Solução_Calculo_Stewart_6e

F.
268 ¤ CHAPTER 6 APPLICATIONS OF INTEGRATION
TX.10
41. 1 second = 1
1
3600
hour, so 10 s =
360
h. With the given data, we can take n =5to use the Midpoint Rule.
∆t = 1/360−0
5
= 1
1800 ,so
distance Kelly − distance Chris = 1/360
0
v K dt − 1/360
0
v C dt = 1/360
0
(v K − v C) dt
43. Let h(x) denote the height of the wing at x cm from the left end.
A ≈ M 5 = 200 − 0
5
≈ M 5 = 1
1800 [(v K − v C )(1) + (v K − v C )(3) + (v K − v C )(5)
+(v K − v C )(7) + (v K − v C )(9)]
= 1 [(22 − 20) + (52 − 46) + (71 − 62) + (86 − 75) + (98 − 86)]
1800
= 1
1800 (2+6+9+11+12)= 1
1800 (40) = 1 45 mile, or 117 1 3 feet
[h(20) + h(60) + h(100) + h(140) + h(180)]
=40(20.3+29.0+27.3+20.5+8.7) = 40(105.8) = 4232 cm 2
45. We know that the area under curve A between t =0and t = x is x
vA(t) dt = sA(x),wherevA(t) is the velocity of car A
0
and s A is its displacement. Similarly, the area under curve B between t =0and t = x is x
0 v B(t) dt = s B (x).
(a)Afteroneminute,theareaundercurveA is greater than the area under curve B. So car A is ahead after one minute.
(b) The area of the shaded region has numerical value s A(1) − s B(1), which is the distance by which A is ahead of B after
1 minute.
(c) After two minutes, car B is traveling faster than car A and has gained some ground, but the area under curve A from t =0
to t =2is still greater than the corresponding area for curve B, so car A is still ahead.
(d) From the graph, it appears that the area between curves A and B for 0 ≤ t ≤ 1 (when car A is going faster), which
corresponds to the distance by which car A is ahead, seems to be about 3 squares. Therefore, the cars will be side by side
at the time x where the area between the curves for 1 ≤ t ≤ x (when car B is going faster) is the same as the area for
0 ≤ t ≤ 1. From the graph, it appears that this time is x ≈ 2.2. So the cars are side by side when t ≈ 2.2 minutes.
47. To graph this function, we must first express it as a combination of explicit
functions of y;namely,y = ±x √ x +3. We can see from the graph that the loop
extends from x = −3 to x =0, 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 y = −x √ x +3. So the area is A =2 0
√
−3 −x x +3 dx. We
substitute u = x +3,sodu = dx and the limits change to 0 and 3,andweget
A = −2 3
[(u − 3)√ u ] du = −2 3
0 0 (u3/2 − 3u 1/2 ) du
= −2
2
5 u5/2 − 2u 3/2 3
0
= −2 √ 2
5 3
2
3 − 2 3 √ 3 √
= 24
5 3