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Unformatted text preview: Now, for a > 1. Therefore, 6.1 I The Definite Integral 291 [af(x)dx <0
1 a l a l
/ f(.x)dx =f f(x)dx +f f(x)dx </ f(x)dx (6.3)
0 o 1 o &—w_.__4
<0 Combining (6.2) and (6.3) shows that a l
f f(x)dx </ f(x)dx
‘ o o for all a z 0 and a 75 1. Hence, a = 1 maximizes the integral f0"(1 —— x2) dx. l 3' Section 6.1 Problems _  ’ I 6.1.1 1. Approximate the area under the parabola y = x2 from 0 to 1,
using four equal subintervals with left endpoints. 2. Approximate the area under the parabola y = x2 from 0 to 1
using ﬁve equal subintervals with midpoints. 3. Approximate the area under the parabola y = x2 from 0 to 1
using four equal subintervals with right endpoints. 4. Approximate the area under the parabola y = 1 — x2 from 0 to 1, using ﬁve equal subintervals with (a) left endpoints and (b) right
endpoints. » a t In Problems 5—14, write each sum in expanded form. 4 5
5. [J]? 6. Ztk— 1)2
k=l k=3
6 3
k2
7. 3" ~— k2+1
3 4
9. Z(x+1)" 10. Zk"
lt=0 k=0
.1 n
11. Z} 1%“ 12. Zf<CkIAxk
i=0 lt=i In Problems 15—22, write each sum in sigma notation. is. 2+4+6+8+~~+2n 16 _1_+1+1+1
ﬁ ﬁ ﬁ J3
17. ln2+ln3+ln4+ln5 18 2+i+§+6+Z
'5 6 7 s 9 l9.—l+l+g+:_3
4 6 7 8
20.1+.1.+.1.+1+_}_+...+_I.
1 2 4 8 16 2" 21.1+q+qz+q3+q“++q"‘l
22.1—a+a2—a3+a‘—a5++(—1)"a" In Problems 2330, use the algebraic rules for sums to evaluate each
sum. Recall that " n(n + 1)
k = 2
and
:k2 = n(n +1)(2n +1)
k=l 6
15 5
23. Zak + 3) 24. 2(4 — k2)
k=l k=1
6 n
25. Zk(k +1) 26. 24k
k=0 k=l
27. Z 4(k — 1)2 28. Ea + 2)(k — 2)
k=l It=1
10 10
29. 2(4)‘ 30. ZHV
k=l It: 31. The steps that follow will show that :kz _ n(n +1)(2n +1)
k=1 6 (a) Show that 210+ 1:13 — 181 = (2‘ —13)+(3’ ‘ 2") + (43 ‘ 33)
k=l +...+[(1+n)3—"3]
= (1 + n)3 ~ 1“ (Sums that “collapse” like this due to cancellation of terms are
called telescoping or collapsing sums.) (b) Use Example 3 and the algebraic rules for sums to show that Em +1.13 — k3] = 321:2 + 3"(" +1) +n
k=l k=l 2 292 Chapter6 I Integration (c) In (a) and (b), we found two expressions for the sum ll Z‘tu + k)3  k3] k=l Those two expressions are therefore equal; that is, n(n +1) 3_ 3___ n 2
(1+n) 1 32k+3 2 k=1 +n Solve this equation for 2L. k2, and show that :k1_ n(n+1)(2n+1)
’__6—_'_ ks] I 6.1 .2
32. Approximate l
f (l — x2)dx
—1 using ﬁve equal subintervals and left endpoints. 33. Approximate 1
f (1—x2)dx
—l using ﬁve equal subintervals and midpoints.
34. Approximate t
f (2 + x1) dx
—1 using ﬁve equal subintervals and right endpoints.
35. Approximate 2
f (2 + x2) dx
2 using four equal subintervals and left endpoints.
36. Approximate
2
f e“ dx
«1 using three equal subintervals and midpoints. 37. Approximate
air/2
/ sin x dx
0 using three equal subintervals and right endpoints. 38. (3) Assume that a > 0. Evaluate foax dx. using the fact that the region bounded by y = x and the xaxis between 0 to a is a
triangle. (See Figure 6.23.) 0
0 a x Figure 6.23 The region for Problem 38. (b) Assume that a > 0. Evaluate ﬁxdx by approximating the region bounded by y = x and the xaxis from O to a
with rectangles. Use equal subintervals and take right endpoints. (Hinn Use the result in Example 3 to evaluate the sum of the areas
of the rectangles.) 39. Assume that 0 < a < b < 00. Use a geometric argument to show that
b 2 2
b —a
d =
fa x x 2 40. Assume thatO < a < b < 00. Use a geometric argument and
Example 1 to show that b 3__3
fxzdx=b 3a Express the limits in Problems 41 4 7 as deﬁnite integrals. Note that
(I) P = [x0, x1, . . . , x,,] is a partition of the indicated interval,
(2) Ci 6 [xk—la It]. t"111(3) Axt = XI: " xkt ll
41. lim 2 2c: An, where P is a partition of [1. 2]
"PM’0 it.) . II
42. lim 2 Jank, where P is a partition of [1, 4]
“PH*0 i=1 n
43. lim Zoe. — 1)Ax., where P is a partition of[—3, 21 “I’ll—>0 3:1 II
1
44. Iim Ax ,where Pisa artition of 1, 2
"who; Ck +1 k p [ ] II . Ck — 1 . . .
45. hm E Ax ,where P is a artmon of 2. 3
"Pu—.0 h, Ck + 2 k p [ J II
46. lim 2(sin ck)Ax,,, where P is a partition of [0, 11']
"I’ll’0 [i=1 II
47. lim 2 e” Ax,“ where P is a partition of [—5, 2]
urn—w 1.31 In Problems 48—53, express the deﬁnite integrals as limits of
Riemann sums. —l 2
X
48. d
/:2 1+x2 x 3 e 7r 2x
50. f e'z'dx 51. f lnxdx 52. [ cos—dx
l 1 o 71' 6
49 f (x+1)‘/3dx
2 s
53. / g(x) dx, where g(x) is a continuous function on [0. 5]
0 In Problems 5 4—60, use a graph to interpret the deﬁnite integral in
terms of areas. Do not compute the integrals J 2 54. / (Zx+l)dx 55. / (x2—1)dx
O —l
5 2 1
56./ «xjdx 57. / e"dx
~22 o n ' 4
58. / cosx dx 59. f In x dx
n 1 /2
2 t
6. l——. .
0'/:J( 2r)dr Mh—wmvm ~ v ln’ Problems 61—67, use an area formula from geometry to ﬁnd the
value of each integral by interpreting it as the (signed) area under the graph of an appropriately chosen function.
3 3 61. x dx 62. 2 o
o
67. f (4 — ‘/9 — x2)dx
—3
‘1 6.1.3 ’
68. Given that evaluate the following: 2 1 2
(11)] in:de (b) 3x2dx
o
3 1 1
(c) / 3xzdx (d) 3x2dx
—1 3
(e) / (x+1)2dx
—2 3 l
73. Find/ tanxdx.
1 ’4. Explain geometrically why 2 2 l
fxzdx=f xzdx—f xzdx
1 o o (6.4)
1nd show that (6.4) can be written as
2 o 2
/ xzdx =f xzdx +/ xzdx (6.5)
1 1 o lelate (6.5) to addition property (5). 6.2 I The Fundamental Theorem of Calculus 293 In Problems 75‘79, verify each inequality without evaluating the integrals.
2 2
76./xdxsf xzdx
1 1 1 1
75. f xde/ xzdx
o o 4 ,
1
77. 05/ ﬁdng 78. is ./1—x2dx51
0 . Sir/6 2
79. E 5] sinxdx 5 l
3 "/6 3 80. Find the value of a z 0 that maximizes [0"(4 — x2) dx. 81. Find the value of a e [0, 271'] that maximizes j: cosx dx. 82. Finda e (0. 271'] such that a
[sinxdxzO
o 83. Find a > 1 such that /(x—2)3dx=0
1 84. Find a > 0 such that /a(1—x)dx=0 85. To determine agespecific mortality, a group of individuals,
all born at the same time, is followed over time. If N (t) denotes
the number still alive at time t, then N (t)/N (0) is the fraction
surviving at time t. The quantity r(t), called the hazard rate
function, measures the rate at which individuals die at time t;
that is, r(t) dt is the probability that an individual who is alive at
time t dies during the inﬁnitesimal time interval (t. t + dt). The
cumulative hazard during the time interval [0, t], j; r(s) ds, can be estimated as — ln Show that the cumulative hazard during the time interval [t, t+1], ﬁ'+1r(s)ds. can be estimated as  ln i 6.2 The Fundamental Theorem of Calculus In Section 6.1, we used the deﬁnition of deﬁnite integrals to compute [0" x2 dx. This
required the summation of a large number of terms, which was facilitated by the explicit summation formula for Z 2:. k2. Fermat and others were able to carry out similar calculations for the area under curves of the form y = x', where r was a
rational number different from ——1. The solution to the case r = 1 was found by the
Belgian mathematician Gregory of St. Vincent (15841667) and published in 1647. At ~2 1 6.2 I The Fundamental Theorem of Ca1culus 305 0 l x Figura 6.26 The graph ofy = between x = —2 and
x = 1. The function is discontinuous at x = 0. I 6.2.1
In Problems 1—14, ﬁnd Sf. .r x “4 l.y=/ 2u3du 2.y=/(l——)du
o o 2 3, y:/ (4u23)du
o S. y=/ ‘/1+2udu.x>0
o 6. y=f V1+u2du.x>0
0 7. y=/ ‘/1+sin2udu,x>0
o X f‘/2+csc2udu, x >0
0
I 9. yzf ue‘"du
3 ‘ 1
11. = d, 2
y £2u+3 14.!)
2 4. y=f (3+u‘)du
o 8.y ll 10. y=f lee—"zdu
1 X 12. y: du 424142 1
13. y=/ sin(u2+1)du 14. y=/ cosz(u —3)du
"/2 "/4 In Problems 1538, use Leibniz‘s rule toﬁnd 31 2.1—!
15. y = (1 +1511: 16. y = f (13 ~ 21.1:
n 0 _41 3.t+2
17. y=/ (212+1)dr 18. y=f (1+1‘)d1
0 0 x1+1 .xZ—z
l9.y=/ £41,190 20.y= ./3+udu,x>0
4 2
3x 21. y = (1 +e’)d! 212—1
22. y =/ (e‘2‘+ez’)dl
0 0 311+: In:
23.y=f (1+1e‘)dt 24. 112/ e"d(.x>0
1 2 ' Section 6.2 Problems; ' 25 3 5
y=f(1+t)dt 26. y=/(1+e’)d! 3 6
27. yzf (1+sint)dt 28. y: (1+tant)dt
2x 2x2
5 1 3 1
29. azdu,x>0 30.y=£2mdt,.‘>0
1
31. y =f sectdt,—1 < .1: <1
1
12 2"
32. y: f cottd! 33. y: (1+12)dz
2+1:z ‘ " 71
34. y:/ tanudu,0<x<z I
35. y=/ ln(t—3)dt.x>0
K
4 \’
36. y=/ n(1+t2)dt,x > 0
x3 x+13
37. y = / sintdt
2 32 1311
38. y =/ costdt
1 +12 I 6.2.2
In Problems 39—96, compute the indeﬁnite integrals. 39. f(1+3x2)dx 40. [(1.3 — 4) dx 1 1
41. /(§x2 — 5.011.: 42. [(4):3 + 5x2)dx l 1 l
43. — 2 —— . .. 5 3~
[(zx +3.1 3) dx 44 [(2.1 +2.1: 1) dx 2x2—x x3+3x
45. d 46. d
J} x / zﬁ x
47. [1.2.5.111 4s. (1+x‘)./.?dx 49. f (1.7/2 + 12/711111 l
5" / (“1+ W) "’
53. [(x — l)(x +1)dx 50. [(xJ/s +x5’3)dx 1
1/3
52. [(3.1 +3xm) dx
s4. [(1  112(1): 306 Chapter6 I Integration 55. [(x — 2)(3 — x)dx
57. f e?" dx 59. / 3e“ dx 61. [xeﬂz/‘zdx 63. f sin(2x) dx 65. / cos(3x)dx 67. / sec2 (3x) dx 5.
69. f—%—dx
1—sm x 71. / tan(2x) dx 73. f (sec2 x + tan x) dx 4
7. d
5 f1+x2 x 1
79. [x +2dx
s1. 2" ‘ 1 dx
3x
x + 3
83. x2 __ 9 dx
85. f 32"“ dx
x _
5.1:2
87. x2 + 1 dx
89. 3‘ dx
91. [S‘Z‘dx 93. [(x2 + 2‘) dx
95. [(JE + Jim: I 6.2.3 56. fax + 3)2 dx
58. f2e3‘ dx 60. f 2e""/3 dx 62. fe"(1— e“) dx
64. [sin 1 ;x dx —4
66. fcosz 5 xdx 68. / csc2(2x)dx , 70. JE— dx 1 cos2 x 72. f cot(3x) dx 74. [(cotx — c302 x) dx x2
7. — d
6 [(1 1+x2) x 78. 5 dx
./1—x2
1
80./x_3dx
82. 2"6+de
x
x+4
84. fxz_16dx
4—x
86. fx2_16dx
ZxZ
88. 1+xzdx In Problems 97—122, evaluate the deﬁnite integrals. 4
97. f (3 — 2):) dx
2 3
98. / (2x2—1)dx
1 s
101. f 11'”3 dx
1
2
103. f (2! — 1)(t +3) d!
0
Jr/4
105. f sin(2x)dx
u n/B '
107. f sec2 (2x) dx
0 2
100. f x”2 dx
1 91+J§
102.] dx
4 J? 2
104. / (2+3:)2d:
—1 3/3 x
106. / 2cos (—) dx
_,,/3 2 n/4
108. f tanxdx n/4
1 ‘1 4
109. f 1 2 dx 110. f 2 dx
0 1 + x _JS 1 + x
1/2 1 1/2 2
111. / dx 112. dx
0 ‘/1— x2 1/2 ‘/1— x2
n/6 "/15
113. / tan(2x) dx 114. ‘ sec(5x) tan(5x) dx
0 , x/zo
o 2 z
115. f e” dx 116. f 21e' d!
1 0
1 1
117. f M dx 118. / e'“l d:
1 —1
e 3
119. / ldx 120. f —1—— dz
1 x 2 Z + 1
~1 1 3 2
121. / du 122. —— dt
2 1 '— u 2 t — 1
123. Use I'Hospital’s rule to compute
1
lim —2— sint d!
xvO x o 124. Use l’Hospital’s rule to compute lim
I140 125. Suppose that i.
f e" dx
0 fxf(t)dt=2x2
0 Find f (x).
126. Suppose that [x f(t)dt = ltan(2x)
0 2 Find f(x). '2 I 6.3 Applications of Integration In this section, we will discuss a number of applications of integrals In the ﬁrst »
application, we will revisit the interpretation of integrals as areas; the second
application interprets integrals as cumulative (or net) change; the third will allow ‘
us to compute averages using integrals; and, ﬁnally, we will use integrals to compute volumes. In each application, you will see that integrals can be interpreted as “sums
of many small increments.” EXAMPLE l3 6.3 l Applications oflntegration 321 Set up, but do not evaluate, the length of the curve of the hyperbola f (x) = between a = 1 and b = 2. Solution. To determine the length of the curve, we need to ﬁnd f ’(x) ﬁrst. f'(X) =l~ x2 Then the length of the curve is given by the integral software packages th
these approaches on
is approximately 1.13. Section 6.3 Problems
I 6.3.1 Find the areas of the region: bounded by the lines and curves in Problems 1—12.
1. y=x2—4,y=x+2 rt
4. y=cosx,y=1,x=0,x= E
5. y = x2 + 1, y = 4x — 2 (in the ﬁrst quadrant)
6. y = x2. y = 2 — x, y = 0 (in the ﬁrst quadrant) l
7. y = x2, y = . y = 4 (in the ﬁrst quadrant)
X It
8. y=sinx,y=cosxfromx=0tox= Z 9. y=sinx,y=lfromx=0tox=g
10.y=x2.y=(x—2)2,y=0fromx=0tox=2
ll.y=x2,y=x3fromx=0tox=2 12. y=e",y=x+lfromx=1tox=1 In Problems 13—16, ﬁnd the areas of the region: bounded by the lines and curves by expressing x as a function of y and integrating
with respect to y. 13. y=x2,y=(x—2)2,y=0fromx=0tox=2
14. y=x,yx = 1,y= § (in the ﬁrst quadrant) l5. x=(y— l)2+3,x =1—(y—1)2fromy=0toy=2(in
the ﬁrst quadrant) l6. x=(y—l)2—l..r=(yl)2+1fromy=0toy=2 I 6.3.2 17. Consider a population whose size at time t is N (t) and whose
lynamics are given by the initialvalue problem dN dt 4: Vlth = (8) Find N (t) by solving the initialvalue problem. (b) Compute the cumulative change in population size between
t = 0 and t = 5. (c) Express the cumulative change in population size between time 0 and time t as an integral. Give a geometric interpretation
of this quantity. 18. Suppose that a change in biomass 80) at time t during the
interval [0, 12] follows the equation d It
$80) = cos for 0 5 t s 12.
(a) Graph 5? as a function of t. (b) Suppose that 3(0) = 80. Express the cumulative change in
biomass during the interval [0, t] as an integral. Give a geometric
interpretation. What is the value of the biomass at the end of the
interval [0, 12] compared with the value at time 0? How are these two quantities related to the cumulative change in the biomass
during the interval [0. 12]? 19. A particle moves along the xaxis with velocity v0) = —(z 2)2+1 for 0 5 t s 5. Assume that the particle is at the origin at time 0.
(9) Graph 1:0) as a function oft. (b) Use the graph of u(t) to determine when the particle moves
to the left and when it moves to the right. (c) Find the location s(t) of the particle at time t for 0 5 t 5 5. Give a geometric interpretation of s(t) in terms of the graph of
v(t). (1!) Graph 5(1) and ﬁnd the leftmost and rightmost positions of
the particle. .1
l
l
i 322 Chapter6 I integration 20. Recall that the acceleration (1(2) of a particle moving along
a straight line is the instantaneous rate of change of the velocity
v(t); that is,
(r) d via
a = ——
dt Assume that a(t) = 32 ft/sz. Express the cumulative change
in velocity during the interval [0, t] as a deﬁnite integral, and
compute the integral. 21. If % represents the growth rate of an organism at time t
(measured in months), explain what 7 d1
— dt
j; dr
represents. 22. If £1 represents the rate of change of the weight of an
organism of age x, explain what sdw
——d
3 (IX x means.
23. If 4f represents the rate of change of biomass at time I, explain what
6 dB
—~ d:
l .1,
means. 24. Let N (t) denote the size of a population at time t, and assume
that dN
I =f(t) Express the cumulative change of the population size in the
interval [0. 3] as an integral. I 6.3.3 25. Let f(x) = x2 — 2. Compute the average value of f(x) over
the interval [0, 2]. 26. Let g(t) = sin(rrt). Compute the average value of g(t) over
the interval [—1, 1].
27. Suppose that the temperature T (measured in degrees Fahrenheit) in a growing chamber varies over a 24—hour period
according to , 7r
T(t) = 68 +sm forO 5 t _<_ 24. (9) Graph the temperature T as a function of time t. (b) Find the average temperature and explain your answer
graphically. 28. Suppose that the concentration (measured in gm'”) of
nitrogen in the soil along a transect in moist tundra yields data
points that follow a straight line with equation y = 673.8 — 34.7x for O 5 x 5 10, where x is the distance to the beginning of the transect. What is the average concentration of nitrogen in the soil
along this transect? 29. Let f(x) = tan x. Give a geometric argument to explain why
the average value of f(x) over [—1, 1] is equal to 0. 30. Suppose that you drive from St. Paul to Duluth and you
average 50 mph. Explain why there must be a time during your
trip at which your speed is exactly 50 mph. 31. Let f (x) = 2x,0 s x 5 2. Use a geometric argument to find
the average value of f over the interval [0, 2], and ﬁnd x such that
f (x) is equal to this average value. 32. A particle moves along the x—axis with velocity vo) = —(z — 3)2 + 5 forO 5 t 5 6.
(a) Graph v(t) as a function oft for 0 5 t 5 6. (b) Find the average velocity of this particle during the interval
[0,6]. (c) Find a time t‘ e [0, 6] such that the velocity at time t‘ is equal
to the average velocity during the interval [0. 6]. Is it clear that
such a point exists? Is there more than one such point in this case?
Use your graph in (a) to explain how you would ﬁnd 2' graphically. I 6.3.4 33. Find the volume of a right circular cone with base radius r and
height h. 34. Find the volume of a pyramid with square base of side length
a and height h. In Problems 35—40, ﬁnd the volumes of the solids obtained by
rotating the region bounded by the given curves about the xaxis.
In each case, sketch the region and a typical disk element. 35, y =4—x2,y=0,x =0(in the ﬁrst quadrant)
36 y=./2x,y=0,x=2 37. y=./sinx,05x5n,y=0 38.y=e‘,y=0,x=0,x=ln2
7r Jr 39. r: ,——< <—, :0
y secx 3_x_3y 40. y=./1—x2,05x51,y=0 In Problems 41—46, ﬁnd the volumes of the solids obtained by
rotating the region bounded by the given curves about thexaxis.
In each case, sketch the region together with a typical disk element. 41, y=x2,y=x,0§x§1
42. y=2—x3.y=2+x3,05x§1
43. =e’,y=e”‘,0§x52 44. y = (/1 — x2, y =1,x = 1 (in the ﬁrst quadrant)
45. y = ,/cosx,y = 1.x =7—2r 1 46. y = ;,x = 0, y = 1, y = 2 (in the ﬁrst quadrant) In Problems 47—52, ﬁnd the volumes of the solids obtained by
rotating the region bounded by the given curves about the yaxis.
In each case, sketch the region together with a typical disk element. 47. y:ﬁ,y=2,x=0
48 Y=x2.y=4,x =0(in the ﬁrstquadrant)
49' Y=ln(x+l).y=ln3,x=0 .50.y=./:€,y=x,0_<_x<1 _ Sl.y=x2,y=ﬁ,05x5
52. =—,. =0, =1
y x r y 2 I 6.3.5
53. Find the length of the straight line y=2x from x = 0 to x = 2 by each of the following methods:
(a) planar geometry mwru” (b) the integral formula for the lengths of curves, derived in
Subsection 6.3.5 54. Find the length of the straight line
y = mx from .r = 0 to x = a, where m and a are positive constants, by
each of the following methods:
(a) planar geometry (b) the integral formula for the lengths of curves, derived in
Subsection 6.3.5 55. Find the length of the curve
3 y2=x fromx =1tox = 4.
56. Find the length of the curve 2y2 = 3xJ fromx =Otox =1.
57. Find the length of the curve _ x3 + 1
y " 6 2x
fromx =1tox =3.
58. Find the length of the curve
__ x‘ + 1
y — 4 8x2 fromx=2tox =4. Chapters I Review Problems 323 In Problems 5962, setup, but do not evaluate, the integrals for the
lengths of the following curves: 59. y=x2,~15x51 60. y =sinx,05.r 5% 61 y=e“.05xsl
62. y=lnx,i _<_x_<_e
63. Find the length of the quartercircle y=‘/1—nx2 for 0 5 x _<_ 1, by each of the following methods:
(a) a formula from geometry (b) the integral formula from Subsection 6.3.5 64. A cable that hangs between two poles at x = —M and x = M
takes the shape of a catenary. with equation 1 ax —ax
y—Ete +e ) where a is a positive constant. Compute the length of the cable
whena = land M = ln2. 65. Show that if
e" + e“
2 then the length of the curve f (x) between x = 0 and x = a for
any a > 0 is given by f’(a). f(x) = Chapter 6 Key Terms Discuss the following deﬁnitions and
concepts: 1. Area 2. Summation notation intervals 3. Algebraic rules for sums 4. A partition of an interval and the mtegrals norm of a partition 5. Riemann sum calculus, part I 6. Deﬁnite integral 13. Leibniz’s rule 7. Riemann integrable l4. Antiderivatives 8. Geometric interpretation of deﬁnite integrals calculus, part II Chapter 6 Review Problems: 1. Discharge of a River In studying the ﬂow of water in an open
channel, such as a river in its bed, the amount of water passing
through a cross section per second—~the discharge (Q)——is of
interest. The following formula is used to compute the discharge: B
Q= / U(b)h(b)db (6.18)
0 In this formula. b is the distance from one bank of the river to the
point where the depth h(b) of the river and the average velocity
Nb) of the vertical velocity proﬁle of the river at b were measured.
The total width of the cross section is B. (See Figure 6.48.) 9. The constantvalue and
constantmultiple rules for integrals
10. The deﬁnite integral over a union of 11. Comparison rules for deﬁnite 12. The fundamental theorem of 15. The fundamental theorem of 16. Evaluating deﬁnite integrals by using
the FTC, part II 17. Computing the area between curves
by using deﬁnite integrals 18. Cumulative change and deﬁnite
integrals 19. The meanvalue theorem for deﬁnite
integrals 20. The volume of a solid and deﬁnite
integrals 21. Rectiﬁcation of curves
22. Length of a curve
23. Are length differential Figure 6.48 The river for Problem 1. To evaluate the integral in (6.18), we would need to know
3(1)) and h(b) at every location b along the cross section. In
practice. the cross section is divided into a ﬁnite number of
subintervals and measurements of U and h are taken at, say, the Q 324 Chapter6 I Integration n'ght endpoints of each subinterval. The following table contains
an example of such measurements: Location )3 i O O 0
1 0.28 0.172
3 0.76 0.213
5 1.34 0.230
7 1.57 0.256
9 1.42 0.241
1 1 1.21 0.206
13 0.83 0.187
15 0.42 0.116
16 0 0 The location 0 corresponds to the left bank, and the location
B = 16 to the right bank, of the river. The units of the location
and of h are meters, and of ‘17. meters per second. Approximate
the integral in (6.18) by a Riemann sum, using the locations in the table, and ﬁnd the approximate discharge, using the data from the
table. 2. Biomass Growth Suppose that you grow plants in several
study plots and wish to measure the response of total biomass
to the treatment in each plot. One way to measure this response
would be to determine the average speciﬁc growth rate of the
biomass for each plot over the course of the growing season. We denote by 8(1) the biomass in a given plot at time t. Then
the speciﬁc growth rate of the biomass at timer is given by 1 dB
B(r) d:
(3) Explain why
1 j" 1 d 8(3)
 —— ds
t 0 (is
is a way to express the average speciﬁc growth rate over the
interval [0, t].
(b) Use the chain rule to show that 1 dB (1 (c) Use the results in (a) and (b) to show that the average speciﬁc
growth rate of B(s) over the interval [0, t] is given by
1 [3(1) 1 ’d provided that B(s) > 0 for s e [0. t]. (d) Explain the measurements that you would need to take if you
wanted to determine the average speciﬁc growth rate of biomass
in a given plot over the interval [0. r]. Problems 3—6 discuss stream speed proﬁles and provide a
jusnﬁcation for the two measurement methods described next.
(Adapted from Herschy. 1995) The speed of water in a channel
varies considerably with depth. Due to friction, the speed reaches
zero at the bottom and along the sides of the channel. The speed
is greatest near the surface of the stream. To ﬁnd the average
speed for the vertical speed proﬁle, two methods are frequently
employed in practice: 1. The 0.6 depth method: The speed is measured at 0.6 of the
depth from the surface, and this value is taken as the average
speed. 2. The 0.2 and 0.8 depth method: The speed is measured at 0.2
and 0.8 of the depth from the surface, and the average of the two
readings is taken as the average speed. The theoretical speed distribution of water ﬂowing in an open
channel is given approximately by __ l/c
Md): (D d) a (6.19) where v(d) is the speed at depth d below the water surface, c is a
constant varying from 5 for coarse beds to 7 for smooth beds, D is
the total depth of the channel, and a is a constant that is equal to the distance above the bottom of the channel at which the speed
has unit value. 3. (3) Sketch the graph of v(d) as a function of d for D = 3 m
anda =1mfor(i)c=5and(ii)c=7. (b) Show that the speed is equal to 0 at the bottom (d = D) and
is maximal at the surface (d = 0). 4. (a) Show by integration that the average speed 5 in the vertical proﬁle is given by
c D 1/ c
a = —
c+1(a) (b) What fraction of the maximum speed is the average speed D“? (c) If you knew that the maximum speed occurred at the surface
of the river [as predicted in the approximate formula for v(d)],
how could you ﬁnd '17? (In practice, the maximum speed may occur
quite a bit below the surface due to friction between the water
on the surface and the atmosphere. Therefore, the speed at the
surface would not be an accurate measure of the maximum speed.) 5. Explain why the depth til, at which 12 =' '17, is given by the
equation
D — d1 ”‘
v = ( )
a
We. can ﬁnd all by equating (6.20) and (6.21). Show that (111 C )c
D" c+1 and that dl/D is approximately 0.6 for values of c between 5 and
7, thus resulting in the rule (6.20) (6.21) 17 7V" U05 where v0.5 is the speed at depth 0.6D. (Hint: Graph 1—(c/(c+1))‘ as a function of c for c e [5, 7], and investigate the range of this
function.) 6. We denote by v0.2 the speed at depth 0.2D. We will now ﬁnd
the depth d; such that
U = %(vo.2 + 0.11) (a) Show that dz satisﬁes
1 (D — o.2o)”‘+ (D 42)” _ c (D)””
2 a a _ c + 1 a
[Hint Use (6.19) and (620).] (b) Show that
d2 __ 1 2C
D ‘ c +1 — (0.8)”‘] and conﬁrm that dz/ D is approximately 0.8 for values of c between ' 5 and 7, thus resulting in the rule — t
v z 5010.2 + U03) ...
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 Winter '07
 Kouba

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