The Area Function and Its Characteristics The Big Picture Our two fundamental interpretations of the definite integralb a f x dx are: the signed area between the graph of f and the horiz
Trang 122.4 Properties of the Definite Integral 741
c
a f (t )dt =
b
a f (t )dt +
c b
f (t )dt
f
f
f
f
c
f
c
f
Figure 22.33
6.a
−af (t ) dt = 0 if f is odd; −aa f (t ) dt = 20af (t ) dtif f is even
f odd
t
f even
t
Figure 22.34
P R O B L E M S F O R S E C T I O N 2 2 4
For Problems 1 and 2, evaluate the following.
1 (a)0
4 (x + 1) dx
2 (a)1
0
√
1 − x2dx (b) −1
1
√
1 − x2dx
(Hint:Think about circles.)
3 Choose the correct answer and explain your reasoning
(a)a
−a
1
1+x 2 dx =
(i) 0 (ii) a
0 1 1+x 2 dx (iii) 2a
0 1 1+x 2 dx (iv) −a
0 1 1+x 2 dx
Trang 2742 CHAPTER 22 Net Change in Amount and Area: Introducing the Definite Integral
(b) −aa x 1+x 2 dx = (i) 0 (ii) a
0 x 1+x 2 dx (iii) 2a
0 x 1+x 2 dx (iv) −a
0 x 1+x 2 dx
4 (a) Explain why3
0.5 1−lnx
x 2 +1 dx >4
0.5 1−ln x
x 2 +1 dx
(Hint: Look at the sign of the integrand.)
(b) Put in ascending order:
0,
1 1/e
1 − ln x
x2+ 1 dx,
2 1/e
1 − ln x
x2+ 1 dx,
e 1/e
1 − ln x
x2+ 1 dx,
4 1/e
1 − ln x
x2+ 1 dx.
5 (a) Assume that a < b Insert a “≥” or “≤” sign between the expressions below as appropriate Explain your reasoning clearly
b
a |f (t)| dt and
b a
f (t ) dt
(b) Under what circumstances will the two expressions be equal?
6 For each of the following, sketch a graph of the indicated region and write a definite integral (or, if you prefer, the sum and/or differences of definite integrals) that gives the area of the region
(a) The area between the horizontal line y = 4 and the parabola y = x2 (b) The area between the line y = x + 1 and the parabola y = x2− 1
7 Put the following four integrals in ascending order (from smallest to largest) Explain, using graphs, how you can be sure that the order you gave is correct
π 0 sin t dt,
π 0
2 sin t dt,
π 0 sin(2t) dt,
0 π sin t dt
8 Suppose05f (t )dt = 10 Evaluate four out of the five expressions that follow One of them you do not have enough information to evaluate
(a) 057f (t) dt (b) 05(f (t ) + 7) dt (c)05f (t ) dt + 7 (d) 5
0 7f (t + 7) dt (e)−2
−7 7f (t + 7) dt
9 If05f (t ) dt = 10 and05g(t ) dt = 3 does it follow that (a) f (t) > g(t) for all t between 0 and 5? Explain
Mathematicians might write this statement in mathematical symbols as follows:
f (t ) > g(t ) ∀ t in [0, 5] The symbol “∀” is read “for all.”
(b) f (t) > g(t) for some t between 0 and 5? Explain
Mathematicians might write this statement in mathematical symbols as follows:
∃ t in [0, 5] such that f (t) > g(t) The symbol “∃” is read “there exists.”
Trang 3The Area Function and Its Characteristics
The Big Picture
Our two fundamental interpretations of the definite integralb
a f (x) dx are:
the signed area between the graph of f and the horizontal axis on the interval [a, b]; the net change in amount A(x) between x = a and x = b if f (x) is the rate of change
of A(x)
Note the parallel in construct between our interpretations of the definite integral and our interpretations of the derivative evaluated at a point We have a geometric interpretation and a physical interpretation (For the derivative at a point these were (i) the slope of a curve at a point and (ii) instantaneous rate of change of a quantity.) Not only did we study
the derivative evaluated at a point, and the significance of this number, but we studied the derivative function, and interpreted it as the slope function or instantaneous rate of change
function The definite integralb
a f (x) dx, where a and b are constants, is a number— whether interpreted as signed area or net gain (or loss) in amount In this chapter we look
at the area (or amount) function We’ll introduce this function via some examples.
x
EXAMPLE 23.1 Consider−1x f (t ) dt, where f is the constant function f (t) = 4 Is−1x f (t ) dt itself a
function? If so, can it be expressed by an algebraic formula?
743
Trang 4744 CHAPTER 23 The Area Function and Its Characteristics
SOLUTION f
t
4
4
3
f (t) = 4
f
t
4
4
4
f (t) = 4
f
t
4
4
π + 1
f
t x
4
4 –1
x + 1
4dt
–1
x
∫
4dt
–1
π
∫
4dt
–1
3
∫
4dt
–1
2
∫
Figure 23.1
Indeed,x
−14 dt is a function It is the function that gives the signed area under f (t) = 4
between t = −1 and t = x This area depends upon x and is uniquely determined by x In fact, because the area is a rectangle, we can compute it by multiplying the height, 4, by the base, x + 1, to obtain the function 4 · (x + 1) = 4x + 4
We could name this area function A(x), but that wouldn’t tell us that it’s the area function for f with anchor point −1, so we dress it up a little more and write−1Af(x).1 The formula−1Af(x) = 4x + 4 holds for x ≤ −1 as well as x > 1
For x = −1 we have−1Af(−1) =−1
−1 4 dt = 0 Any area function will be zero at its anchor point
For x < −1 we use the endpoint reversal property of definite integrals to write
x
−14 dt = −
−1
x 4 dt = −4(−1 − x)
= 4 + 4x
x
4 3 2 1 –1
–1A f (x)
Figure 23.2
For example, we evaluate−1Af(−2) as follows:
−1Af(−2) =
−2
−1 4dt = −
−1
−2 4 dt = −(4)(1) = −4
Computing Distances: General Principles
In the previous example, and in many applications of integration to come, you will be computing both vertical distances and horizontal distances
From Graphical, Numerical, and Symbolic Points of View, Saunders College Publishing, 1997 The notation Afis theirs.
Trang 523.1 An Introduction to the Area Function a f (t ) dt 745
vertical distance = (high y-value) − (low y-value) horizontal distance = (right x-value) − (left x-value)
It is as simple as that The location of the x- and y-axes is completely irrelevant Be sure this makes sense to you as you look at Figure 23.3
2
–1
y
x
2 – (–1) = 3
–7 –3
–3
y
y
–3 – (–7) = –3 + 7 = 4
–1 – (–3) = –1 + 3 = 2
f (x) = y
f (x1) – g(x1)
y1
q (y1) – r ( y1)
x = q( y)
x = r( y)
y
x y
x
x1
g (x) = y
–1 –2
1 – (–2) = 1 + 2 = 3
y
x
1
Figure 23.3
Let’s return to the area function from Example 23.1 but change the anchor point from
−1 Below we look at the area functions−2Af(x),0Af(x), and1Af(x), where f (t) = 4 and the anchor points are −2, 0, and 1, respectively
f
x
t
4
f
x x
t
4
f
4
8
–4
A f
–2A f
–2A f (x) = 4dt = 4 • (x + 2)
1A f
0A f
The graph of A f is
a straight line with slope of 4.
–2
x
1
x
∫
0A f (x) = 4dt = 4x
0
x
∫
Figure 23.4
Notice that changing the anchor point only changes the area function by an additive constant
EXERCISE 23.1 Argue that the formulas given for the functions−2Af(x),0Af(x), and1Af(x)above are
valid for x to the left of the anchor point
D e f i n i t i o n
For a continuous function f and a constant “a” in the domain of f we defineaAf(x)
to beaxf (t ) dt
aAf(x)gives the signed area between the graph of f and the t-axis from t = a to t = x
Therefore, we will refer to it as the area function.
Trang 6746 CHAPTER 23 The Area Function and Its Characteristics
Amount Added, Accumulation, Accruement,
In this chapter we will look at the characteristics of the functionaAf(x) =axf (t ) dt While our predominant interpretation ofaAf(x)will be as the area function, we could give it other interpretations as well For example, suppose t = 0 corresponds to noon, t is measured in hours, and f (t) gives the rate, in gallons per hour, at which water is leaving a storage tank Then−1Af(x) =x
−1f (t ) dtis the amount of water that leaves the tank between 11:00 a.m and time x For instance,−1Af(3) =3
−1f (t ) dtis the amount of water that has left the tank between 11:00 a.m and 3:00 p.m When f gives a rate, Af(x)represents the net change
in amount So, we can think of the A as standing for Area (signed area), or as standing for
Amount where by “amount” we mean “amount added.” The “A” can stand for accretion
(an increase by addition), accumulation (a collection over time), or accruement (an amount
added) Certainly “A” is a convenient letter for this function
P R O B L E M S F O R S E C T I O N 2 3 1
1 Let f (t) = 7 We’ll define three area functions The difference between their definitions
is the anchor point We’ll denote them as follows
0Af(x) =
x 0
f (t ) dt 2Af(x) =
x 2
f (t ) dt 3Af =
x 3
f (t ) dt (a) Find a formula (not involving an integral) for 0Af(x), where x ≥ 0 What is
0Af(−3)? Does your formula work in general for negative x?
(b) Find a formula (not involving an integral) for2Af(x), where x ≥ 2 Does your formula work for x < 2 as well?
(c) Find a formula (not involving an integral) for 3Af(x), where x ≥ 3 Does your formula work for x < 3 as well?
(d) Graph the functions0Af(x),2Af(x), and3Af(x)
(e) Differentiate0Af(x),2Af(x), and3Af(x)
2 Let f (t) = 2t We’ll define three area functions The difference between their defini-tions is the anchor point We’ll denote them as follows:
0Af(x) =
x 0
f (t ) dt 2Af(x) =
x 2
f (t ) dt 3Af =
x 3
f (t ) dt
(a) Find a formula (not involving an integral) for 0Af(x), where x ≥ 0 What is
0Af(−3)? Does your formula work in general for negative x?
(b) Find a formula (not involving an integral) for2Af(x), where x ≥ 2 Does your formula work for x < 2 as well?
(c) Find a formula (not involving an integral) for 3Af(x), where x ≥ 3 Does your formula work for x < 3 as well?
(d) Graph the functions0Af(x),2Af(x), and3Af(x)
(e) Differentiate0Af(x),2Af(x), and3Af(x)
3 Let f (t) = sin t Let Af(x) =0xsin t dt, where x ≥ 0
Trang 723.2 Characteristics of the Area Function 747
(a) Put the following in ascending order, with “<” or “=” signs between them
Af(0), Af π
2
, Af(π ), Af 3π
2
, Af(2π ) (b) For what values of x is Af(x) = 0?
(c) For what values of x is Af(x)negative?
(d) Fow what values of x is Af(x)maximum?
4 The rate that water is entering a tank is given by f (t) = 40 − 2t gallons/minute where
tis measured in minutes past noon
(a) Interpret0Af(x)in words, for x ≥ 0
(b) For what values of x is0Af (x)increasing?
(c) At what time is the water level in the tank the same as it was at noon? At this time, what is the value of0Af(x)?
EXAMPLE 23.2 Let1Af(x) =1xf (t ) dt, where the graph of f is drawn below and the domain of Af is
restricted to 1 ≤ x ≤ 12
f
t
1 –1 –2
Figure 23.5
(a) On what interval(s) is the function1Af(x)increasing? What characteristic of f ensures that1Af(x)is increasing?
(b) On what interval(s) is the function1Af(x)decreasing? What characteristic of f ensures that1Af(x)is decreasing?
(c) Where on [1, 12] does1Af(x)have its local maxima and minima?
Where on [1, 12] does1Af(x)attain its absolute maximum and minimum values? (d) On what interval(s) is the function1Af(x)concave up? Concave down?
What characteristic of f ensures that1Af(x)is concave up and what characteristic ensures that1Af(x)is concave down?
SOLUTIONS (a)1Af(x)is increasing on [1, 6] and [10, 12] Wherever f is positive, the area function
increases with x, because a positive contribution to area is being made
(b)1Af(x)is decreasing on [6, 10] Wherever f is negative, the accumulated area de-creases with x, because a negative contribution is being made
Trang 8748 CHAPTER 23 The Area Function and Its Characteristics
(c) On the number line below we indicate where1Af(x)is increasing and where it is decreasing
graph of 1A f
sign of f
1 + 6 – 10 +12
The local maximum occurs at x = 6; the local minimum at x = 10
To identify where1Af(x)takes on its absolute maximum and minimum values,
we must also consider the endpoints of the domain, x = 1 and x = 12
The absolute minimum occurs either at x = 1 or x = 10 We must compare the value
of the function at these two points
1Af(1) =11f (t ) dt = 0 For 1 < x ≤ 12,1Af(x)is always positive, since between
t = 1 and any value greater than 1 there is always more positive accumulated area than there is negative accumulated area Therefore the absolute minimum value of1Af on [1, 12] is 0 and it is attained at x = 1
To find out where1Afattains its absolute maximum value, we must compare the values
at x = 6 and x = 12 The area between 6 and 10 gives a negative contribution and is larger in magnitude than the positive contribution between 10 and 12, so1Af(12) is smaller than1Af(6) The absolute maximum is attained at x = 6
f
t
1 –1 –2
Figure 23.6
(d) We’ll break this discussion into cases
We begin by looking at1Af on regions where f is positive Where f is constant, area is accumulating at a constant rate; thus,1Af is increasing at a constant rate and is neither concave up nor concave down
Where f is positive and increasing, the area function is increasing at an increasing rate;
1Af is concave up
Where f is positive and decreasing, the area function is increasing at a decreasing rate;
1Af is concave down
Now let’s look at1Af on regions where f is negative Where f is constant, the area function is decreasing at a constant rate and is neither concave up nor concave down
Where f is negative and increasing (i.e., f is getting closer to zero),1Afis decreasing
by smaller and smaller amounts, so1Af is concave up
Where f is negative and decreasing,1Af is decreasing more and more steeply, so1Af
is concave down
Trang 923.2 Characteristics of the Area Function 749
We conclude that where f is increasing,1Af is concave up;
where f is decreasing,1Af is concave down
1Af is concave up on [1, 2), (4, 5), (8, 12] and concave down on (2, 4), (5, 8)
EXAMPLE 23.3 Let1Afbe the area function given byx
1 f (t )as in Example 23.2, but this time let’s enlarge the domain to [−3, 12] We already know the behavior of1Afon [1, 12], so we’ll concentrate
on the interval [−3, 1] We’re interested in determining whether the conclusions we drew above when considering1Af(x)for x to the right of the anchor point x = 1 also hold to the left of this anchor point In particular, we’d like to answer the following questions (a) Where f is positive, is1Af increasing? Where f is negative, is1Af decreasing? (b) Where f is increasing, is1Af concave up? Where f is decreasing, is1Af concave down?
The graph of f is reproduced below
f
t
1 –1 –2
Figure 23.7
SOLUTION Direct analysis of the area function to the left of the anchor point can sometimes make your
head spin.2Instead, it will be simpler just to move the anchor point so that there are no values of the domain to the left of it We’ll look at the function−3Af and then show that
1Af and−3Af differ only by an additive constant;−3Af and1Af are vertical translates
−3Af is simple to analyze using the same logic as in Example 23.2
Where f is positive,−3Af is increasing; where f is negative,−3Af is decreasing Where f is increasing,−3Af is concave up; where f is decreasing,−3Af is concave down
How are the functions−3Afand1Af related? How can we apply our knowledge of the former to the latter?
−3Af(x) =
x
−3
f (t )while1Af(x) =
x 1
f (t ) dt
From the splitting interval property we know thatb
a f (t ) dt =c
a f (t ) dt +b
c f (t ) dt Therefore,
2 You need to use the fact that 1 A f (x) = x
f (t ) dt = − 1
is increasing (e.g., 1, 2, 3), then when negated the sequence is decreasing (−1, −2, −3).
Trang 10750 CHAPTER 23 The Area Function and Its Characteristics
x
−3f (t ) dt =
1
−3f (t ) dt +
x 1
f (t ) dt
−3Af(x) =
1
−3f (t ) dt +1Af(x)
−3Af(x) −
1
−3f (t ) dt =1Af(x)
f
t
1 –1 –2
–3A f(1)
Figure 23.8
1
−3f (t ) dt is a constant, so1Af(x) =−3Af(x) + K for some constant K.1Af(x)and
−3Af(x)differ by a constant, so their graphs are vertical translates of one another Therefore, where−3Af(x)is increasing, 1Af(x)is also increasing Where−3Af(x)is decreasing,
1Af(x)is also decreasing Similarly, where−3Af(x)is concave up,1Af(x)is also concave
up, and likewise for concave down We summarize below
Where f is positive,1Af(x)is increasing; where f is negative,1Af(x)is decreasing Where f is increasing,1Af(x)is concave up; where f is decreasing,1Af(x)is concave down
In fact, any anchor point “a” in the interval [−3, 12] could be chosen andaAf(x)would simply be a vertical translate of−3Af(x) For any constant “a” we know that
x
−3
f (t ) dt =
a
−3
f (t ) dt +
x a
f (t ) dt
x
−3f (t ) dt = K +
x a
f (t ) dt
Let’s look at a case that’s more complicated than the constant function of Example 23.1 but more concrete and computational than Example 23.2
EXAMPLE 23.4 Let f (t) = 2 − t Find formulas for −2Af(x), 0Af(x), and 1Af(x), the area functions
anchored at −2, 0, and 1, respectively How are these area functions related? Restrict the domains of these functions to the interval [−2, 6]