have to locate the local maximum point ( ) of sin( ) using the methods of Chapter 4. Then we would have to solve the equation sin for in terms of to obtain the functions () and () shown in the second figure. Finally we would find the volume using [()] [ ()]. Using shells is definitely preferable to slicing. c 6 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

32 CHAPTER 6 APPLICATIONS OF INTEGRATION Let. Thus,,so ( ) ( ) or. [(4 ) ] ( + ) ( ) or. [(6 ) ] ( +6 ) ( + 6) 8 c 6 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

33 8. By slicing: ( ) ( 4 ) SECTION 6. VOLUMES BY CYLINDRICAL SHELLS By cylindrical shells: ( ) The shell has radius, circumference, and height,so ( ) 4.. The shell has radius, circumference, and height,so ( ) 4 4 (4) 8.. The shell has radius, circumference, and height,so 8 ( ) c 6 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

34 4 CHAPTER 6 APPLICATIONS OF INTEGRATION. The shell has radius, circumference,and height + 9,so ( + 9) ( + 9) 6 ( 4 +) The shell has radius, circumference, and height +( ) ( ) ,so ( +4 ) ( +4 ) The curves intersect when ( ) or. The shell has radius, circumference, and height (4 ) ( 4 +4) +,so ( +) ( ) The shell has radius, circumference ( ), and height 8. ( )(8 ) (4 8 +4) c 6 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

35 6. The shell has radius ( ) +, circumference ( +), and height 4. ( + )(4 ) 4 ( + )( ) 4 ( + +) SECTION 6. VOLUMES BY CYLINDRICAL SHELLS 5 7. The shell has radius, circumference ( ), and height (4 ) +4. ( )( +4 ) ( ) The shell has radius 5, circumference (5 ), and height. 4 (5 ) 4 (5 5 + ) The shell has radius, circumference ( ), and height. ( )( ) 4 ( )( ) 4 ( +) c 6 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

36 6 CHAPTER 6 APPLICATIONS OF INTEGRATION. The shell has radius ( ) +, circumference ( +), and height ( +). ( + )( ) ( + +) 4 ( +) [by Theorem 5.5.7] 8 ( ) (a) (b) 46 ( ). (a) 4 (b) 5 tan. (a) 4 ( )[cos 4 ( cos 4 )] ( ) cos 4 [or 8 cos 4 using Theorem 5.5.7] (b) (a) ( ) ( )( + +) or [ ( )] + (b) 664 c 6 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

37 5. (a) (4 ) sin (b) SECTION 6. VOLUMES BY CYLINDRICAL SHELLS 7 6. (a) 4 (5 ) +7 (b) Let() +. Then the Midpoint Rule with 5gives () 5 [() + () + (5) + (7) + (9)] (99) Multiplying by gives (). Let() (), where the values of are obtained from the graph. Using the Midpoint Rule with 5gives 9.. Multiplying by gives () 5 [() + () + (5) + (7) + (9)] [() + () + 5(5) + 7(7) + 9(9)] [(4 ) + (5 ) + 5(4 ) + 7(4 ) + 9(4 )] ( ) (6) 5 (4 ). The solid is obtained by rotating the region 4, about the y-axis using cylindrical shells. ln. The solid is obtained by rotating the region ln, about the -axis using cylindrical shells. c 6 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

38 8 CHAPTER 6 APPLICATIONS OF INTEGRATION ( +) the line using cylindrical shells.. The solid is obtained by rotating the region, 4 about. ( )( ). The solid is obtained by rotating the region, about the line using cylindrical shells.. From the graph, the curves intersect at and 75,with + on the interval (). So the volume of the solid obtained by rotating the region about the -axis is + ( ) From the graph, the curves intersect at 96 and 75, with sin 4 +5on the interval ( ). So the volume of the solid obtained by rotating the region about the -axis is sin ( 4 +5) 5 5. CAS sin sin 4 6. [ ( )]( sin ) CAS ( ) Use shells: 4 ( +6 8) 4 ( +6 8) [( ) ( 4+6 6)] (4) 8 c 6 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

39 8. Use disks: 4 ( +6 8) 4 ( ) SECTION 6. VOLUMES BY CYLINDRICAL SHELLS 9 9. Use washers: ± ± ( ) + [4 ( +)] [by symmetry] ( ) 4 4. Use disks: ± ( ) Use disks: +( ) ± ( ) ( ) ( ) Use shells: 5 ( )[4 ( ) ] 5 ( )( +6 5) 5 ( +7 +5) c 6 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

40 4 CHAPTER 6 APPLICATIONS OF INTEGRATION 4. +( ) + + ( ) or. Use disks: [( +) ( )] [( ) ( )] [( +) ( +) ] [( +4 +4) ( )] Use cylindrical shells to find the volume. ( )() 4 ( ) 4 4 ( ) Now solve for in terms of : Use shells: ( ) ( ) ( ) 4 ( ) ( ) 4( + ) [let ] 4 +4 The first integral is the area of a semicircle of radius,thatis,, and the second is zero since the integrand is an odd function. Thus, c 6 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

41 48. By symmetry, the volume of a napkin ring obtained by drilling a hole of radius through a sphere with radius is twice the volume obtained by rotating the area above the -axis and below the curve (the equation of the top half of the cross-section of the sphere), between SECTION 6.4 WORK 4 and, about the -axis. This volume is equal to outer radius 4 4 ( ) inner radius But by the Pythagorean Theorem,, so the volume of the napkin ring is 4 6,whichis independent of both and ; that is, the amount of wood in a napkin ring of height is the same regardless of the size of the sphere used. Note that most of this calculation has been done already, but with more difficulty, in Exercise Another solution: The height of the missing cap is the radius of the sphere minus half the height of the cut-out cylinder, that is,. Using Exercise 6..49, napkin ring sphere cylinder cap Work. (a) The work done by the gorilla in lifting its weight of 6 pounds to a heightoffeet is (6lb)( ft) 7 ft-lb. (b) The amount of time it takes the gorilla to climb the tree doesn t change the amount of work done, so the work done is still 7 ft-lb.. () [( kg)(98 ms )]( m) (96 N)( m) 588 J. () ft-lb 4. cos sin N m J. Interpretation: From to, the force does work equal to cos J in accelerating the particle and increasing its kinetic energy. From to, the force opposes the motion of the particle, decreasing its kinetic energy. This is negative work, equal in magnitude but opposite in sign to the work done from to. 5. The force function is given by () (in newtons) and the work (in joules) is the area under the curve, given by 8 () 4 () + 8 () 4 (4)() + (4)() 8 J () 4 [(6) + () + (4) + (8)] 4 4 [ ] 4(8) J 7. According to Hooke s Law, the force required to maintain a spring stretched units beyond its natural length (or compressed units less than its natural length) is proportional to,thatis,(). Here, the amount stretched is 4 in. ft and c 6 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

42 4 CHAPTER 6 APPLICATIONS OF INTEGRATION the force is lb. Thus, lbft, and (). The work done in stretching the spring from its natural length to 6 in. ft beyond its natural length is 5 5 ft-lb According to Hooke s Law, the force required to maintain a spring stretched units beyond its natural length (or compressed units less than its natural length) is proportional to, thatis,(). Here, the amount compressed is 4 cm mand the force is 6 N. Thus, 6 () 6Nm,and() 6. Thework required to compress the spring mis 6 () N-m (or J). The work required to compress the spring cm 5 m is (5) 675 J. 9. (a) If J, then (44) 7 and Nm. 7 9 Thus, the work needed to stretch the spring from 5 cm to 4 cm is J. 4 (b) (),so 5 7 and m 8 cm 9 5. If,then 4lbft and the work required is ft-lb.. Thedistancefrom cm to cm is m, so with (),weget. Now 4.Thus,.. Let be the natural length of the spring in meters. Then 6 ( ) ( ) and 4 4 (4 ) ( ). Simplifying gives us (44 4) and (5 4). Subtracting the first equation from the second gives 88,so,. Now the second equation becomes 5 4,so 4 m 8cm. In Exercises, is the number of subintervals of length, and isasamplepointintheth subinterval [ ].. (a) The portion of the rope from ft to ( + ) ft below the top of the building weighs lb and must be lifted ft, so its contribution to the total work is ft-lb. The total work is lim ft-lb 4 4 Notice that the exact height of the building does not matter (as long as it is more than 5 ft). (b) When half the rope is pulled to the top of the building, the work to lift the top half of the rope is ft-lb. The bottom half of the rope is lifted 5 ft and the work needed to accomplish 4 4 c 6 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

43 that is 5 5 SECTION 6.4 WORK ft-lb. The total work done in pulling half the rope to the top of the building is ft-lb (a) The 6 ft cable weighs 8 lb,orlbft. If we divide the cable into equal parts of length 6 ft, then for large, all points in the th part are lifted by approximately the same amount. Choose a representative distance from the winch in the th part of the cable, say.if 5 ft, then the th part has to be lifted roughly ft. If 5 ft,thentheth part has to be lifted 5 ft. Theth part weighs ( lbft)( ft) lb, so the work done in lifting it is ( ) if 5 ft and ( )(5) 75 if 5 ft. The work of lifting the top 5 ft of the cable is lim 5 5 (65) 975ft-lb. Here represents the number of parts of the cable in the top 5 ft. The work of lifting the bottom 5 ft of the cable is lim (6 5) 65 ft-lb,where represents the number of small parts in the 5 bottom 5 feet of the cable. The total work done is ft-lb. (b) Once feet of cable have been wound up by the winch, there is (6 )ftof cable still hanging from the winch. That portion of the cable weighs (6 )lb. Lifting it feet requires (6 ) ft-lb of work. Thus, the total work needed to lift the cable 5 ft is 5 (6 ) ft-lb. 5. Theworkneededtoliftthecableis lim 5 5 5, ft-lb.theworkneededtolift the coal is 8 lb 5 ft 4, ft-lb. Thus, the total work required is 5, + 4, 65, ft-lb. 6. Assumptions:. After lifting, the chain is L-shaped, with 4 m of the chain lying along the ground.. The chain slides effortlessly and without friction along the ground while its end is lifted.. The weight density of the chain is constant throughout its length and therefore equals (8 kgm)(98ms )784Nm. The part of the chain m from the lifted end is raised 6 m if 6 m, and it is lifted mif6m. Thus, the work needed is lim (6 ) (6 ) (784)(8) 4 J 7. At a height of meters ( ), the mass of the rope is (8 kgm)( m) (96 8) kg and the mass of the water is 6 kgm ( m) (6 ) kg. The mass of the bucket is kg, so the total mass is (96 8)+(6 )+(556 8) kg, and hence, the total force is 98(556 8) N.Theworkneededtolift the bucket m through the th subinterval of [ ] is 98(556 8 ), so the total work is lim 98(556 8 ) (98)(556 8) (96) 857 J c 6 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

44 44 CHAPTER 6 APPLICATIONS OF INTEGRATION 8. Theworkneededtoliftthebucketitselfis4 lb 8 ft ft-lb. At time (in seconds) the bucket is ft above its original 8 ft depth, but it now holds only (4 ) lb of water. In terms of distance, the bucket holds 4 lb of water when it is ft above its original 8 ft depth. Moving this amount of water a distance requires 4 ft-lb of work. Thus, the work needed to lift the water is lim 4 Adding the work of lifting the bucket gives a total of ft-lb of work. 9. The chain s weight density is ( ) ft-lb 5 lb 5lbft. The part of the chain ft below the ceiling (for 5 ) has to be lifted ft ( 5) ft, so the work needed to lift the th subinterval of the chain is ( 5)(5 ). The total work needed is lim ( 5)(5) [( 5)(5)] 5 ( 5) (5 5) ft-lb. A horizontal cylindrical slice of water ft thick has a volume of ft and weighs about 65 lbft 44 ft 9 lb. If the slice lies ft below the edge of the pool (where 5), then the work needed to pump it out is about 9. Thus, lim (5 ) 8, ft-lb. A slice ofwater m thick and lying at a depth of m(where )hasvolume( ) m,amassof kg, weighs about (98)( ) 9,6 N, and thus requires about 9,6 J of work for its removal. So lim 9,6 9, J.. We use a vertical coodinate measured from the center of the water tank. The top and bottom of the tank have coordinates ft and ft, respectively. A thin horizontal slice of water at coordinate is a disk of radius as shown in the figure. The disk has area ( ), so if the slice has thickness,theslicehasvolume( ) and weight 65( ). The work needed to raise this water from ground level (coordinate 7) to coordinate, a distance of (7 )ft,is 65( )(7 ) ft-lb. The total work needed to fill the tank is c 6 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

45 approximated by a Riemann sum 65[( ( ) )](7 ). Thus, the total work is lim 65[( ( ) )](7 ) 65( )(7 ) SECTION 6.4 WORK [7( ) ( )] 65() 7( ) [by Theorem 5.5.7] even function odd function 5(7) 9 9,68, ft-lb The 5 horsepower pump does 5(55) 85 ft-lb of work per second. To fill the tank, it will take,68, ft-lb 85 ft-lbs 9,48 s 97 hours.. A rectangular slice of water m thick and lying m above the bottom has width mandvolume8 m.itweighs about (98 )(8 ) N, and must be lifted (5 ) m by the pump, so the work needed is about (98 )(5 )(8 ) J. The total work required is (98 )(5 )8(98 ) (4 8 ) (98 ) 8 (98 )(8 7) (98 )(8) J 4. Let measure depth (in meters) below the center of the spherical tank, so that at the top of the tank and 4 at the spigot. A horizontal disk-shaped slice of water m thick and lying at coordinate has radius 9 mandvolume (9 ) m. It weighs about (98 )(9 ) N and must be lifted ( +4)m by the pump, so the work needed to pump it out is about (98 )( +4)(9 ) J. The total work required is (98 )( +4)(9 ) (98 ) [(9 )+4(9 )] (98 )()(4) (9 ) [by Theorem 5.5.7] (784 ) 9 (784 )(8),4, 44 6 J 5. Let measure depth (in feet) below the spout at the top of the tank. A horizontal disk-shaped slice of water ft thick and lying at coordinate has radius (6 ) ft () 8 andvolume 9 (6 64 ) ft.itweighs about (65) 9 64 (6 ) lb and must be lifted ft by the pump, so the work needed to pump it out is about (65) 9 (6 64 ) ft-lb. The total work required is 8 9 (65) (6 64 ) (65) 9 64 (65) 9 64 (65) (56 + ) 8 (56 + ) (65) ,64, 4 5 ft-lb () From similar triangles, 8. 8 So (8 ) (8) 8 8 (6 ) + (8 ) 8 c 6 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

46 46 CHAPTER 6 APPLICATIONS OF INTEGRATION 6. Let measure the distance (in feet) above the bottom of the tank. A horizontal slice of water ft thick and lying at coordinate has volume () ft. It weighs about (65) lb and must be lifted (6 ) ft by the pump, so the work needed to pump it out is about (65)(6 ) ft-lb. The total work required is 6 (65)(6 )5 6 (6 ) 5 6 5(6) 45, ft-lb. 7. If only 47 5 J of work is done, then only the water above a certain level (call it ) will be pumped out. So we use the same formula as in Exercise, except that the work is fixed, and we are trying to find the lower limit of integration: 47 5 (98 )(5 ) To find the solution of this equation, we plot 5 +45between and. We see that the equation is satisfied for.sothedepthofwaterremaininginthetankisabout m. 8. The only changes needed in the solution for Exercise 4 are: () change the lower limit from to and () change to 9. (98 9)( +4)(9 ) (98 9) ( ) (98 9) (98 9)(95) 8, J [about 58% oftheworkinexercise4] 9.,so is a function of and canalsoberegardedasafunctionof. If and,then () ( ()) ( ()) () [Let (),so ().] ( ) by the Substitution Rule.. 6 lbin 6 44 lbft, in 78 ft,and8 in 8 78 ft. 4 (6 44) 4, Therefore, and (a) () () () (()) () () () () () (465)(5) ft-lb. 5 c 6 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

47 (b) The mass of the bowling ball is mi h 58 ft mi is 8 88 lb fts 8 slug. Converting mih to fts gives us SECTION 6.4 WORK 47 h 6 s 88 fts. From part (a) with and 88, the work required to hurl the bowling ball 8 () 484 6ft-lb.. The work required to move the 8 kg roller coaster car is 6 6 (57 +5) , , J. (4,) Using Exercise (a) with,weget 4 ms. 8. (a) () (b) By part (a), where mass of the earth in kg, radius of the earth in m, +,, and mass of satellite in kg. (Note that km,, m.) Thus, (667 )(598 4 )() 4. (a) Assume the pyramid has smooth sides. From the figure for 78, an equation for the side is ( 48). The horizontal length of a cross-section is 48 and the area of a cross-section is () ( 48). A slice of thickness at height has volume ft and weight 5 lb, so the work needed to build the pyramid was (b) Work done h day ( 48) J ( ) ft-lb 4 days year gives us about,56 laborers yr ft-lb 6 laborer 7 ft-lb.dividing by hour laborer c 6 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

48 48 CHAPTER 6 APPLICATIONS OF INTEGRATION 6.5 Average Value of a Function. ave () ( ) ( +8) [ +4 ] [(8 + 6) ( +4)]7. ave () 4. ave sin 6 4. ave () cos ( ) 6 ( ) 6 cos [by Theorem 5.5.7] () ( + + ) 5. ave () sin cos [sin ] ( ) 6. ave () ( ) ( +) ave 8. ave 4 () cos4 sin 4 4 [by Theorem 5.5.7] 9. (a) ave 5 () 5 ln 5 4 ln 5 8 (ln 5) 5 ( ) 9 ( ) 9 (8 + ) 5 ( ) ln 5 4 ( ) [ cos, sin ] ln (c) (b) () ave ( ) ± or 4. (a) ave ln (ln ln ) ln (c) (b) () ave ln ln 8 c 6 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

49 SECTION 6.5 AVERAGE VALUE OF A FUNCTION 49. (a) ave ( sin sin ) cos + cos (c) (b) () ave sin sin 4 8 or 88. (a) ave ( 4 +) (c) (b) () ave ( 4 ) 6 or 87. is continuous on [ ], so by the Mean Value Theorem for Integrals there exists a number in [ ] such that () ()( ) 8(); that is, there is a number such that () The requirement is that (). The LHS of this equation is equal to ,sowesolvetheequation+ + ± ( ) 4 ± 5. Both roots are valid since they are positive. 5. Use geometric interpretations to find the values of the integrals. 8 () () + () + () + 4 () + 6 () + 7 () + 8 () Thus, the average value of on [ 8] ave 8 () (9) (a) ave (). Use the Midpoint Rule with and 4to estimate. 4[() + (6) + ()] 4[ ] 4(7) 548. Thus, ave (548) 45 kmh. (b) Estimating from the graph, () 45 when 5 s. 7. Let and correspond to 9 AM and 9 PM, respectively. 8. ave () ave sin 5 4 cos F 59 F ( ) Since () is decreasing on (], max () 4. Thus, ave max. c 6 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.