/IndefiniteIntegral(II)

THE INDEfinite INTEGRAL AN INTRODUCTION TO THE INDEFINITE INTEGRAL (PART II)

EXAMPLE 25 Evaluate the integral

∫

2

(x

2

0

– | x – 1|) dx

Solution 2

The integrand is f(x) = x

– | x – 1|, whose graph is displayed on the following page

From the figure, we see that the area we are looking for is entirely on the positive side of the x-axis, unlike the functions in examples 22, 23, and 24. Also, from the definition of the absolute function,

f(x)

=

x – |x – 1| 2

=

x2 – (x – 1) x2 – [–(x – 1)]

if x ≥ 0 if x < 0

which equals

f(x)

=

x – |x – 1| 2

=

x2 – x + 1

if x ≥ 0

x2 + x – 1

if x < 0

Thus, the graph of the integrand equals the graph of x2 – x + 1 2

x

+ x–1

as illustrated below:

drawn on the interval [1, ∞] drawn on the interval [–∞, 1]

So, based on the area we are looking for, we find that

∫

2 0

(x

2

– | x – 1|) dx

∫

=

1 0

(x

2

+ x – 1) dx

+

∫

2 1

(x

2

– x + 1) dx

Using FTC 2 to evaluate the integrals on the RHS, we have

∫

2 0

(x

2

– | x – 1|) dx

=

– 1/6

=

5/3

+

11/6

EXAMPLE 26 There are 6 questions below, each solution illustrates the Total Change Theorem. Study each problem and solution carefully.

EXAMPLE 26.1 If w'(t) is the rate of growth of a child in pounds per year, What does

∫

10 5

w'(t) dt

represent?

Solution Observe the unit given for w'(t): Pounds per year. Since w'(t) is the derivative of w(t), it follows that w(t) represents the “weight function”, i.e. The weight (in pounds) attained by a child at the age of t years. So, from the definition of the Total Change Theorem,

∫

10 5

w'(t) dt

=

w(10)

–

w(5)

Since w(t) is defined as the “weight function”, clearly, w(10) – w(5) represents the weight difference in 5 years. In terms of rate of change, we say it represents the change in the child's

weight with respect to his/her age. Again, we can interpret it as the difference in the child's weight between ages 5 and 10.

EXAMPLE 26.2 The current in a wire is defined as the derivative of the charge: I(t) = Q'(t). What does represent?

∫

b a

I(t) dt

Solution When electric charges move through a wire, current is said to exist. In this case, let's consider one section of a wire.

Suppose ∆Q represents the net charge (i.e., Q2 – Q1, measured in Coulombs) that moves through a section of wire in a given time period (∆t). Then, the average current passing through is given by ∆Q / ∆t. However, the current passing through at a particular time is given by the limit of ∆Q / ∆t over increasingly smaller time intervals. Thus, we define current as the rate at which charge flows through a surface, typically measured in Coulombs/Second or Amperes.

So, let's go back to the question: We know that I(t) = Q'(t) This means

∫

b a

I(t) dt

∫

=

b a

=

Q' (t) dt

Q(b)

–

Q(a)

From the definitions given above, Q(b) – Q(a) represents the total quantity of charge (in Coulombs) that passed through the wire within the time period [a,b] (in seconds).

EXAMPLE 26.3 If oil leaks at a rate of r(t) gallons per minute at time t, what does

∫

120 0

r(t) dt

represent?

Solution Let's solve this problem using the “rate of change” concept: From the question, the function y = r(t) represents the rate at which water leaks from a tank at a time t, which is measured in Gallons per minute. So based on the definition of rate of change, the function can be expressed as

r(t) where y

=

C(t2) – C(t1) t2 – t1

=

C (t) represents the quantity of water (in gallons) leaking from the tank at a time t (in minutes).

Therefore,in this case,

∫

120 0

r(t) dt

=

C(120) – C(0) 120 – 0

From the equation above, the integral can be interpreted as the quantity of water that leaked from the tank within

120 minutes (i.e., 2 hours).

EXAMPLE 26.4 A honeybee population starts with 100 bees and increases at a rate of n'(t) bees per week. What does

100 + represent?

∫

15 0

n'(t) dt

Solution The function n'(t) is given in bees per week, which suggests that it is a derivative of n(t), which would represent the number of bees in the population in a particular week. So, from the Total change Theorem,

∫

15 0

n'(t) dt

=

n(15)

–

n(0)

Where n(15) – n(0) represents the increase in the number of bees in the population after the 15 th week (or we could say during the first 15 weeks). Don't forget that the population started with 100 bees, which means that the integral

100 +

∫

15 0

n'(t) dt

represents the TOTAL number of bees in that colony.

EXAMPLE 26.5 The Marginal Revenue Function R'(x) is defined as the derivative of the Revenue Function R(x), where x is the number of units sold. What does

∫

5000 1000

R'(x) dx

represent?

Solution Let's briefly discuss the applications of Calculus to Marketing: If a company sells x units of an item, and P(x) is the price per unit, then the TOTAL REVENUE derived is given by the product of the number of units sold and the price per unit:

R(x) = xP(x) Where R(x) is the SALES FUNCTION or REVENUE FUNCTION, and P(x) is called the DEMAND FUNCTION or the PRICE FUNCTION. We therefore define R(x) as the MARGINAL REVENUE FUNCTION; that is, the rate of change of revenue with respect to the number of units sold.

Now, let's go back to the question: From the Total Change Theorem,

∫

5000 1000

R'(x) dx

=

R(5000)

–

R(1000)

The expression R(5000) – R(1000) represents the net increase in revenue when the production level is raised from an initial value of 1000 units to 5000 units.

Note:

The PROFIT FUNCTION P(x) is given by P(x)

=

R(x)

–

C(x)

Thus, P'(x) represents the MARGINAL PROFIT FUNCTION. To maximize profit, Marginal Profit must equal ZERO. Also, if R'(x) – C'(x) = 0 then

R'(x)

=

C'(x)

Which means that profit is also maximized when marginal cost equals marginal revenue.

EXAMPLE 27 The velocity function (in meters/sec) is given for a particle moving along a line. Find (a) the displacement and (b) the distance covered by the particle during the given time interval. (For examples 27.1 and 27.2). ●

v(t)

=

3t – 5,

0 ≤ t ≤ 3

●

v(t)

=

t2 – 2t – 8 ,

1 ≤ t ≤ 6

The acceleration function (m/s2)and the initial velocity are given for a particle moving along a line. Find (a) the velocity at time t and the distance traveled during the time interval. (For examples 27.3 and 27.4). ●

a(t)

=

t + 4,

●

a(t)

= 2t + 3,

v(0) = 5,

0 ≤ t ≤ 10

v(0) = -5,

0 ≤ t ≤ 3

Before we solve the four problems, let's take a quick look at how the Total Change Theorem is applied towards motion: In Differential Calculus, we define ACCELERATION as the rate of change of VELOCITY with respect to time.

Thus, if a(t) represents acceleration and the velocity is represented by v(t), then, by the Total Change Theorem, we have

∫

t2

v'(t) dt

τ1

=

v(t2)

–

From the definition given above, we say that a(t)

∫

t2 τ1

a(t) dt

=

v(t2)

–

1

v(t1)

= '(t). Thus, we can re-write equation 1 as

v(t1)

2

The Velocity v(t) of an object is defined as the rate of change of DISPLACEMENT s(t) with respect to time. In other words, v(t)

OR

∫

τ1

∫

τ1

t2

t2

=

s'(t). Thus, the Total Change Theorem gives

s'(t) dt

=

s(t2)

–

s(t1)

3

v(t) dt

=

s(t2)

–

s(t1)

4

The Displacement function s(t) is also called the POSITION FUNCTION, which is defined as the change of position of a particle.

When we talk of linear motion, there are two possible directions a particle can move: backward and forward.

Thus, when calculating the distance traveled by the particle during a time period, we consider two intervals: I.

when v(t) ≤ 0,(which means the particle is moving towards the left), and

II.

when v(t) ≥ 0,(which means the particle is moving towards the right).

Bringing these two intervals together results in |v(t)|. Therefore, to

calculate the distance covered by the

particle, we evaluate the integral

∫

t2 τ1

| v(t)| dt

The figure below is a graphical illustration of how displacement and distance traveled is calculated:

EXAMPLE 27.1 v(t)

=

0 ≤ t ≤ 3

3t – 5,

To obtain the displacement of the particle, we integrate the velocity function v(t)

= 3t – 5 with respect to the given

time interval [0,3]. We express this as

∫

3 0

v(t) dt

=

v(t) dt

=

∫

3

(3t – 5) dt

0

Using FTC 2, we have

∫

3 0

]

1.5t2 – 5t

3 0

which equals –1.5. (See the graph of the velocity function below). This implies that the object moved 1.5 meters to the left (that's that significance of the minus sign). Now, to calculate the distance covered, we compute the value of

∫

3 0

|3t – 5|

dt

From the graph of v(t), we find that v(t) ≤ 0 on the interval [0, 5/3] and v(t) ≥ 0 on [5/3, 3]. Carefully compare the graphs of v(t) and |v(t)| below. Therefore,

∫

3 0

|3t – 5|

dt

= =

∫ ∫

5/3 0 5/3 0

–[v(t)]

–(3t – 5)

dt dt

+

∫

+

∫

3 5/3 3 5/3

[v(t)]

(3t – 5)

dt dt

Using FTC 2, we have

∫

3 0

|3t – 5|

dt

=

25/6

=

41/6m

+

8/3 ≈

6.833 m

Summary: given the velocity function v(t) = 3t – 5, the object moved or in other words, “displaced itself” 1.5m to the left in the first three seconds, and at the same time, covered a total distance of approximately 6.883m.

EXAMPLE 27.2 v(t)

1 ≤ t ≤ 6

t2 – 2t – 8 ,

=

The displacement of the particle is given by

∫

6 1

(t

2

– 2t – 8) dt

which when evaluated, using FTC 2, gives roughly –2.667m. This implies that the particle moved approximately 2.667m to the left. From the graph of v(t) (on the next page), we see that we find that v(t) ≤ 0 on the interval [1,4] and v(t) ≥ 0 on [4,6]. Therefore, the distance covered by the particle is given by

∫

6 1

|t

2

– 2t – 8| dt

=

∫

4 1

–(t

2

– 2t – 8) dt

+

∫

6 4

(t

2

– 2t – 8) dt

Using FTC 2, we have

∫

6 1

|t

2

– 2t – 8| dt

= =

18

+

(44/3)

98/3m

≈

32.667 m

Thus, the particle covered a total distance of approximately 32.667m in 5 seconds.

EXAMPLE 27.3 a(t) We know that

a(t)

=

=

t + 4,

v'(t),

0 ≤ t ≤ 10

v(0) = 5,

which means that, to find the velocity function of the particle, we integrate the

acceleration function a(t), i.e.

v(t)

=

∫

a(t) dt

=

∫

( t + 4)

dt

Which equals

v(t)

=

½ t2 + 4t + C

(from the general antiderivative formula)

To find the value of C, we use the fact that the initial velocity was 5m/s (i.e.,v(0) = 5). Putting t = 0 gives

½ (0)2 + 4(0) + C =

5

When we solve for C, we find that C = 5. This means that the actual velocity function for this particle is

v(t)

=

½ t2 + 4t + 5

Now that we have a velocity function, we can easily obtain the distance covered by the particle during the time period [0,10], which is given by

∫

10

|½ t

0

2

+ 4t + 5| dt

From the graph of v(t) below, we see that v(t) ≥ 0 on the entire interval [0,10]. Therefore,

∫

10 0

|½ t

2

+ 4t + 5| dt

=

∫

10 0

(½ t

2

+ 4t + 5) dt

which is roughly equal to 416.6667m.

So, the distance covered by the particle in 10 seconds is approximately 416.6667m.

EXAMPLE 27.4 a(t)

= 2t + 3,

We integrate a(t) to obtain v(t):

v(0) = -5,

0 ≤ t ≤ 3

v(t)

∫

=

∫

=

a(t) dt

(2t + 3)

dt

Using the general antiderivative formula, we have v(t) = t2 + 3t + C. To find the value of C, we apply the given fact that v(0) = -4,which implies that

t2 + 3t + C

= –4

Putting t = 0 gives

(0)2 + 3(0) + C

=

–4

So, C = – 4. The velocity function of the particle is therefore given by

v(t)

=

t2 + 3t – 4

From the graph of v(t), v(t) ≤ 0 on the interval [0,1] and v(t) ≥ 0 on [1,3]. Therefore, the distance covered by the particle is given by

∫

3 0

|t

2

+ 3t – 4| dt

=

∫

1 0

– (t

2

+ 3t – 4) dt

+

∫

3 1

(t

2

+ 3t – 4) dt

which equals

∫

3 0

|t

2

+ 3t – 4| dt =

(13/6)

+

(38/3)

≈

14.833 m

Therefore, the particle covered a distance of approximately 14.83m in 3 seconds.

EXAMPLE 28 The linear density of a rod of length 4m is given by ρ(x) = 9 + 2√x measured in kilograms per meter, where x is measured in meters from one end of the rod.

Solution

First, understand that the linear density ρ(x) of a given mass is calculated as the derivative of its mass m(x) with respect to its length (which is measured from the end to a particular point x. In other words, ρ(x) = m'(x) Therefore, from the Total Change Theorem,

∫ OR

b a

∫

b a

ρ(x) dx

=

m(b)

–

m(a)

m'(x) dx

=

m(b)

–

m(a)

In this case the linear density function is given as ρ(x) = 9 + 2√x

(in kg/m). The length of the rod (measured from

one end to the other) is 4m, and we want to find the mass of the rod. Again, since the length of the rod is measured from end to end, this suggests the interval [0,4], and thus,

∫

This gives

∫

4

ρ(x) dx

0

4 0

=

ρ(x) dx

mass of rod

9x – 4/3 (√x3)

=

From the equation above, the mass of the rod is given by m(x)

]

4 0

= 9x – 4/3 (√x3).

(where m(x) is obtained by computing the antiderivative of ρ (x)). So now, to find the mass of the rod, we have to evaluate

9x – 4/3 (√x3)

]

4 0

which equals 46.6667m. Thus, the mass of the rod is approximately 46.6667m.

EXAMPLE 29 An animal population is increasing at a rate of 200 + 5t per year.(where t is measured in years). By how much does the animal population increase between the 4th and 10th years?

Solution Let's say that the population function is given by p(t). Then p'(t) would represent the rate at which the animal population is increasing with respect to time. This implies that p'(t)

=

200 + 5t.

The question is asking us to find the increase in the population between the 4 th and 10th years, which suggests an interval [4,10]. To solve this problem, we need and equation that appropriately represents the population function. To get that, we find the antiderivative of p'(t).In other words, we evaluate

∫ which equals p(t)

=

p'(t) dt

∫

(200 + 5t)

dt

2

= 200t + 2.5t .

To find the increase in the animal population, we evaluate p(10) – p(4), which equals

[200(10) + 2.5(10)2] =

[2000 + 250]

=

1410

–

–

[200(4) + 2.5(4)2]

[800 + 40]

=

2250 – 840

This means that, between the 4th and 10th years, the animal population would have increased by 1410. What I've done

above is simplify the entire process. Regardless, my point is, given the information, all we need to do to solve the problem is to evaluate the integral

∫

10 4

(200 + 5t)

dt

which equals 1410.

EXAMPLE 30 The marginal cost of manufacturing x yards of a certain fabric is

C'(x)

=

3 – 0.01x + 0.000006x2

(in dollars per yard). Find the increase in the cost of production is the production level is raised from 2000 yards to 4000 yards.

Solution Whenever x units of a commodity are produced by a company, a certain cost is always incurred. This cost is called the COST FUNCTION, which in this case, we'll denote by C(x). Obviously, ∆x will represent the net increase in the number of units produced, and ∆C will represent the extra cost incurred. Thus, ∆C / ∆x = C'(x) represents the rate of change of cost with respect to the number of units/items produced. This rate is called MARGINAL COST. In most cases, the cost function is often represented by a polynomial: C(x) = a + bx + cx2 + dx3 where a represents rent, heat, maintenance and other costs, while the other terms represent labor costs, cost of raw materials, and others.

Let's return to the question: The marginal cost function is given as C'(x)

=

3 – 0.01x + 0.000006x2.

To find the increase in production cost, we need the cost function, which is given by evaluating the antiderivative of the marginal cost function, i.e.:

∫ which results in C(x)

C'(x) dx

= 3x – 0.005x2 + 0.0000002x3.

Now that we have a cost function, we can now find the increase in production cost when the production level is raised from an initial value of 2000 yards to 4000 yards. This is given by

C(4000) – C(2000) which equals

[3(4000) – 0.005(4000)2 + 0.0000002(4000)3] – [3(2000) – 0.005(2000)2 + 0.0000002(2000)3] =

[12000 – 80000 + 128000]

=

58,000

–

[6000 – 20000 + 16000]

=

60,000 – 2,000

Therefore, the company has to incur a cost of $58,000 when it increases production from 2000 to 4000 yards!

Again, I have simplified the process. To solve the problem, all we had to do was to apply the Total Change Theorem and evaluate

∫

4000 2000

C'(x) dx

which equals $58,000.

EXAMPLE 31 On May 7, 1992, the space shuttle Endeavour was launched on mission STS-49, the purpose of which was to install a new perigee kick motor in an Intelsat communications satellite. The table (actual figures given) below gives the velocity data for the shuttle between liftoff and the jettisoning of the solid rocket boosters.

Event Launch Begin roll maneuver End roll maneuver Throttle to 89% Throttle to 67% Throttle to 104% Maximum dynamic pressure Solid rocket booster separation

Time (s) 0 10 15 20 32 59 62 125

Velocity (ft/s) 0 185 319 447 742 1325 1445 4151

(a)

Use a graphing calculator or computer to model these data by a third-degree polynomial.

(b)

Use the model in part (a) to estimate the height reached by the Endeavour ,125 seconds after liftoff.

Solution Using the cubic regression function on my graphing calculator, I obtained

v(t)

=

0.0014613t3 – 0.1155339t2 + 24.9816919t – 21.268723

where v(t) represents the velocity of the shuttle t seconds after liftoff. You can approximate the terms in the cubic polynomial, but in this case, we'll leave it as it is, so that the calculation we are about to perform will be considerably

accurate. The task here is to estimate the height reached by the rocket 125 seconds after liftoff. That piece information indicates the interval we'll be working on: [0,125].

From the Total Change Theorem,

∫

125 0

|v(t)| dt

=

height attained by rocket in 125 seconds

From the graph of v(t), we see that we see that v(t) ≥ 0 on the entire interval [0,125]. This means

∫

125 0

|v(t)| dt

=

]

125

0.000365325t4 – 0.0385113t3 + 12.49045958t2 – 21.268723t

0

Since one endpoint is zero, we'll simply evaluate the antiderivative at the other endpoint (at t = 125):

=

0.000365325(125)4 – 0.0385113(125)3 + 12.49045958(125)2 – 21.268723(125)

=

89190.67383 – 75217.38281 + 195169.468 – 2658.59075

=

206,484.1686 ft.

Thus, 125 seconds after liftoff, the shuttle reached a height of approximately 206,484.2 feet OR 39.11 miles.

EXAMPLE 32 The velocity of a car was read from its speedometer at ten-second intervals and recorded in the table. Use the Midpoint Rule to estimate the distance traveled by the car.

t (s)

v (mi/h)

t (s)

v (mi/h)

0 10 20 30 40 50

0 38 52 58 55 51

60 70 80 90 100

56 53 50 47 45

Solution On the table, observe that the units do not match; the time is given in seconds, and the velocity is given in miles per hour. So, one possible alternative is to convert the given velocities to feet/sec. Using the fact that 1mile/hour

=

5280/3600 feet/sec,

we have the following in table (the velocities are expressed as fractions, for the sake of accuracy):

t (s) 0 10 20 30 40 50

v (ft/s) 0 836/15 1144/15 1276/15 242/3 374/5

t (s) 60 70 80 90 100

v (ft/s) 1232/15 1166/15 220/3 1034/15 66

These values are plotted and we have a graph (on the next page). From the graph, it's clear that we have 10 subintervals: [0,10], [10,20],[20,30],[30,40],[40,50],[50,60],[60,70],[70,80],[80,90], and [90,100].Since we don't have an explicit formula, we'll have to estimate the distance traveled by using the midpoint of each interval, which in this case are: 5, 15, 25, 35, 45, 55, 65, 75, 85, and 95. The second graph (also on the next page) shows the area partitioned into 10 subintervals.

The respective coordinates of the midpoints of the subintervals are: [5, 418/15], [15, 66], [25, 242/3], [35, 1243/15], [45, 1166/15], [55, 1177/15], [65, 1199/15], [75, 1133/15], [85, 1067/15], AND [95, 1012/15].

From the Total Change Theorem, the distance covered by the car in 100 seconds is given by the definite integral

∫

100 0

|v(t)| dt

Using the Midpoint Rule, we have a = 0, b = 100, n = 10, which means ∆t = 10.

∫

100 0

|v(t)| dt

=

[v(5) + v(15) + v(25) + v(35) +.......... ........+ v(85) + v(95)]∆t

=

[(418/15) + (66) + (242/2) + (1242/15) + (1166/15) + (1177/15) + (1199/15) + (1133/15) + (1067/15) + (1012/15)]10

=

[2123/3]10

≈

7076.67 ft.

If we want to change the unit of the result to miles, we have approximately 1.34 miles. Therefore, the distance covered by the car in 100 seconds is approximately 1.34 miles.

EXAMPLE 33 The inflation rate is defined as the derivative of the Consumer Price Index (CPI), which is published by the U.S. Bureau of Labor Statistics and measures prices of items in a “representative market basket” of typical urban consumers. The table (actual figures given) gives the inflation rate in the United States from 1981 to 1997. Write the total percentage increase in the CPI from 1981 to 1997 as a definite integral. Then use the Midpoint Rule to estimate it.

t

r(t)

t

r(t)

1981 1982 1983 1984 1985 1986 1987 1988 1989

10.3 6.2 3.2 4.3 3.6 1.9 3.6 4.1 4.8

1990 1991 1992 1993 1994 1995 1996 1997

5.4 4.2 3.0 3.0 2.6 2.8 2.9 2.3

Solution Plotting the points in the table above produces this:

Let's assume that the CPI function is given by C(t), and the Inflation function is given by r(t). So, from the question,

r(t)

=

C'(t)

Therefore, using the Total Change Theorem, we write the total percentage increase in the CPI from 1981 to 1997 as

∫

OR

∫

1997 1981

1997 1981

C'(t) dt

r(t) dt

=

=

C(1997)

C(1997)

–

–

R(1981)

R(1981)

So, a = 1981, b = 1997, and from the graph below, n = 16. This means ∆t = 1.

Thus,

∫

1997 1981

r(t) dt

=

[r(1981.5)

=

[8.25

+ r(1982.5) + r(1983.5) + ............. + r(1996.5)]∆t

+ 4.70 + 3.75 + 3.95 + 2.75 + 2.75 + 3.85 + 4.45 +

5.10 + 4.80 + 3.60 + 3.00 + 2.80 + 2.70 + 2,85 + 2.60 ]1

=

61.9 %

Therefore, the total percentage increase in the CPI from 1981 to 1997 is approximately 61.9%.

EXAMPLE 34 The area of the region that lies to the right of the y-axis and to the left of the parabola x = 2y – y2 (the shaded region in the figure below) is given by the integral

∫

2 0

( 2y – y ) 2

dy

(You could turn you head clockwise and think of the region as lying below the curve x = 2y – y2 from y = 0 to y = 2).

Find the area of the region.

y

x = 2y – y2

2

0 1

x

Solution Before we tackle this problem, let's look at some important properties of the parabola: Generally, a parabola is defined by the second-degree equation y = ax2 + bx + c. The four figures below show the most general positions that a parabola can take:

y = ax2

x = ay2

y = ax2

(a>0)

x = ay2

(a>0)

(a<0)

(a<0)

Properties of a Parabola : A parabola y = ax2 opens upward if a>0 A parabola y = ax2 opens upward if a<0 A parabola x = ay2 opens to the right if a>0 A parabola x = ay2 opens to the left if a<0 Explanation: Let's assume that we interchange x and y in the equation y = ax2. Then the result is x = ay2, which also happens to be a parabola. However, if a<0, the parabola x = ay2 opens to the left, and if a>0, it opens to the right. See the figures above.

Let's go back to the question. At this point, there's no need to be confused about the integral

∫

2 0

( 2y – y ) 2

dy

It's just the same as

∫

2

( 2x – x ) 2

0

dx

except that x and y have been interchanged. We therefore compute the integral the same way we compute regular integrals. So, using FTC 2,

∫

2 0

( 2y – y ) 2

dy

=

Evaluating the antiderivative at points 0 and 2 gives:

∫

2 0

( 2y – y ) 2

dy

=

4 – (8/3) – 0

y2 – (y3/3)

]

2 0

∫

2 0

( 2y – y ) 2

dy

=

4 /3

EXAMPLE 35 Economists use a cumulative distribution called a Lorenz curve to describe the distribution of income between households in a given country. Typically, a Lorenz curve is defined on [0,1], passes through (0,0) and (1,1), and is continuous, increasing, and concave upward.

The points on this curve are determined by ranking all households by income and then computing the percentage of households whose income is less than or equal to a given percentage of the total income of the country. For example, the point (a/100, b/100) is on the Lorenz curve if the bottom a% of the households receive less than or equal to b% of the total income. Absolute Equality of income distribution would occur if the bottom a% of the households receive a % of the total income, in which the Lorenz curve would be the line y = x. The area between the Lorenz curve and the line y = x (the orange area) measures how much income distribution differs from absolute equality. The Coefficient of Inequality is the ratio of the area between the Lorenz curve and y = x to the area under y = x. (a).

Show that the coefficient of inequality is twice the area between the Lorenz curve, and the line y = x, that is, show that COEFFICIENT OF INEQUALITY

(b).

=

2

∫

1 0

(x

– L(x)) dx

The income distribution for a certain country, is represented by the Lorenz curve defined by the equation

L(x)

=

5x2/12 + 7x/12

What is the percentage of total income received by the bottom 50% of the households? Find the coefficient of inequality.

Solution (a).

By definition,

COEFFICIENT OF INEQUALITY

=

Area between y = L(x) and y = x Area under y = x

which can be interpreted as

∫

1 0

(x

∫

You can see that

– L(x)) dx 1 0

x

dx

COEFFICIENT OF INEQUALITY

i the graph above appropriately illustrates equation 1. From the graph,

=

ORANGE AREA GREEN AREA + ORANGE AREA

Bear in mind that, at this point, we do not have an explicit equation for the Lorenz curve (i.e, L(x)). The question is also saying that the coefficient of inequality is twice the area between y = L(x) and y = x):

2

∫

1 0

(x

– L(x)) dx

ii

Thus, the task is to prove that (i) and (ii) are equal. In other words, we are to show that

∫

1 0

(x

∫

– L(x)) dx 1 0

x

dx

=

2

∫

1 0

(x

– L(x)) dx

So, let's evaluate both sides and see what we get:

∫

1 0

x

–

dx

∫ =

1

x

0

∫

–

0.5

∫

1

1

L(x) dx

0

=

dx

L(x) dx

0

2

=

∫

1 0

x dx

–

2

–

2

∫

1

–

2

∫

0.5

–

2(0.5)

0.5

1 0

∫

1 0

L(x) dx

L(x) dx

We multiply both sides by 0.5 to get

0.5

∫

–

0.5

1

L(x) dx

0

=

0.5

=

0.5

–

∫

1 0

0.5

=

L(x) dx

∫

1 0

1 0

L(x) dx

L(x) dx

Both sides are equal, which proves that the coefficient of inequality equals

2

∫

1 0

(x

– L(x)) dx

It also shows that the term “coefficient of inequality” can be graphically can be defined in two ways: It is the ratio of the area between the Lorenz curve and the straight line, and It is twice the area between the Lorenz curve and the Straight line y = x.

(b).

Here, the Lorenz curve is defined by the equation

L(x)

=

5x2/12 + 7x/12

The question is asking us to find out what percentage of income is received by the bottom 50% of the households. Note that on the “Lorenz scale”, 50% = 0.5. Therefore,

L(0.5)

=

[5(0.5)2/12]

=

[1.25/12]

=

0.105

+

+ +

[7(0.5)/12] [3.5/12]

0.292

= =

[1.25/12]

+

[3.5/12]

0.397

Changing the result back to percentage form gives 39.7% or approximately 40%. This means that 50% of the households received approximately 40% of the total income of the country.

In the following section, we learn a method for evaluating relatively complex integrals, a technique called the substitution rule.

calculus4engineeringstudents.com