6  Antiderivatives and the Rules of Integration  Integration by Substitution  Area and the Definite Integral  The Fundamental Theorem of Calculus  Evaluating Definite Integrals  Area Between Two Curves  Applications of the Definite Integral to Business and Economics Integration

6.1 Antiderivatives and the Rules of Integration

Antiderivatives  Recall the Maglev problem discussed in chapter 2.  The question asked then was: ✦ If we know the position of the maglev at any time t, can we find its velocity at that time? ✦ The position was described by f(t), and the velocity by f ′ (t).  Now, in Chapters 6 and 7 we will consider precisely the opposite problem: ✦ If we know the velocity of the maglev at any time t, can we find its position at that time? ✦ That is, knowing its velocity function f ′ (t), can we find its position function f(t)?

Antiderivatives  To solve this kind of problems, we need the concept of the antiderivative of a function. ✦ A function F is an antiderivative of f on an interval I if F ′ (t) = f(t) for all of t in I.

Examples  Let  Show that F is an antiderivative of Solution  Differentiating the function F, we obtain and the desired result follows. Example 1, page 398

Examples  Let F(x) = x, G(x) = x + 2, H(x) = x + C, where C is a constant.  Show that F, G, and H are all antiderivatives of the function f defined by f(x) = 1. Solution  Since we see that F, G, and H are indeed antiderivatives of f. Example 2, page 398

Theorem 1  Let G be an antiderivative of a function f.  Then, every antiderivative F of f must be of the form F(x) = G(x) + C where C is a constant.

Example  Prove that the function G(x) = x 2 is an antiderivative of the function f(x) = 2x.  Write a general expression for the antiderivatives of f. Solution  Since G ′ (x) = 2x = f(x), we have shown that G(x) = x 2 is an antiderivative of f(x) = 2x.  By Theorem 1, every antiderivative of the function f(x) = 2x has the form F(x) = x 2 + C, where C is a constant. Example 3, page 399

The Indefinite Integral  The process of finding all the antiderivatives of a function is called antidifferentiation or integration.  We use the symbol ∫, called an integral sign, to indicate that the operation of integration is to be performed on some function f.  Thus, where C and K are arbitrary constants.

Basic Integration Rules  Rule 1: The Indefinite Integral of a Constant

Example  Find each of the following indefinite integrals: a.b. Solution  Each of the integrals had the form f(x) = k, where k is a constant.  Applying Rule 1 in each case yields: a.b. Example 4, page 400

Basic Integration Rules  From the rule of differentiation, we obtain the following rule of integration: ✦ Rule 2: The Power Rule

Examples  Find the indefinite integral: Example 5, page 401

Examples  Find the indefinite integral: Example 5, page 401

Examples  Find the indefinite integral: Example 5, page 401

Basic Integration Rules ✦ Rule 3: The Indefinite Integral of a Constant Multiple of a Function where c is a constant.

Examples  Find the indefinite integral: Example 6, page 402

Examples  Find the indefinite integral: Example 6, page 402

Basic Integration Rules ✦ Rule 4: The Sum Rule

Examples  Find the indefinite integral: Example 7, page 402

Basic Integration Rules ✦ Rule 5: The Indefinite Integral of the Exponential Function

Examples  Find the indefinite integral: Example 8, page 403

Basic Integration Rules ✦ Rule 6: The Indefinite Integral of the Function f(x) = x –1

Examples  Find the indefinite integral: Example 9, page 403

Differential Equations  Given the derivative of a function, f ′, can we find the function f ?  Consider the function f ′(x) = 2x – 1 from which we want to find f(x).  We can find f by integrating the equation: where C is an arbitrary constant.  Thus, infinitely many functions have the derivative f ′, each differing from the other by a constant.

Differential Equations  Equation f ′(x) = 2x – 1 is called a differential equation.  In general, a differential equation involves the derivative of an unknown function.  A solution of a differential equation is any function that satisfies the differential equation.  For the case of f ′(x) = 2x – 1, we find that f(x) = x 2 – x + C gives all the solutions of the differential equation, and it is therefore called the general solution of the differential equation.

Differential Equations   Different values of C yield different functions f(x).   But all these functions have the same slope for any given value of x.   For example, for any value of C, we always find that f ′(1) = 1. – –1 f ′ (1) = 1 f(x) = x 2 – x – 1 x y f ′ (1) = 1 f(x) = x 2 – x + 0 f ′ (1) = 1 f(x) = x 2 – x + 1 f ′ (1) = 1 f(x) = x 2 – x + 2 f ′ (1) = 1 f(x) = x 2 – x + 3

Differential Equations  It is possible to obtain a particular solution by specifying the value the function must assume for a given value of x.  For example, suppose we know the function f must pass through the point (1, 2), which means f(1) = 2.  Using this condition on the general solution we can find the value of C: f(1) = 1 2 – 1 + C= 2 C= 2  Thus, the particular solution is f(x) = x 2 – x + 2

(1, 2) Differential Equations  Here is the graph of the particular solution of f when C = 2.  Note that this graph does go through the point (1, 2). – –1 x y f(x) = x 2 – x + 2

Initial Value Problems  The problem we just discussed is of a type called initial value problem.  In this type of problem we are required to find a function satisfying 1.A differential equation. 2.One or more initial conditions.

Example  Find the function f if it is known that Solution  Integrating the function f ′, we find Example 10, page 404

Example  Find the function f if it is known that Solution  Using the condition f(1) = 9, we have  Therefore, the required function f is Example 10, page 404

Applied Example: Velocity of Maglev  In a test run of a maglev, data obtained from reading its speedometer indicate that the velocity of the maglev at time t can be described by the velocity function  Find the position function of the maglev.  Assume that initially the maglev is located at the origin of a coordinate line. Applied Example 11, page 404

Applied Example: Velocity of Maglev Solution  Let s(t) denote the position of the maglev at any given time t (0  t  30). Then, s ′ (t) = v(t).  So, we have the initial value problem  Integrating the function s ′, we find Applied Example 11, page 404

Applied Example: Velocity of Maglev Solution  Using the condition s(0) = 0, we have  Therefore, the required function s is Applied Example 11, page 404

6.2 Integration by Substitution

 The method of substitution is related to the chain rule for differentiating functions.  It is a powerful tool for integrating a large class of functions.

How the Method of Substitution Works  Consider the indefinite integral  One way to solve this integral is to expand the expression and integrate the resulting integrand term by term.  An alternative approach simplifies the integral by making a change of variable.  Write u = 2x + 4 with differential du = 2dx

How the Method of Substitution Works  Substitute u = 2x + 4 and du = 2dx in the original expression:  Now it’s easy to integrate:  Replacing u by u = 2x + 4, we obtain:

How the Method of Substitution Works  We can verify the result by finding its derivative:  The derivative is indeed the original integrand expression.

The Method of Integration by Substitution Step 1Let u = g(x), where g(x) is part of the integrand, usually the “inside function” of the composite function f(g(x)). Step 2Find du = g ′ (x)dx. Step 3Use the substitution u = g(x) and du = g ′ (x)dx to convert the entire integral into one involving only u. Step 4Evaluate the resulting integrand. Step 5Replace u by g(x) to obtain the final solution as a function of x.

Examples  Find Solution Step 1The integrand involves the composite function with “inside function” So, we choose Example 1, page 413

Examples  Find Solution Step 2Find du/dx and solve for du: Example 1, page 413

Examples  Find Solution Step 3Substitute u = x and du = 2xdx, to obtain an integral involving only u: Example 1, page 413

Examples  Find Solution Step 4Evaluate the integral: Step 5Replace u by x to find the solution: Example 1, page 413

Examples  Find Solution  Let u = –3x, so that du = –3dx, or dx = – ⅓ du.  Substitute to express the integrand in terms of u:  Evaluate the integral:  Replace u by –3x to find the solution: Example 4, page 414

Examples  Find Solution  Let u = 3x 2 + 1, so that du = 6xdx, or xdx = du.  Substitute to express the integrand in terms of u:  Evaluate the integral:  Replace u by 3x to find the solution: Example 5, page 415

Examples  Find Solution  Let u = ln x, so that du = 1/x dx, or dx/x = du.  Substitute to express the integrand in terms of u:  Evaluate the integral:  Replace u by ln x to find the solution: Example 6, page 415

6.3 Area and the Definite Integral

The Area Under the Graph of a Function  Let f be a nonnegative continuous function on [a, b]. Then, the area of the region under the graph of f is where x 1, x 2, …, x n are arbitrary points in the n subintervals of [a, b] of equal width  x = (b – a)/n.

The Definite Integral Let f be a continuous function defined on [a, b]. If exists for all choices of representative points x 1, x 2, …, x n in the n subintervals of [a, b] of equal width  x = (b – a)/n, then the limit is called the definite integral of f from a to b and is denoted by Thus, The number a is the lower limit of integration, and the number b is the upper limit of integration.

Integrability of a Function  Let f be a continuous on [a, b]. Then, f is integrable on [a, b] ; that is, the definite integral exists.

Geometric Interpretation of the Definite Integral  If f is nonnegative and integrable on [a, b], then is equal to the area of the region under the graph of f on [a, b].

Geometric Interpretation of the Definite Integral  The definite integral is equal to the area of the region under the graph of f on [a, b] : x y a b

Geometric Interpretation of the Definite Integral  If f is continuous on [a, b], then is equal to the area of the region above [a, b] minus the region below [a, b].

Geometric Interpretation of the Definite Integral  The definite integral is equal to the area of the region above [a, b] minus the region below [a, b] : x y a b R1R1R1R1 R2R2R2R2 R3R3R3R3

6.4 The Fundamental Theorem of Calculus

Theorem 2 The Fundamental Theorem of Calculus  Let f be continuous on [a, b]. Then, where F is any antiderivative of f; that is, F ′ (x) = f(x).

Example  Let R be the region under the graph of f(x) = x on the interval [1, 3].  Use the fundamental theorem of calculus to find the area A of R and verify your result by elementary means. Example 1, page 431

ExampleSolution  The graph shows the region to be evaluated.  Since f is nonnegative on [1, 3], the area of R is given by the definite integral of f from 1 to 3. x R x = 3 x = y Example 1, page 431

ExampleSolution  By the fundamental theorem of calculus, we have  Thus, the area A of region R is 4 square units.  Note that the constant of integration C dropped out. This is true in general. Example 1, page 431

4321ExampleSolution  Using elementary means, note that area A is equal to R 1 (base ☓ height) plus R 2 ( ☓ base ☓ height).  Thus, A = R 1 + R 2 = 2(1) + (2)(2) = 4 x y R1R1R1R1 2 2 R2R2R2R2 Example 1, page 431

–2 – Example  Find the area of the region under the graph of y = x from x = –1 to x = 2. Solution  Note below that the full region R under consideration lies above the x axis. x y f(x) = x R Example 3, page 432

Example  Find the area of the region under the graph of y = x from x = –1 to x = 2. Solution  Using the fundamental theorem of calculus, we find that the required area is or 6 square units. Example 3, page 432

Net Change Formula  The net change in a function f over an interval [a, b] is given by provided f ′ is continuous on [a, b].

Applied Example: Population Growth in Clark County  Clark County, Nevada, (dominated by Las Vegas) is the fastest growing metropolitan area in the United States.  From 1970 through 2000, the population was growing at a rate of people per decade, where t = 0 corresponds to the beginning of  What was the net change in population over the decade from 1980 to 1990? Applied Example 6, page 433

Applied Example: Population Growth in Clark County Solution  The net change in population over the decade from 1980 to 1990 is given by P(2) – P(1), where P denotes the population in the county at time t.  But P ′ = R, and so the net change in population is Applied Example 6, page 433

Applied Example: Assembly Time of Workers  An efficiency study conducted for Elektra Electronics showed that the rate at which Space Commander walkie- talkies are assembled by the average worker t hours after starting work at 8:00 a.m. is given by the function  Determine how many walkie-talkies can be assembled by the average worker in the first hour of the morning shift. Applied Example 8, page 435

Applied Example: Assembly Time of Workers Solution  Let N(t) denote the number of walkie-talkies assembled by the average worker t hours after starting work in the morning shift.  Then, we have  Therefore, the number of units assembled by the average worker in the first hour of the morning shift is or 20 units. Applied Example 8, page 435

6.5 Evaluating Definite Integrals

Properties of the Definite Integral Let f and g be integrable functions, then

Examples: Using the Method of Substitution  Evaluate Solution  First, find the indefinite integral: ✦ Let u = 9 + x 2 so that Example 1, page 442

Examples: Using the Method of Substitution  Evaluate Solution  First, find the indefinite integral: ✦ Then, integrate by substitution using xdx = du: Example 1, page 442

Examples: Using the Method of Substitution  Evaluate Solution  Using the results, we evaluate the definite integral: Example 1, page 442

Examples: Using the Method of Substitution  Evaluate Solution  Let u = x so that Example 3, page 444

Examples: Using the Method of Substitution  Evaluate Solution  Find the lower and upper limits of integration with respect to u: ✦ When x = 0, the lower limit is u = (0) = 1. ✦ When x = 1, the upper limit is u = (1) = 2.  Substitute x 2 dx = 2 du, along with the limits of integration: Example 3, page 444

Examples: Using the Method of Substitution  Find the area of the region R under the graph of f(x) = e (1/2)x from x = –1 to x = 1. f(x) = e (1/2)x from x = –1 to x = 1.Solution  The graph shows region R: –2 – x y f(x) = e (1/2)x R Example 4, page 444

Examples: Using the Method of Substitution  Find the area of the region R under the graph of f(x) = e (1/2)x from x = –1 to x = 1. f(x) = e (1/2)x from x = –1 to x = 1.Solution  Since f(x) is always greater than zero, the area is given by  To evaluate this integral, we substitute so that Example 4, page 444

Examples: Using the Method of Substitution  Find the area of the region R under the graph of f(x) = e (1/2)x from x = –1 to x = 1. f(x) = e (1/2)x from x = –1 to x = 1.Solution  When x = –1, u = –, and when x = 1, u =.  Substitute dx = 2du, along with the limits of integration: or approximately 2.08 square units. Example 4, page 444

Average Value of a Function  Suppose f is integrable on [a, b].  Then, the average value of f over [a, b] is

Applied Example: Automobile Financing  The interest rates changed by Madison Finance on auto loans for used cars over a certain 6-month period in 2008 are approximated by the function where t is measured in months and r(t) is the annual percentage rate.  What is the average rate on auto loans extended by Madison over the 6-month period? Applied Example 6, page 446

Applied Example: Automobile Financing Solution  The average rate over the 6-month period is given by or 9% per year. Applied Example 6, page 446

6.6 Area Between Two Curves

The Area Between Two Curves  Let f and g be continuous functions such that f(x)  g(x) on the interval [a, b].  Then, the area of the region bounded above by y = f(x) and below by y = g(x) on [a, b] is given by

Examples  Find the area of the region bounded by the x-axis, the graph of y = –x 2 + 4x – 8, and the lines x = –1 and x = 4. Solution  The region R is being bounded above by the graph f(x) = 0 and below by the graph of g(x) = y = –x 2 + 4x – 8 on [–1, 4]: – 2 246– 2 246– 2 246– – 4 – 8 – 12 x y y = –x 2 + 4x – 8 R x = 4 x = – 1 Example 1, page 454

Examples  Find the area of the region bounded by the x-axis, the graph of y = –x 2 + 4x – 8, and the lines x = –1 and x = 4. Solution  Therefore, the area of R is given by Example 1, page 454

Examples  Find the area of the region bounded by f(x) = 2x – 1, g(x) = x 2 – 4, x = 1, and x = 2. f(x) = 2x – 1, g(x) = x 2 – 4, x = 1, and x = 2.Solution  Note that the graph of f always lies above that of g for all x in the interval [1, 2]: – 4 – – 2– 2– 4– 44422– 2– 2– 4– 4 x y y = x 2 – 4 R y = 2x – 1 x = 2 x = 1 Example 2, page 454

Examples  Find the area of the region bounded by f(x) = 2x – 1, g(x) = x 2 – 4, x = 1, and x = 2. f(x) = 2x – 1, g(x) = x 2 – 4, x = 1, and x = 2.Solution  Since the graph of f always lies above that of g for all x in the interval [1, 2], the required area is given by Example 2, page 454

Examples  Find the area of the region that is completely enclosed by the graphs of f(x) = 2x – 1 and g(x) = x 2 – 4. Solution  First, find the points of intersection of the two curves.  To do this, you can set g(x) = f(x) and solve for x: so, the graphs intersect at x = – 1 and at x = 3. Example 3, page 455

Examples  Find the area of the region that is completely enclosed by the graphs of f(x) = 2x – 1 and g(x) = x 2 – 4. Solution  The graph of f always lies above that of g for all x in the interval [– 1, 3] between the two intersection points: – 4 – – 442– 4 x y y = x 2 – 4 R y = 2x – 1 (– 1, – 3) (3, 5) Example 3, page 455

Examples  Find the area of the region that is completely enclosed by the graphs of f(x) = 2x – 1 and g(x) = x 2 – 4. Solution  Since the graph of f always lies above that of g for all x in the interval [– 1, 3], the required area is given by Example 3, page 455

Examples  Find the area of the region bounded by f(x) = x 2 – 2x – 1, g(x) = – e x – 1, x = – 1, and x = 1. f(x) = x 2 – 2x – 1, g(x) = – e x – 1, x = – 1, and x = 1.Solution  Note that the graph of f always lies above that of g for all x in the interval [– 1, 1]: – 3 – – 3– – 3– 3 x y y = x 2 – 2x – 1 R y = – e x – 1 x = – 1 x = 1 Example 4, page 455

Examples  Find the area of the region bounded by f(x) = x 2 – 2x – 1, g(x) = – e x – 1, x = – 1, and x = 1. f(x) = x 2 – 2x – 1, g(x) = – e x – 1, x = – 1, and x = 1.Solution  Since the graph of f always lies above that of g for all x in the interval [– 1, 1], the required area is given by Example 4, page 455

Examples  Find the area of the region bounded by f(x) = x 3, the x-axis, x = – 1, and x = 1. f(x) = x 3, the x-axis, x = – 1, and x = 1.Solution  The region being considered is composed of two subregions R 1 and R 2 : 1 1– 11– 1 x y y = x 3 R1R1 x = – 1 x = 1 R2R2 – 1– 1 Example 5, page 456

Examples  Find the area of the region bounded by f(x) = x 3, the x-axis, x = – 1, and x = 1. f(x) = x 3, the x-axis, x = – 1, and x = 1.Solution  To find R 1 and R 2 consider the x-axis as g(x) = 0.  Since g(x)  f(x) on [– 1, 0], the area of R 1 is given by 1 1– 11– 1 x y y = x 3 R1R1 x = – 1 x = 1 – 1– 1 Example 5, page 456

Examples  Find the area of the region bounded by f(x) = x 3, the x-axis, x = – 1, and x = 1. f(x) = x 3, the x-axis, x = – 1, and x = 1.Solution  To find R 1 and R 2 consider the x-axis as g(x) = 0.  Since g(x)  f(x) on [0, 1], the area of R 2 is given by – 1– 111– 1– 1 x y y = x 3 x = – 1 x = 1 R2R2R2R2 – 1– 1– 1– 1 Example 5, page 456

R1R1R1R1Examples  Find the area of the region bounded by f(x) = x 3, the x-axis, x = – 1, and x = 1. f(x) = x 3, the x-axis, x = – 1, and x = 1.Solution  Therefore, the required area R is square units – 1– 111– 1– 1 x y y = x 3 x = – 1 x = 1 R2R2R2R2 – 1– 1– 1– 1 Example 5, page 456

 Find the area of the region bounded by f(x) = x 3 – 3x + 3 and g(x) = x + 3. f(x) = x 3 – 3x + 3 and g(x) = x + 3.Solution  The region R being considered is composed of two subregions R 1 and R 2 : Examples531 y y = x 3 – 3x + 3 y = x + 3 – 3 – x R1R1R1R1 R2R2R2R2 Example 6, page 457

 Find the area of the region bounded by f(x) = x 3 – 3x + 3 and g(x) = x + 3. f(x) = x 3 – 3x + 3 and g(x) = x + 3.Solution  To find the points of intersection, we solve simultaneously the equations y = x 3 – 3x + 3 and y = x + 3. Examples531 y y = x 3 – 3x + 3 R1R1R1R1 R2R2R2R2 – 3 – x  So, x = 0, x = – 2, and x = 2.  The points of intersection of the two curves are (– 2, 1), (0, 3), and (2, 5). (2, 5) y = x + 3 (– 2, 1) (0, 3) Example 6, page 457

 Find the area of the region bounded by f(x) = x 3 – 3x + 3 and g(x) = x + 3. f(x) = x 3 – 3x + 3 and g(x) = x + 3.Solution  Note that f(x)  g(x) for [– 2, 0], so the area of region R 1 is given by Examples531 y y = x 3 – 3x + 3 R1R1R1R1 – 3 – x (2, 5) y = x + 3 (– 2, 1) (0, 3) Example 6, page 457

 Find the area of the region bounded by f(x) = x 3 – 3x + 3 and g(x) = x + 3. f(x) = x 3 – 3x + 3 and g(x) = x + 3.Solution  Note that g(x)  f(x) for [0, 2], so the area of region R 2 is given by Examples531 y y = x 3 – 3x + 3 R2R2R2R2 – 3 – x (2, 5) y = x + 3 (– 2, 1) (0, 3) Example 6, page 457

 Find the area of the region bounded by f(x) = x 3 – 3x + 3 and g(x) = x + 3. f(x) = x 3 – 3x + 3 and g(x) = x + 3.Solution  Therefore, the required area R is square units. Examples531 y y = x 3 – 3x + 3 R1R1R1R1 R2R2R2R2 – 3 – x (2, 5) y = x + 3 (– 2, 1) (0, 3) Example 6, page 457

6.7 Applications of the Definite Integral to Business and Economics

Consumers’ and Producers’ Surplus  Suppose p = D(x) is the demand function that relates the price p of a commodity to the quantity x demanded of it.  Now suppose a unit market price ` p has been established, along with a corresponding quantity demanded ` x.  Those consumers who would be willing to pay a unit price higher than ` p for the commodity would in effect experience a savings.  This difference between what the consumer would be willing to pay and what they actually have to pay is called the consumers’ surplus.

Consumers’ and Producers’ Surplus  To derive a formula for computing the consumers’ surplus, divide the interval [0, ` x] into n subintervals, each of length  x = ` x/n, and denote the right endpoints of these intervals by x 1, x 2, …, x n = ` x: `p`p`p`p p = ` p x p x1x1x1x1 x2x2x2x2 x n –1 xnxnxnxn x3x3x3x3 x4x4x4x4 D(x1)D(x1)D(x1)D(x1) D(x2)D(x2)D(x2)D(x2) D(x3)D(x3)D(x3)D(x3) D(x4)D(x4)D(x4)D(x4) D(x n –1 ) D(x n ) = ` p

r1r1r1r1 Consumers’ and Producers’ Surplus  There are consumers who would pay a price of at least D(x 1 ) for the first  x units instead of the market price of ` p.  The savings to these consumers is approximated by which is the area of the rectangle r 1 : p = ` p x p D(x1)D(x1)D(x1)D(x1) x1x1x1x1 `p`p`p`p

r1r1r1r1 Consumers’ and Producers’ Surplus  Similarly, the savings the consumer experiences for the consecutive increments of  x are depicted by the areas of rectangles r 2, r 3, r 4, …, r n : r2r2r2r2 r n –1 p = ` p x p r3r3r3r3 r4r4r4r4 rnrnrnrn xnxnxnxn `p`p`p`p

`x`x`x`x Consumers’ and Producers’ Surplus  Adding r 1 + r 2 + r 3 + … + r n, and letting n approach infinity, we obtain the consumers’ surplus CS formula: where D(x) is the demand function, ` p is the unit market price, and ` x is the quantity demanded. x p `p`p`p`p p = D(x) CS

Consumers’ and Producers’ Surplus  Similarly, we can derive a formula for the producers’ surplus.  Suppose p = S(x) is the supply function that relates the price p of a commodity to the quantity x supplied of it.  Again, suppose a unit market price ` p has been established, along with a corresponding quantity supplied ` x.  Those sellers who would be willing to sell at unit price lower than ` p for the commodity would in effect experience a gain or profit.  This difference between what the seller would be willing to sell for and what they actually can sell for is called the producers’ surplus.

`x`x`x`x Consumers’ and Producers’ Surplus  Geometrically, the producers’ surplus is given by the area of the region bounded above the straight line p = ` p and below the supply curve p = S(x) from x = 0 to x = ` x: x p `p`p`p`p p = S(x) PS

Consumers’ and Producers’ Surplus  The producers’ surplus PS is given by where S(x) is the supply function, ` p is the unit market price, and ` x is the quantity supplied. `x`x`x`x x p `p`p`p`p p = S(x) PS

Example  The demand function for a certain make of 10-speed bicycle is given by where p is the unit price in dollars and x is the quantity demanded in units of a thousand.  The supply function for these bicycles is given by where p stands for the price in dollars and x stands for the number of bicycles that the supplier will want to sell.  Determine the consumers’ surplus and the producers’ surplus if the market price of a bicycle is set at the equilibrium price. Example 1, page 466

ExampleSolution  To find the equilibrium point, equate S(x) and D(x) to solve the system of equations and find the point of intersection of the demand and supply curves:  Thus, x = – 625/2 or x = 300. The first number is discarded for being negative, so the solution is x = 300. Example 1, page 466

ExampleSolution  Substitute x = 300 to find the equilibrium value of p:  Thus, the equilibrium point is (300, 160).  That is, the equilibrium quantity is 300,000 bicycles, and the equilibrium price is $160 per bicycle. Example 1, page 466

ExampleSolution  To find the consumers’ surplus, we set ` x = 300 and ` p = 160 in the consumers’ surplus formula: or $18,000,000. Example 1, page 466

ExampleSolution  To find the producers’ surplus, we set ` x = 300 and ` p = 160 in the producers’ surplus formula: or $11,700,000. Example 1, page 466

ExampleSolution  Consumers’ surplus and producers’ surplus when the market is in equilibrium: x p CSPS p = ` p = 160 Example 1, page 466

Accumulated or Total Future Value of an Income Stream  The accumulated, or total, future value after T years of an income stream of R(t) dollars per year, earning interest rate of r per year compounded continuously, is given by

Applied Example: Income Stream  Crystal Car Wash recently bought an automatic car- washing machine that is expected to generate $40,000 in revenue per year, t years from now, for the next 5 years.  If the income is reinvested in a business earning interest at the rate of 12% per year compounded continuously, find the total accumulated value of this income stream at the end of 5 years. Applied Example 2, page 468

Applied Example: Income Stream Solution  We are required to find the total future value of the given income stream after 5 years.  Setting R(t) = 40,000, r = 0.12, and T = 5 in the accumulated income stream formula we get or approximately $274,040. Applied Example 2, page 468

Present Value of an Income Stream  The present value of an income stream of R(t) dollars in a year, earning interest at the rate of r per year compounded continuously, is given by

Applied Example: Investment Analysis  The owner of a local cinema is considering two alternative plans for renovating and improving the theater.  Plan A calls for an immediate cash outlay of $250,000, whereas plan B requires an immediate cash outlay of $180,000.  It has been estimated that adopting plan A would result in a net income stream generated at the rate of f(t) = 630,000 dollars per year, whereas adopting plan B would result in a net income stream generated at the rate of g(t) = 580,000 for the next three years.  If the prevailing interest rate for the next five years is 10% per year, which plan will generate a higher net income by the end of year 3?

Applied Example: Investment Analysis Solution  We can find the present value of the net income NI for plan A setting R(t) = 630,000, r = 0.1, and T = 3, using the present value formula: or approximately $1,382,845. Applied Example 2, page 468

Applied Example: Investment Analysis Solution  To find the present value of the net income NI for plan B setting R(t) = 580,000, r = 0.1, and T = 3, using the present value formula: or approximately $1,323,254. Applied Example 2, page 468

Applied Example: Investment Analysis Solution  Thus, we conclude that plan A will generate a higher present value of net income by the end of the third year ($1,382,845), than plan B ($1,323,254). Applied Example 2, page 468

Amount of an Annuity  The amount of an annuity is where P, r, T, and m are as defined earlier.

Applied Example: IRAs  On January 1, 1990, Marcus Chapman deposited $2000 into an Individual Retirement Account (IRA) paying interest at the rate of 10% per year compounded continuously.  Assuming that he deposited $2000 annually into the account, how much did he have in his IRA at the beginning of 2006? Applied Example 4, page 470

Applied Example: IRAs Solution  We set P = 2000, r = 0.1, T = 16, and m = 1 in the amount of annuity formula, obtaining  Thus, Marcus had approximately $79,061 in his account at the beginning of Applied Example 4, page 470

Present Value of an Annuity  The present value of an annuity is given by where P, r, T, and m are as defined earlier.

Applied Example: Sinking Funds  Tomas Perez, the proprietor of a hardware store, wants to establish a fund from which he will withdraw $1000 per month for the next ten years.  If the fund earns interest at a rate of 6% per year compounded continuously, how much money does he need to establish the fund? Applied Example 5, page 471

Applied Example: Sinking Funds Solution  We want to find the present value of an annuity with P = 1000, r = 0.06, T = 10, and m = 12.  Using the present value of an annuity formula, we find  Thus, Tomas needs approximately $90,238 to establish the fund. Applied Example 5, page 471

End of Chapter