We begin by generalizing the observation that area functions of f with different anchor points differ only by an additive constant.. Where f is positive, Af is increasing; where f is neg
Trang 1t
1
1 –1
2
Figure 23.9
SOLUTION Let’s begin with−2Af(x)because none of the domain is to the left of the anchor point
−2Af(x)is increasing on [−2, 2] and decreasing on [2, 6]
−2Af(x)is concave down on [−2, 6] because f is decreasing
graph of –2A f sign of f
– +
6
Finding a Formula for−2Af(x). First consider−2Af(x)for x on the interval [−2, 2]
−2Af(x)is the area of the trapezoid bounded by f (t) = 2 − t, the t-axis, t = −2, and
t = x (See Figure 23.10.) Its area is
1
2(h1+ h2) · b =1
2[4 + (2 − x)] · (x − (−2))
=1
2(6 − x) · (x + 2)
=1
2(12 − x2+ 4x)
= −x
2
2 + 2x + 6
−2Af(x) = −x22 + 2x + 6 on [−2, 2]
f
t
x 1
–1
2
3 4
4
5 6
2 + x
length = 2–x
f
t
–1
2 1
3 4
4
5 6
2 + x
length = 2–x (where x is
negative)
x is negative
Figure 23.10
For x on (2, 6] to find−2Af(x)we must subtract from the area of the triangle above the t-axis the area of the appropriate triangle below (See Figure 23.11.)
Trang 2t
2 4 4
length = 0 – (2–x) = x – 2
f (x) = 2–x
x – 2
x
Figure 23.11
(area of triangle above the t-axis) − (area of triangle below the t-axis)
= 1
2· 4 · 4
− 1
2 · (x − 2) · (x − 2)
= 8 −1
2(x
2
− 4x + 4)
= 8 −x
2
2 + 2x − 2
= −x
2
2 + 2x + 6
So−2Af(x) = −x22 + 2x + 6 on the entire domain given
Once we have found−2Af(x), finding0Af(x)and1Af(x)is not difficult because the functions differ only by additive constants
Finding a Formula for0Af(x). 0Af(x) =0xf (t ) dt To relate0Af(x)to−2Af(x)we use the splitting interval property,abf (t ) dt =acf (t ) dt +cbf (t ) dt
x
−2f (t ) dt =
0
−2f (t ) dt +
x
0
f (t ) dt
−2Af(x) =
0
−2f (t ) dt + 0Af(x), where−20 f (t ) dtis the area of the trapezoid shaded in Figure 23.12
0
−2f (t ) dt =12· (4 + 2) · 2 = 6
−2Af(x) = 6 + 0Af(x) So
0Af(x) =−2Af(x) − 6
−2Af(x) = −x2 + 2x + 6; therefore,0Af(x) = −x2 + 2x
Trang 3t
–2 2
–2A f(0) 4
Figure 23.12
Finding a Formula for1Af(x). Similarly,−2x f (t ) dt =−21 f (t ) dt +1xf (t ) dt
−2Af(x) =
1
−2
f (t ) dt + 1Af(x), where−21 f (t ) dtis the area shaded in Figure 23.13
t f
–2 –1 1 2 –2A f(0)
Figure 23.13
−2Af(1) = 12(4 + 1) · 3 =152
−2Af(x) = 15
2 +1Af(x) So
1Af(x) =−2Af(x) − 15
2 .
−2Af(x) = −x22 + 2x + 6; therefore,1Af(x) = −x22 + 2x − 1.5
x
–2A f
1A f
0A f
A f
–1 –1
1 2 3 4 5 6
6
–2
Figure 23.14
Alternatively, we could have found1Af(x)by starting with1Af(x) = −x22 + 2x + K Knowing that1Af(1) = 0 we can solve for K
Trang 40 = −1
2+ 2 + K
K = −3 2 Therefore,
1Af(x) = −x
2
2 + 2x − 1.5
Notice that
Af(x) = −x
2
2 + 2x + K, where the value of K depends upon the anchor point The area function graphs are vertical translates of one another In fact,
d
dxAf(x) = d
dx
−x
2
2 + 2x + K
= −x + 2 = f (x)
Interesting How much of this can we generalize?
We begin by generalizing the observation that area functions of f with different anchor points differ only by an additive constant This can be shown using the splitting interval property of integrals Let a and c be constants in the domain of f Then
x
a f (t ) dt =
c
a f (t ) dt +
x
c
f (t ) dt, so
aAf(x) =
c
a f (t ) dt +cAf(x), whereacf (t ) dt, the area under f from a to c, is a constant.3
How are Af and f related?
Let’s gather our observations from the previous examples and see what we can make of them
Observations
Area functions for f are vertical translates of one another
Where f is positive, Af is increasing;
where f is negative, Af is decreasing
Where f is increasing, Af is concave up;
where f is decreasing, Af is concave down
These statements are true regardless of the anchor point (so we’ve omitted it from the notation)
In addition, from the examples in which we were able to arrive at algebraic expressions for the area functions (Examples 23.1 and 23.4), we made the following intriguing observation
3c
f (t ) dt can also be written as A (c).
Trang 5In two examples we find the derivative of the area function Af is f itself.
For the time being this is only an observation of a phenomenon that presented itself in
two examples It does not have the status of a fact.
This last observation ties together the first three
Where the derivative of a function is positive, the function is increasing;
where the derivative is negative, the function is decreasing
Where the derivative of a function is increasing, the function is concave up;
where the derivative is decreasing, the function is concave down
If two functions have the same derivative, then they differ by a constant.4
We have a conjecture in the making
Conjecture
The derivative of the area function Af is f itself
We will prove this conjecture in the next section and find that it has beautiful and far-reaching consequences
P R O B L E M S F O R S E C T I O N 2 3 2
1 Let Af(x)be the area function given by Af(x) =0xf (t ) dt, where 0 ≤ x ≤ 11 The graph of f is given below
f
t
(a) On what interval(s) is the function Af(x)increasing?
(b) On what interval(s) is the function Af(x)decreasing?
(c) What is Af(0)?
(d) Is Af(x)ever negative on the interval [0, 11]?
(e) On [0, 11], where is Af(x)maximum? Minimum?
4
Trang 62 Below is the graph of f (t) f is an odd function.
f
t
(a) Which of the graphs below could be the graph of0Af(x) =0xf (t ) dt? Explain your reasoning
(b) Which of the graphs below could be the graph of2Af(x) =2xf (t ) dt? Explain
x
x
x
x
2
3 Below is the graph of f (t)
f
t
1 –1
(a) If F (x) =0xf (t ) dt, which of the graphs given below could be a graph of F (x)? Explain your criterion
(b) If G(x) =−2x f (t ) dt, which of the graphs given below could be a graph of G(x)?
A
t
1 –1
B
t
1 –1
C
t
1 –1
Trang 723.3 THE FUNDAMENTAL THEOREM OF CALCULUS
We are now on the threshold of obtaining a truly wonderful and important result At the end of the previous section, we conjectured that the derivative of the area function Afis f
In this section, we give a geometric justification of this conjecture Then we’ll see what an amazing and extremely useful result it is
Proof that if f is a Continuous Function, then dxd Af(x) = f (x)
Let f be a continuous bounded function and c a constant in the domain of f Consider the function Af(x) =cxf (t ) dt, the signed area between the graph of f and the t-axis between t = c and t = x
To calculate dxdAf(x)let’s begin by looking at
Af(x + x) − Af(x), the numerator of the difference quotient We will show that
lim
x→0
Af(x + x) − Af(x)
by using the “Squeeze Theorem” or “Sandwich Theorem.”
For simplicity’s sake, let’s begin by assuming that f is positive and increasing on the interval [x, x + x], where x is positive Then we can represent Af(x + x) − Af(x)
by the region shaded in Figure 23.15(b) (Later we will drop the requirements that f is positive and increasing; our argument can be adapted easily to the general case.)
f
t f
t
A f (x)
A f (x + ∆ x)
A f (x + ∆ x) – A f (x)
x + ∆ x x + ∆ x
Figure 23.15
We can write the following inequalities (Refer to Figure 23.15b.)
It’s legal to divide by x because x is positive
Take the limit as x tends toward zero
limx→0f (x) ≤ limx→0 Af (x+x)−A f (x)
The middle expression is the definition of derivative
Evaluate the limits
Trang 8dxAf(x)is being “squeezed” on both sides by f (x); it must be equal to f (x)
This argument can be adapted to deal with functions that are not necessarily positive and increasing on [x, x + x] The fact that [x, x + x] is a closed, bounded interval and
fis continuous means we can always find M and m, the absolute maximum and minimum values of f on [x, x + x], respectively We can set up the inequality
minimum value
of f on [x, x + x]
x ≤ Af(x + x) − Af(x) ≤
maximum value
of f on [x, x + x]
x
and proceed as above.5We obtain
lim
x→0
minimum value
of f on [x, x + x]
≤ dxd Af(x) ≤ lim
x→0
maximum value
of f on [x, x + x]
f is continuous, so as x → 0 the minimum and maximum values of f on [x + x] both approach f (x)
The conclusion is that for any continuous function f , dxdAf(x) = f (x) In words, this means that the rate of change of the area function at any point is the height of the function
f at that point Think about this geometrically for a minute; it makes sense
T h e F u n d a m e n t a l T h e o r e m o f C a l c u l u s , v e r s i o n I
If f is continuous on [a, b] and c ∈ [a, b], then the function
Af(x) =
x
c
f (t ) dt x ∈ [a, b]
is differentiable on (a, b) and d
dx Af(x) = f (x)
We can rewrite the Fundamental Theorem of Calculus without explicit reference to the area function:
If f is continuous on [a, b], thenaxf (t ) dtis differentiable on (a, b) and
d dx
x
a f (t ) dt = d(x)
In the next chapter we will see that this result is amazingly useful.
5 This inequality assumes x > 0 If x < 0, simply switch the inequalities to ≥ and switch again after dividing by x.
Trang 9P R O B L E M S F O R S E C T I O N 2 3 3
1 The graph of f (t) is given below Let F (x) =−6x f (t ) dt and G(x) =1xf (t ) dt
f
t
–6 –4 –2
2
1 3.5
4 6
(a) Where on [−6, 6] is F (x) increasing? Decreasing? Concave up? Concave down? (b) Where on [−6, 6] is G(x) increasing? Decreasing?
(c) Where on [−6, 6] does F have a local minimum?
(d) Explain why the local extrema of F and G occur at the same values of x (e) How are the graphs of F and G related?
2 Let C(q) be the cost of producing q widgets The graph below gives dCdq, the mar-ginal cost of producing widgets Economists interpret the term “marmar-ginal cost” as the additional cost of producing an additional item
dC dq
q
Assume that fixed costs, the cost when no widgets are being produced, is F Graph C(x) = F +0x
dC
dq dq
3 Below is a graph of r(t), the rate of flow of liquid in and out of a tank t = 0 corresponds
to noon
r (t) in liters/hr.
t
Trang 10Suppose you know the following.
7
−2r(t ) dt = 4
6
2 r(t ) dt = 0
7
2 r(t ) dt = 1
(a) What is the net change in the amount of liquid in the tank between 10:00 a.m and 6:00 p.m.?
(b) If there are 50 liters in the tank at 2:00 p.m., how many liters are in the tank at 7:00 p.m.?
4 A jug of cold lemonade is loaded into a cooler to be brought on a summer picnic The lemonade is 40 degrees at the start of the trip During the hour and a half drive to the park the lemonade gets steadily warmer at a rate of 1 degree every 15 minutes Upon arriving
at the park it is carried (in the cooler) to the river via a 40-minute hike During the hike the lemonade gains 1 degree every 10 minutes Once it reaches the river it is poured into cups Now that it is out of the cooler the lemonade warms at a rate proportional to the difference between the temperature of the air and that of the liquid Half an hour later it has warmed 12 degrees
Let L(t) be the temperature of the lemonade at time t, where t is the number of minutes into the trip
(a) Sketch dLdt versus time Label important points
(b) Sketch L(t) versus time Label important points Be sure your pictures of L and dL/dtare consistent