IndefiniteIntegral(II)

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

EXAMPLE 25Evaluate the integral

SolutionThe integrand is f(x) = x2 – | 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,

which equals

Thus, the graph of the integrand equals the graph of x2 – x + 1 drawn on the interval [1, ∞]

x2 + x – 1 drawn on the interval [–∞, 1]

as illustrated below:

(x2 – | x – 1|) dx∫ 0

2

f(x) = x2 – |x – 1| = x2 – (x – 1) if x ≥ 0

x2 – [–(x – 1)] if x < 0

f(x) = x2 – |x – 1| = x2 – x + 1 if x ≥ 0 x2 + x – 1 if x < 0

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

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

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

carefully.

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

SolutionObserve 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,

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

= – 1/6 + 11/6

= 5/3

(x2 – | x – 1|) dx∫ 0

2(x2 + x – 1) dx∫ 0

1 (x2 – x + 1) dx∫ 1

2 + =

(x2 – | x – 1|) dx∫ 0

2

w'(t) dt∫ 5

10

= w(10) – w(5) w'(t) dt∫ 5

10

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

SolutionWhen 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

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

SolutionLet'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

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

Therefore,in this case,

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

I(t) dt∫ a

b

=

r(t) = C(t2) – C(t1)

t2 – t1

= Q(b) – Q(a) I(t) dt∫ a

b Q' (t) dt∫ a

b

r(t) dt∫ 0120

= r(t) dt∫ 0120 C(120) – C(0)

120 – 0

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

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

represent?

SolutionThe 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,

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

represents the TOTAL number of bees in that colony.

EXAMPLE 26.5The 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

represent?

SolutionLet'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,

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:

100 + n'(t) dt∫ 015

= n(15) – n(0) n'(t) dt∫ 015

100 + n'(t) dt∫ 015

R'(x) dx∫ 1000 5000

= R(5000) – R(1000) R'(x) dx∫ 1000 5000

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 27The 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, v(0) = 5, 0 ≤ t ≤ 10

● a(t) = 2t + 3, 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

From the definition given above, we say that a(t) = '(t). Thus, we can re-write equation 1 as

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

OR

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

2

1

3

4

= v(t2) – v(t1) v'(t) dt∫ τ1

t2

= v(t2) – v(t1) a(t) dt∫ τ1

t2

= s(t2) – s(t1) s'(t) dt∫ τ1

t2

= s(t2) – s(t1) v(t) dt∫ τ1

t2

particle, we evaluate the integral

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

EXAMPLE 27.1v(t) = 3t – 5, 0 ≤ t ≤ 3

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

Using FTC 2, we have

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

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,

| v(t)| dt∫ τ1

t2

=v(t) dt∫ 0

3 (3t – 5) dt∫ 0

3

1.5t2 – 5t ] 3 0

|3t – 5| dt∫ 0

3

=

= –(3t – 5) dt∫ 0

5/3

(3t – 5) dt∫ 5/3

3 +

=v(t) dt∫ 0

3

|3t – 5| dt∫ 0

3 –[v(t)] dt∫ 0

5/3 + [v(t)] dt∫ 5/33

Using FTC 2, we have

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.

= 25/6 + 8/3

= 41/6m ≈ 6.833 m

|3t – 5| dt∫ 0

3

EXAMPLE 27.2v(t) = t2 – 2t – 8 , 1 ≤ t ≤ 6

The displacement of the particle is given by

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

Using FTC 2, we have

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

EXAMPLE 27.3a(t) = t + 4, v(0) = 5, 0 ≤ t ≤ 10

We know that a(t) = v'(t), which means that, to find the velocity function of the particle, we integrate the

acceleration function a(t), i.e.

(t2 – 2t – 8) dt∫ 1

6

|t2 – 2t – 8| dt∫ 1

6 –(t2 – 2t – 8) dt∫ 1

4 (t2 – 2t – 8) dt∫ 4

6 = +

= 18 + (44/3) |t2 – 2t – 8| dt∫ 1

6

= 98/3m ≈ 32.667 m

a(t) dt∫ (t + 4) dt∫ v(t) = =

Which equals

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

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

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

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

which is roughly equal to 416.6667m.

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

EXAMPLE 27.4a(t) = 2t + 3, v(0) = -5, 0 ≤ t ≤ 3

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

v(t) = ½ t2 + 4t + C (from the general antiderivative formula)

½ (0)2 + 4(0) + C = 5

v(t) = ½ t2 + 4t + 5

|½ t2 + 4t + 5| dt∫ 0

10

= |½ t2 + 4t + 5| dt∫ 0

10 (½ t2 + 4t + 5) dt∫ 0

10

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

Putting t = 0 gives

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

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

which equals

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

EXAMPLE 28The 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

t2 + 3t + C = – 4

(0)2 + 3(0) + C = – 4

v(t) = t2 + 3t – 4

a(t) dt∫ (2t + 3) dt∫ v(t) = =

|t2 + 3t – 4| dt∫ 0

3 – (t2 + 3t – 4) dt∫ 01 = + (t2 + 3t – 4) dt∫ 1

3

|t2 + 3t – 4| dt∫ 0

3 = (13/6) + (38/3) ≈ 14.833 m

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

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

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

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

EXAMPLE 29An 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?

SolutionLet'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 4th 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) = 200t + 2.5t2.

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

[200(10) + 2.5(10)2] – [200(4) + 2.5(4)2]

= [2000 + 250] – [800 + 40] = 2250 – 840

= 1410

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

=

= m(b) – m(a) ρ(x) dx∫ a

b

= m(b) – m(a) m'(x) dx∫ a

b

= mass of rod ρ(x) dx∫ 0

4

9x – 4/3 (√x3) ] 4 0

ρ(x) dx∫ 0

4

9x – 4/3 (√x3) ] 4 0

p'(t) dt∫ (200 + 5t) dt∫=

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

which equals 1410.

EXAMPLE 30The 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.

SolutionWhenever 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) = 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] – [6000 – 20000 + 16000] = 60,000 – 2,000

= 58,000

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

(200 + 5t) dt∫ 4

10

C'(x) dx∫

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

and evaluate

which equals $58,000.

EXAMPLE 31On 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 Time (s) Velocity (ft/s)Launch 0 0Begin roll maneuver 10 185End roll maneuver 15 319Throttle to 89% 20 447Throttle to 67% 32 742Throttle to 104% 59 1325Maximum dynamic pressure 62 1445Solid rocket booster separation 125 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.

SolutionUsing 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

C'(x) dx∫ 2000 4000

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,

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

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

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

EXAMPLE 32The 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.

= 0.000365325t4 – 0.0385113t3 + 12.49045958t2 – 21.268723t ] 125

0

= height attained by rocket in 125 seconds |v(t)| dt∫ 0

125

|v(t)| dt∫ 0

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.

t (s) v (mi/h) t (s) v (mi/h)0 0 60 5610 38 70 5320 52 80 5030 58 90 4740 55 100 4550 51

SolutionOn 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) v (ft/s) t (s) v (ft/s)0 0 60 1232/15

10 836/15 70 1166/1520 1144/15 80 220/330 1276/15 90 1034/1540 242/3 100 6650 374/5

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

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

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 33The 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.

= [v(5) + v(15) + v(25) + v(35) +.......... ........+ v(85) + v(95)]∆t |v(t)| dt∫ 0

100

= [(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.

|v(t)| dt∫ 0

100

t r(t) t r(t) 1981 10.3 1990 5.41982 6.2 1991 4.21983 3.2 1992 3.01984 4.3 1993 3.01985 3.6 1994 2.61986 1.9 1995 2.81987 3.6 1996 2.91988 4.1 1997 2.31989 4.8

SolutionPlotting 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

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

Thus,

= [r(1981.5) + r(1982.5) + r(1983.5) + ............. + r(1996.5)]∆t

= [8.25 + 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 %

= C(1997) – R(1981) C'(t) dt∫ 1981 1997

= C(1997) – R(1981) r(t) dt∫ 1981 1997

r(t) dt∫ 1981 1997

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

EXAMPLE 34The 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

(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.

SolutionBefore 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:

( 2y – y2) dy∫ 0

2

x = 2y – y2

2

01 x

y

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

It's just the same as

except that x and y have been interchanged. We therefore compute the integral the same way we compute regular

integrals. So, using FTC 2,

Evaluating the antiderivative at points 0 and 2 gives:

y = ax2 (a>0) y = ax2 (a<0)

= ] 2 0y2 – (y3/3)

= 4 – (8/3) – 0

x = ay2 (a>0) x = ay2 (a<0)

( 2y – y2) dy∫ 0

2

( 2x – x2) dx∫ 0

2

( 2y – y2) dy∫ 0

2

( 2y – y2) dy∫ 02

EXAMPLE 35Economists 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

(b). 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.

( 2y – y2) dy∫ 02

= 4 /3

(x – L(x)) dx∫ 0

12COEFFICIENT OF INEQUALITY =

Solution(a). By definition,

which can be interpreted as

You can see that the graph above appropriately illustrates equation 1. From the graph,

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):

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

ii

i

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

Area under y = xCOEFFICIENT OF INEQUALITY =

(x – L(x)) dx∫ 0

1

x dx∫ 0

1

ORANGE AREA

GREEN AREA + ORANGE AREA COEFFICIENT OF INEQUALITY =

(x – L(x)) dx∫ 0

12

= (x – L(x)) dx∫ 0

1

x dx∫ 0

1 (x – L(x)) dx∫ 0

12

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

We multiply both sides by 0.5 to get

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

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] + [7(0.5)/12]

= [1.25/12] + [3.5/12] = [1.25/12] + [3.5/12]

= 0.105 + 0.292 = 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

= x dx∫ 0

1

∫ 0

1x dx2

L(x) dx∫ 0

1 –

x dx∫ 0

1 – ∫ 0

1 L(x) dx2

0.5= – L(x) dx∫ 0

1

0.5= 2(0.5) – ∫ 0

1 L(x) dx2

= 1 – ∫ 0

1 L(x) dx2

0.5 – L(x) dx∫ 0

1

0.50.5 0.5

= 0.5 – L(x) dx∫ 01

0.5 – L(x) dx∫ 01

=

(x – L(x)) dx∫ 0

12