AND SIGNED AREA In Section 22.1 we established that the signed area under a rate graph is of particular interest to us; this area corresponds to the net change in amount.. EXAMPLE 22.7 T
Trang 1(a) The lead the chicken has when the goat begins to move
(b) The distance between the goat and the chicken at time t = 1.5
(c) The distance between the goat and the chicken at time t = 3
(d) The time at which the goat is farthest ahead of the chicken
(e) Approximate the time(s) at which the chicken and the goat pass one another on the path
(f ) Approximate the time at which the distance between the goat and the chicken is increasing most rapidly At this time, what is the relationship between v
gand v
c?
6 Suppose we want to approximate02f (x) dxby partitioning the interval [0, 2] into n equal pieces and constructing left- and right-hand sums Let f (x) = x3
(a) Put the following expressions in ascending order
2
0
f (x) dx, L4, R4, L20, R20, L100, R100 (b) Find |R4− L4|
(c) Find |R100− L100|
(d) How large must n be to assure that |Rn− Ln| < 0.05?
(e) Write out R4, once using summation notation, once without Evaluate R4
7 Turn back to Problem 1 of Problems for Section 22.1 on page 723 Express the following quantities using a definite integral or the sum or difference of definite integrals (a) The length of the line at 11:00 a.m
(b) The length of the line at its longest
(c) The number of people who came to the clinic for flu shots
(d) The number of people actually served by the clinic
(e) The length of the line at 3:00 p.m
(f ) The amount of time a person arriving at noon has to wait in line
(g) The amount of time by which the clinic must extend its hours in order to serve everyone who is in line before 4:00 p.m
8 Repeat parts (a) through (e) of Problem 6, letting f (x) =x+11
AND SIGNED AREA
In Section 22.1 we established that the signed area under a rate graph is of particular interest
to us; this area corresponds to the net change in amount In Section 22.2 we established that
if f is a reasonably well-behaved function,10we can, in theory, find the signed area under the graph of f (t) on the interval [a, b] as follows Partition the interval [a, b] into n equal pieces, each of length t =b−an , and label as shown on the following page
10
Trang 2t0 t1 t2 t n–1 t n
The signed area is defined to be limn→∞n
i=1f (ti) t, provided this limit exists,11and is denoted byabf (t ) dt In this section we interpretabf (t ) dtas signed area (positive for
f >0 and negative for f < 0) and use our knowledge of functions and their graphs, along with a bit of basic geometry, to approximate some definite integrals and evaluate others.12 REMARK: The definite integralsb
a f (t )dt,b
a f (x) dx,b
a f (w)dw, andb
a f (s)dsare all equivalent Each gives the signed area between the graph of f and the horizontal axis from a to b The variables t, x, w, and s are all dummy variables
EXAMPLE 22.7 The function r(t) gives the rate at which water is flowing into (or out of) a backyard
swimming pool The graph of r(t) is given below At time t = 0 there are 200 gallons
of water in the pool
120
–150
(5, 120)
t (hrs)
r (t) (gal/hr)
Figure 22.20
(If w(t) is the number of gallons of water in the pool at time t, where t is given in hours, then r(t) =dwdt.)
Express your answers to the questions below using definite integrals wherever appro-priate and evaluate these integrals
(a) When is the amount of water in the pool greatest? At that time, how many gallons of water are in the pool?
(b) What is the net flow in or out of the pool between t = 4 and t = 12?
(c) At what time is the pool’s water level back at 200 gallons?
(d) Sketch a graph of the amount of water in the pool at time t
SOLUTIONS (a) The amount of water is maximum at t = 9 because water is entering the pool from t = 0
to t = 9 and leaving for t greater than 9
11 If f is continuous we can guarantee that this limit exists.
12 We will rely on the following bits of geometry:
The area of a disk is π r 2 The area of a trapezoid is 1 (h1+ h 2 ) · b, where 1 (h1+ h 2 ) is the average of the heights.
We can think of a triangle as a trapezoid with one “height” of length zero and a rectangle as a trapezoid with both heights
Trang 3The amount of water entering the pool on the interval [0, 9] is9
0 r(t )dt, the area under the graph from 0 to 9
9
0 r(t )dt = the area of the rectangle + the area of the triangle
= (5)(120) + (1/2)(4)(120)
= 600 + 240
= 840
r (t)
Figure 22.21
840 gallons of water have entered the pool in those 9 hours Adding this to the original
200 gallons gives a total of 1040 gallons of water in the pool
(b) The net flow in or out of the pool between t = 4 and t = 12 is given by412r(t )dt Compute this by summing the signed area under r(t) from t = 4 to t = 5, from 5 to 9, and from 9 to 12 In other words, compute12
4 r(t )dt by expressing it as follows
12
4 r(t )dt =
5
4 r(t )dt +
9
5 r(t )dt +
12
9
r(t )dt
= area of rectangle + area of triangle + (− area of triangle)
= 120(1) + (1/2)(120)(4) − (1/2)(3)(height of triangle)13
= 120 + 240 − (1/2)(3)(90)
= 360 − 135
= 225 The pool has a net gain of 225 gallons of water
12 90
r (t)
Figure 22.22
13 The slope of the line forming the triangles is (−120/4) = −30, so between t = 9 and t = 12 the value of r(t) drops by 90,
Trang 4(c) There are many ways to approach this question From part (a), we know that by time
t = 9 the pool has gained 840 gallons of water After t = 9 the pool loses water Between
t = 9 and t = 14 the pool has lost (1/2)(5)(150) = 375 gallons of water, so it now has a net gain of 840 gallons − 375 gallons = 465 gallons Subsequently water is flowing out
at a rate of 150 gallons per hour In t more hours the pool will have lost an additional 150t gallons of water For what t will 150t = 465? When t = 465/150 = 3.1 Therefore, at time t = 14 + 3.1 = 17.1, (17 hours and 6 minutes) the pool will be back at its original level of 200 gallons Using integral notation, we have solved the following equation for T
T
0 r(t )dt = 0
Or, equivalently,
9
0 r(t )dt = −
T
9
r(t )dt
r (t)
areas must be equal
Figure 22.23
(d) Here is a graph of w(t), where w(t) is the amount of water in the pool at time t
w (t)
t(hrs)
(5, 800)
(9, 1040)
(14, 665)
(17.1, 200)
1000 800 600 400 200
Figure 22.24
EXAMPLE 22.8 Evaluate the following definite integrals by interpreting each as a signed area
(a)2π
−asin x dx (c)−77 x4 +xx2 +1dx (d) −23 |x| − 2 dx
SOLUTIONS (a)2π
cos x dx = 0 The “positive” area and the “negative” area cancel
Trang 5cos x
x
1
–1
2
3 π 2 π
2
Figure 22.25
(b)−aa sin x dx = 0 Sine is an odd function (sin(−x) = − sin x), so again the negative and positive areas cancel regardless of the value of a
x
sin x
Figure 22.26
(c) The integrand is an odd function: (−x)
(−x) 4 +(−x) 2 +1= −x4 +xx2 +1 Thus, the negative and positive areas cancel and we conclude that−77 x4 +xx2 +1dx = 0
y
x
x
x4 + x2 + 1
–.4
.4
y =
Figure 22.27
(d) Sketch the graph of |x| − 2 The area of the triangle lying below the x-axis is (1/2)(4)(2) = 4 Therefore,−22 (|x| − 2) dx = −4 The area of the triangle lying above the x-axis between x = 2 and x = 3 is (1/2)(1)(1) = 1/2 We conclude that
3
−2(|x| − 2) dx =
2
−2(|x| − 2) dx +
3
2 (|x| − 2) dx
= −4 + 0.5 = −3.5
Trang 62
f (x) = |x| –2
Figure 22.28
EXAMPLE 22.9 Put the following expressions in descending order
A =
−π
−2π
0
−π
2π
0
e−0.1xsin x dx,
D =
3π
0
2π
π
π
π
e−0.1xsin x dx
SOLUTION Sketch the graph of f (x) = e−0.1xsin x
f (x)
x
Figure 22.29
We begin by grouping the expressions above by sign: F =π
π e−0.1xsin x dx = 0:
a
a f (x) dxis always zero; there is no area contained between x = a and x = a
The following expressions are positive:
A =
−π
−2π
e−0.1xsin x dx, C =
2π
0
e−0.1xsin x dx, D =
3π
0
e−0.1xsin x dx The following expressions are negative:
B =
0
−π
2π
π
e−0.1xsin x dx
Convince yourself, by considering the graphical interpretation of each of these definite integrals, that the integrals should be ordered as follows
−π
−2π
e−0.1xsin x dx >
3π
0
e−0.1xsin x dx >
2π
0
e−0.1xsin x dx >
π
π
e−0.1xsin x dx >
2π
π
e−0.1xsin x dx >
0
−π
e−0.1xsin x dx
A > D > C > F > E > B
Trang 7P R O B L E M S F O R S E C T I O N 2 2 3
1 By interpreting the definite integral as signed area, calculate the following
(a)5
−2|x| dx (d) 3
−13 dx (e)02πsin t dt (f) −ππ cos z dz (g) −22 (x + 1) dx (h) −22 |x + 1| dx
2 Let f be the function whose graph is given below
f
x
(3, 2)
3
−6f (x) dxis closest to which of the following? Explain your reasoning
(a) −18 (b) −9 (c) −3 (d) 1.5 (e) 3 (f) 9 (g) 18
3 Put the following integrals in ascending order, placing “<” or “=” signs between
them as appropriate (Strategy: First determine which integrals are positive, which are
negative, and which are zero.)
(a)π
0 sin(t) dt (b) 2π
−π sin(t) dt (c)2π
−πcos(t) dt (d) 0π/2cos(t) dt (e) 0πcos(2t) dt (f)03π/2| sin(t)| dt
4 Put the following in ascending order, placing “<” or “=” signs between them as appropriate
(a)π/2
0 sin(t) dt (b) π/2
−π/2t dt (d) −π/20 sin(t) dt (e) 0π/21 dt (f) π/2
Below is the graph of a function f (x) Problems 5 and 6 refer to this function The curve from ( 0, 0) to (6, 0) is a semicircle.
f
x
(3, –3)
(–2, 4)
4 2 –2 –2 –4
–4 –6 –8
2 4
6
Use this graph to evaluate the following:
Trang 85 (a) −8−2f (x) dx (b) −80 f (x) dx (c)06f (x) dx (d) 36f (x) dx
6 (a) 3
−2f (x) dx (b) 6
−8f (x) dx (c) 6
0 |f (x)| dx (d) 6
−8|f (x)| dx
7 (a) What is the equation of a circle of radius 2 centered at the origin?
(b) Write a function, complete with domain, that gives the equation of the top half of
a circle of radius 2 centered at the origin
(c) Let f (x) =
√
4 − x2 for −2 ≤ x ≤ 0, 2x for x > 0
Evaluate−22 f (x) dx
From the definition of the definite integral and its interpretation as signed area, we obtain the following properties.14
1 Constant Factor Property
For any constant k,abkf (t ) dt = kabf (t ) dt
A constant factor can be “pulled out” of a definite integral
2 Dominance Property
If f ≤ g on the interval [a, b], thenb
a f (t ) dt ≤abg(t ) dt
3 Endpoint Reversal Property
a
b f (t ) = −abf (t ) dt
4 Additive Integrand Property
b
a[f (t) + g(t)] dt =abf (t ) dt +abg(t ) dt
5 Splitting Interval Property
c
a f (t ) dt =abf (t ) dt +bcf (t ) dt (This is true regardless of the relative positions of a, b, and c.)
6 Symmetry Property
a
−af (t ) dt = 0 if f is odd;a
−af (t ) dt = 2a
0 f (t )dtif f is even
These properties are illustrated below
Properties Illustrated
1 For any constant k,b
a kf (t ) dt = kabf (t ) dt
We can pull a constant multiple out of a definite integral This is a property we use repeatedly in our work
(a) Interpretation as the integral of a rate function
Suppose f (t) is the rate that water is entering a pool Doubling the rate at which water enters the pool over a certain time interval will double the amount of water
14 Assume f is bounded on the intervals of integration and all integrals exist.
Trang 9added to the pool in that time interval Similarly, halving the rate will halve the net change in water
(b) Interpretation as area under f from t = a to t = b
If the height of f is doubled, the area it bounds doubles (If this doesn’t make sense intuitively, consider partitioning the region and approximating its area with rectangles; the height of each rectangle will double, so the area of each rectangle will double.)
2 f
f t b a
f (x)
f (x)
Figure 22.30
2 If f ≤ g on the interval [a, b], thenb
a f (t ) dt ≤abg(t ) dt
(a) Interpretation as the integral of a rate function
Suppose f (t) and g(t) give the velocities of two cars If the velocity of the second
is greater than or equal to that of the first, then the second car will have the same or greater net change in position In other words, if the rate of change of position of the second car is greater, the net change in position will also be greater
(b) Interpretation as area under f from t = a to t = b
a
g (t) dt
a b
b
g f
g f
f (t) dt ≤
a
b
g (t) dt
a
b
f (t) dt ≤
a b
more negative
less negative
Figure 22.31
3.a
b f (t ) dt = −b
a f (t ) dt
(a) Interpretation as the integral of a rate function
Suppose f (t) is the rate at which water is entering a pool and t = a corresponds
to 10:00 a.m., while t = b corresponds to noon b
a f (t )dt is the net change in water in the pool between 10:00 a.m and noon Let’s say this is 1000 gallons Then the net change in water between noon and 10:00 a.m (2 hours earlier) is
−1000 because there were 1000 gallons less water at 10:00 a.m than at noon
10
12 f (t ) dt = −12
10 f (t ) dt When you seea
b f (t ) dt, where a < b, you can think
of running a movie backward
Trang 10(b) Interpretation as area under f from t = a to t = b
Suppose a < b Think back to the approximating rectangles; each rectangle has base t, where t =b−an is positive If we switch the roles of a and b, then our measure of t becomes negative, so the value of the integral reverses sign
Advice: Whenever you come across c
b f (t ) dt, where b > c, switch the endpoints and negate immediately For instance, when dealing with −2
0 f (t ) dt, write it as
−0
−2f (t ) dt
4.b
a[f (t) + g(t)] dt =b
a f (t ) dt +b
a g(t ) dt
(a) Interpretation as the integral of a rate function
A pool is being filled with two hoses Suppose f (t) is the rate at which water is entering through one hose and g(t) is the rate that water is entering via another hose Then the total amount of water that enters the pool is the water that enters through one hose plus the water that enters through the other
(b) Interpretation as area under f from t = a to t = b
Think about the approximating rectangles In the bottom graph of Figure 22.32 the rectangles are positioned on top of one another; their positioning relative to one another does not affect their sum
f
f t g
y
g
g
g
g
t
f +g
f +g
f
f
f
g
}
}
}
Figure 22.32
5.c
a f (t ) dt =b
a f (t ) dt +c
b f (t ) dt
(a) Interpretation as the integral of a rate function
Suppose f (t) is the rate at which water is entering a pool The total amount of water that enters the pool from 9:00 a.m to 5:00 p.m is the amount that enters from 9:00 a.m.to noon plus the amount that enters from noon to 5:00 p.m
(b) Interpretation as area under f from t = a to t = c
Using the endpoint reversal property, (3), you can convince yourself that the split-ting interval property holds regardless of whether or not b lies between a and c The figures on the following page illustrate the statement for different relative positions of a, b, and c Areas shaded with gray indicate a positive weighing in; areas shaded with hash marks indicate a negative weighting due to endpoint reversal Work through each scenario slowly to assure yourself that in each case