## Calculus Of One Real Variable – By Pheng Kim Ving Chapter 12: Applications Of The Integral – Section 12.10: Net Change

12.10
Net Change

 1. Net Change

Refer to Fig. 1.1. When x increases from a to b, y = f (x) changes from f (a) to f (b). The quantity y may go from f (a)
straight to f (b), as in Fig. 1.1, where it increases from f (a) straight to f (b), or it may reverse directions a number of times as
it goes from f (a) to f (b), as in Fig. 1.2, where it increases from f (a) to f (b), keeps increasing from f (b) to f (c), then
reverses direction and decreases from f (c) to f (b). In any case, of course the amount f (b) – f (a) depends only on f (a) and
f (b) and not on anything else, as clearly there are only f (a) and f (b) in the expression; it doesn't depend on anything that # Fig. 1.1

Net change of the function is f (b) – f (a), which is positive. # Fig. 1.2

Net change of the function is f (b) – f (a), which is positive.

may happen to f (x) as it journeys from f (a) to f (b). The difference f (b) – f (a) is thus the net change of y = f (x) as x
increases from a to b. It can be thought of as the displacement of y, as opposed to the distance travelled by y; see
Section 12.5.

### Definition 1.1 – Net Change

 Let f (x) be a function continuous on an interval containing a and b with a < b. The net change of f (x) as x increases from a to b is the difference f (b) – f (a).

In Figs. 1.1 and 1.2, f (a) < f (b), and the net change is positive. In Fig. 1.3, f (a) > f (b), and the net change is negative. In
Fig. 1.4, f (a) = f (b), and the net change is zero. Note that in Fig. 1.4, when x increases from a to b, although the net change
of f (x) is zero, f (x) doesn't stay constant, it increases then decreases then increases. # Fig. 1.3

Negative Net Change On [a, b]. # Fig. 1.4

Zero Net Change On [a, b].

 2. The Net-Change Theorem

### The Net Change Equals The Integral Of The Rate Of Change

Consider a linear function y = f (x) = mx. Of course the derivative or rate of change of f (x) is f '(x) = m, a constant. Recall
that the rate of change of a function is the change of the function per 1-unit change of the variable. When x increases from a
to b where a < b, f (x) changes from f (a) to f (b) at the constant rate of m y-units per x-unit, so the net change of f (x) is:

f (b) – f (a) = ((change of f (x)) y-units/x-unit) . (ba) x-units = m(ba) y-units = f '(x)(ba) y-units.

(Example: if speed is constant, then net change in position = displacement = distance = speed . time elapsed.)

The function f (x) = mx is shown graphically in Fig. 2.1, where m is taken as an example to be positive and about 2.5. In # Fig. 2.1

f (b) – f (a) = m(ba) = f '(x)(ba). # Fig. 2.2  # Fig. 2.3 Fig. 2.2, we sketch the graph of the derivative function f '(x) = m on [a, b]. We have: The net change of f (x) over [a, b] equals the integral of the derivative f '(x) of f (x) over [a, b].

### Using The Fundamental Theorem Of Calculus

The equation: Part 2 of the Fundamental Theorem Of Calculus, as presented in Section 9.4 Theorem 2.1, asserts that if f (x) is a continuous
function on [a, b] and F(x) is any antiderivative of f (x), then: Eq. [2.3] says that the net change of a function when the variable changes from a to b equals the definite integral of the
derivative of the function over the interval [a, b]. As for the derivative f '(x), in this context it's appropriate to interpret it as the
rate of change of y = f (x) with respect to x. So Eq. [2.3] declares that the net change of a function equals the integral of the
rate of change of the function. Recall that the rate of change of a function is its change that corresponds to the change of 1 unit
of the variable.

### Sum Of Small Changes

Refer to Fig. 2.4. Let x be an arbitrary point in [a, b] and T the tangent to the graph of f at x. Let y = f (x), so that dy =
f '(x) dx. We have: Thus we can intuitively treat the net change of f as the sum of infinitesimally small changes of f. Because the small changes
dy 's in the sum are infinitesimally small, the sum is the definite integral. In Fig. 2.4, of course the “infinitesimally small”
changes dx 's and dy 's are magnified, so that we can see them. This treatment is useful in the memorization of Eq. [2.3]: net
change f (b) – f (a) equals sum of small changes dy 's, and dy equals f '(x) dx. # Fig. 2.4

Net Change = Sum Of Infinitesimally Small Changes.

### Small Changes Of The Function Are Signed Quantities

As x increases by an amount of dx, we have dx > 0; if y increases, then dy > 0; if y decreases, then dy < 0; if y doesn't
change, then dy = 0. Of course the small changes dy 's of the function y = f (x) are signed quantities. Fig. 2.5 displays a case
where there are times when dy > 0 and times when dy < 0. As a matter of fact, the positive dy 's and the negative dy 's on
the same y-interval cancel each other out (the sum of the positive ones equals the absolute value of the sum of the negative
ones). # Fig. 2.5

Small Changes dy 's Are Signed Quantities.

### The Net-Change Theorem

Eq. [2.3] is presented as Theorem 2.1 below and re-produced as Eq. [2.4]. In this context, “integral” means “definite integral”.

Theorem 2.1 – The Net-Change Theorem ### Remark 2.1

If the expression of f (x) is known, we can calculate f (b) - f (a) directly without using the net-change theorem. The net-change
theorem is tremendously useful when f (x) isn't known and f '(x) is hard to integrate to find f (x), or impossible to integrate in
terms of elementary functions, or when f '(x) has no explicit expression but is obtained as recorded data from experiment or
observation. In these cases, we of course have to resort to approximate numerical integration, and so we have to settle with
the approximate value of the net change. For the case of recorded data, see problems & solutions 7 and 9.

 3. Applications

### Distance And Displacement

Suppose an object moves along a straight line with position function s(t), where t is time. Then its velocity is v(t) = s'(t). So
the net change of its position during the time period from t = t1 to t = t2 is: This net change of position is the displacement of the object. Distance and displacement are investigated in Section 12.5. To
find distance, we integrate |v(t)|.

Note that |s(t2) – s(t1)| isn't the distance. It's the absolute value of the displacement. The distance is: The distance is greater than or equal to the absolute value of the displacement.

### Example 3.1

A particle moves along a line in such a way that its velocity at time t is v(t) = t2 + t – 6, where t is in sec and v in m/sec.
a. Find the displacement of the particle from t = 1 to t = 5.
b. Determine the distance travelled by the particle during this time period.

Solution
a. The displacement is: The  particle has displaced about 29.33 m to the right.

b. v(t) = t2 + t – 6 = (t + 3)(t – 2) = 0 when t = –3 or 2, the distance travelled is: The particle has travelled a distance of about 33.66 m.
EOS

### Volumes Of Liquids

Water is flowing into a reservoir. Let V (t) be the volume of the water in the reservoir at time t. Then the derivative V '(t), the
rate of increase of V (t), is the rate at which water flows into the reservoir, and the increase of the amount of water from time
t1 to time t2 > t1 is: ### Mass Of A Rod Note that if we apply this formula to the homogeneous rod then: the same as obtained earlier, as expected.

### Populations ### Costs Of Productions

In economics, the marginal cost of production is defined to be the cost of producing one more unit, that is, the rate of increase
of cost, cost being a function of the number of units produced. Let C(x) be the cost of producing x units of a commodity. Then
the derivative C '(x) is the marginal cost at production of x units, and the increase of cost when production increases from x1
units to x2 units is: Note that, for example, the cost of increasing production from 1,000 units to 1,001 units may be different from the cost of
increasing production from 100 units to 101 units. In general, the marginal cost is a non-constant function of the number of
units produced.

### Units

In: ### Example 3.2 Solution EOS

 Problems & Solutions Solution

The absolute value of: is the volume of water in litres that has leaked from the tank 60 minutes or 1 hour after water starts to leak. Because there's
no water flowing into the tank during this time period, it's the decrease of the volume of water in the tank during this time
period.  ### Solution represents the number of bacteria 7 days later.  ### Solution

Let h(t) be the height of the child when he/she is t years old. Then: represents the growth of height of the child in cm between the ages of 5 and 10 years old.  Solution  5. Suppose the unit of x is metre and the unit of f (x) is kilogram per metre. What's the unit of:

a. f '(x)? Solution

a. The unit of f '(x) = df (x)/dx is (kilogram per metre) per metre (= (kg/m)/m), or kilogram per metre squared (= kg/m2).  6. A particle moves along a line in such a way that its velocity at time t is v(t) = t2t – 12, where t is in sec and v in m/sec.
a. Find the displacement of the particle from t = 1 to t = 10.
b. Determine the distance travelled by the particle during this time period.

Solution

a. The displacement is: The particle has travelled a distance of 220.50 m.  7. The speed of a car was read from its speedometer at 10-second intervals and recorded in the table below. Use the
Rectangular Rule to determine the approximate distance travelled by the car 120 seconds after it started to move. ## Solution

Let's convert km/h to m/s. As 1 km/h = (1,000 m)/(3,600 s) = (5/18) m/s, the given table yields the following table. Let v (t) be a velocity function of time t on [0, 120] that takes on the values in the above table. We use the 12 sub-intervals [0,
10], [10, 20], ..., [110, 120]. For the value of v (t) at the midpoint of a sub-interval, we utilize the average of the values of v (t)
at the endpoints of that sub-interval, as shown in the table below. The length of each sub-interval is 10 s. The distance travelled by the car 120 seconds after it started to move is: ## Note

See Remark 2.1.  ## Note

As time goes, the rate at which water flows out decreases. The amount of water flowing out per minute starts at 300 litres per
minute and decreases, till it reaches 0 litre per minute at 50 minutes later.

Solution

Let A(t) be the amount of water that has flowed out during the first t minutes. The amount of water that has flowed out during
the first 20 minutes is: ## Note

Another approach for the solution is as follows. During the first 20 minutes, the volume of water in the tank has decreased by 4,800 litres, so the amount of water that has
flowed out is 4,800 litres. 9. Water flows into and out of a storage tank. The rate of change r(t) of the volume of water in the tank, in cubic metres per

day, is measured and graphed below. The amount of water in the tank when the measurement starts is 25 m3. Use the
Rectangular Rule to estimate the amount of water in the tank 5 days later. ### Solution

Let V (t) be the volume of water in the tank in cubic metres at time t. Use the midpoints of the sub-intervals [0, 1], [1, 2], [2,
3], [3, 4], and [4, 5], which are 0.5, 1.5, 2.5, 3.5, and 4.5. The length of each sub-interval is 1. The net change of the amount
of water in the tank 5 days later is: ## Note

See Remark 2.1.  Solution

Let m(x) be the mass of the rod from the selected end of the rod to a point that's x metres from that end. The total mass of
the rod is:  11. The marginal cost of producing L metres of a fabric is C '(L) = 3 – 0.01L + 0.000007L2 dollars per metre. Determine the
increase of cost if the production level is increased from 1,000 metres to 3,000 metres.

Solution

The required increase of cost is: 