Introductory maths analysis chapter 14 official

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Introductory maths analysis chapter 14 official

  1. 1. INTRODUCTORY MATHEMATICALINTRODUCTORY MATHEMATICAL ANALYSISANALYSISFor Business, Economics, and the Life and Social Sciences ©2007 Pearson Education Asia Chapter 14Chapter 14 IntegrationIntegration
  2. 2. ©2007 Pearson Education Asia INTRODUCTORY MATHEMATICAL ANALYSIS 0. Review of Algebra 1. Applications and More Algebra 2. Functions and Graphs 3. Lines, Parabolas, and Systems 4. Exponential and Logarithmic Functions 5. Mathematics of Finance 6. Matrix Algebra 7. Linear Programming 8. Introduction to Probability and Statistics
  3. 3. ©2007 Pearson Education Asia 9. Additional Topics in Probability 10. Limits and Continuity 11. Differentiation 12. Additional Differentiation Topics 13. Curve Sketching 14. Integration 15. Methods and Applications of Integration 16. Continuous Random Variables 17. Multivariable Calculus INTRODUCTORY MATHEMATICAL ANALYSIS
  4. 4. ©2007 Pearson Education Asia • To define the differential. • To define the anti-derivative and the indefinite integral. • To evaluate constants of integration. • To apply the formulas for . • To handle more challenging integration problems. • To evaluate simple definite integrals. • To apply Fundamental Theorem of Integral Calculus. Chapter 14: Integration Chapter ObjectivesChapter Objectives ∫∫∫ du u dueduu nn 1 and,
  5. 5. ©2007 Pearson Education Asia • To use Trapezoidal rule or Simpson’s rule. • To use definite integral to find the area of the region. • To find the area of a region bounded by two or more curves. • To develop concepts of consumers’ surplus and producers’ surplus. Chapter 14: Integration Chapter Objectives
  6. 6. ©2007 Pearson Education Asia Differentials The Indefinite Integral Integration with Initial Conditions More Integration Formulas Techniques of Integration The Definite Integral The Fundamental Theorem of Integral Calculus 14.1) 14.2) 14.3) Chapter 14: Integration Chapter OutlineChapter Outline 14.4) 14.5) 14.6) 14.7)
  7. 7. ©2007 Pearson Education Asia Approximate Integration Area Area between Curves Consumers’ and Producers’ Surplus 14.8) 14.9) 14.10) Chapter 14: Integration Chapter OutlineChapter Outline 14.11)
  8. 8. ©2007 Pearson Education Asia Chapter 14: Integration 14.1 Differentials14.1 Differentials Example 1 – Computing a Differential • The differential of y, denoted dy or d(f(x)), is given by ( ) ( )dxxfdyxxfdy '' =⇒∆= Find the differential of and evaluate it when x = 1 and ∆x = 0.04. Solution: The differential is When x = 1 and ∆x = 0.04, 432 23 −+−= xxxy ( ) ( ) xxxxxxx dx d dy ∆+−=∆−+−= 343432 223 ( ) ( )[ ]( ) 08.004.031413 2 =+−=dy
  9. 9. ©2007 Pearson Education Asia Chapter 14: Integration 14.1 Differentials Example 3 - Using the Differential to Estimate a Change in a Quantity A governmental health agency examined the records of a group of individuals who were hospitalized with a particular illness. It was found that the total proportion P that are discharged at the end of t days of hospitalization is given by Use differentials to approximate the change in the proportion discharged if t changes from 300 to 305. ( ) ( ) 3 2 300 300 31         + −== t tPP
  10. 10. ©2007 Pearson Education Asia Chapter 14: Integration 14.1 Differentials Example 3 - Using the Differential to Estimate a Change in a Quantity Example 5 - Finding dp/dq from dq/dp Solution: We approximate ∆P by dP, ( ) ( ) ( ) dt t dt t tPdPP 4 3 2 300 300 3 300 300 3' + =        + −==≈∆ Solution: .2500ifFind 2 pq dq dp −= p p dp dqdq dp p p dp dq 2 2 25001 2500 − −==⇒ − −=
  11. 11. ©2007 Pearson Education Asia Chapter 14: Integration 14.2 The Infinite Integral14.2 The Infinite Integral • An antiderivative of a function f is a function F such that . In differential notation, . • Integration states that • Basic Integration Properties: ( ) ( )xfxF =' ( )dxxfdF = ( ) ( ) ( ) ( )xfxFCxFdxxf =+=∫ 'onlyif
  12. 12. ©2007 Pearson Education Asia Chapter 14: Integration 14.2 The Infinite Integral Example 1 - Finding an Indefinite Integral Example 3 - Indefinite Integral of a Constant Times a Function Example 5 - Finding Indefinite Integrals Find . Solution: dx∫5 Cxdx +=∫ 55 Find . Solution: dxx∫7 C x dxx +=∫ 2 7 7 2 CtC t dxtdx t +=+== ∫∫ − 2 2/1 1 a. 2/1 2/1 C x C x dx x +−=+      +− = +− ∫ 2 13 3 12 1 136 1 6 1 b.
  13. 13. ©2007 Pearson Education Asia Find . Solution: Chapter 14: Integration 14.2 The Infinite Integral Example 7 - Indefinite Integral of a Sum and Difference ( )dxexx x ∫ −+− 11072 35 4 ( ) ( ) ( ) ( ) Cexx Cxe xx dxexx x x x ++−= +−+−= −+−∫ 10 4 7 9 10 10 4 7 5/9 2 11072 45/9 45/9 35 4
  14. 14. ©2007 Pearson Education Asia Find Solution: Chapter 14: Integration 14.2 The Infinite Integral Example 9 - Using Algebraic Manipulation to Find an Indefinite Integral ( )( )dx xx ∫ +− 6 312 a. dx x x ∫ − 2 3 1 b. ( )( ) ( ) ( ) C xxx Cx xx dx xx +−+= +      −+= +− ∫ 212 5 9 3 2 5 3 2 6 1 6 312 a. 23 23 ( ) C x x dxxx dx x x ++= −= − ∫ ∫ − 1 2 1 b. 2 2 2 3
  15. 15. ©2007 Pearson Education Asia Chapter 14: Integration 14.3 Integration with Initial Conditions14.3 Integration with Initial Conditions Example 1 - Initial-Condition Problem • Use initial conditions to find the constant, C. If y is a function of x such that y’ = 8x − 4 and y(2) = 5, find y. Solution: We find the integral, Using the condition, The equation is ( ) ( ) CxxCx x dxxy +−=+−=−= ∫ 444 2 848 2 2 ( ) ( ) 3 24245 2 −= +−= C C 344 2 −−= xxy
  16. 16. ©2007 Pearson Education Asia Chapter 14: Integration 14.3 Integration with Initial Conditions Example 3 - Income and Education For a particular urban group, sociologists studied the current average yearly income y (in dollars) that a person can expect to receive with x years of education before seeking regular employment. They estimated that the rate at which income changes with respect to education is given by where y = 28,720 when x = 9. Find y. 164100 2/3 ≤≤= xx dx dy
  17. 17. ©2007 Pearson Education Asia Chapter 14: Integration 14.3 Integration with Initial Conditions Example 3 - Income and Education Solution: We have When x = 9, Therefore, Cxdxxy +== ∫ 2/52/3 40100 ( ) 000,19 940720,28 2/5 = += C C 000,1940 2/5 += xy
  18. 18. ©2007 Pearson Education Asia Chapter 14: Integration 14.3 Integration with Initial Conditions Example 5 - Finding Cost from Marginal Cost In the manufacture of a product, fixed costs per week are $4000. (Fixed costs are costs, such as rent and insurance, that remain constant at all levels of production during a given time period.) If the marginal-cost function is where c is the total cost (in dollars) of producing q pounds of product per week, find the cost of producing 10,000 lb in 1 week. ( ) 2.02500200000010 2 +−= qq.. dq dc
  19. 19. ©2007 Pearson Education Asia Chapter 14: Integration 14.3 Integration with Initial Conditions Example 5 - Finding Cost from Marginal Cost Solution: The total cost c is When q = 0, c = 4000. Cost of 10,000 lb in one week, ( ) ( )[ ] Cq qq . dqqq..qc ++      −= +−= ∫ 2.0 2 25 3 002.0 0000010 2.02500200000010 23 2 ( ) ( ) 67.5416$10000 40002.0 2 25 3 0020 0000010 23 = ++      −= c q qq. .qc
  20. 20. ©2007 Pearson Education Asia Chapter 14: Integration 14.4 More Integration Formulas14.4 More Integration Formulas Power Rule for Integration Integrating Natural Exponential Functions Integrals Involving Logarithmic Functions 1if 1 1 −≠+ + =∫ + nC n u dxu n n ∫ += Cedue uu 0forln 1 ≠+=∫ xCxdx x
  21. 21. ©2007 Pearson Education Asia Chapter 14: Integration 14.4 More Integration Formulas Basic Integration Formulas
  22. 22. ©2007 Pearson Education Asia Chapter 14: Integration 14.4 More Integration Formulas Example 1 - Applying the Power Rule for Integration Find the integral of Solution: ( ) ( ) ( ) C x C u duudxx + + =+==+ ∫∫ 21 1 21 1a. 2121 2020 ( ) dxx 20 1a. ∫ + ( ) dxxx∫ + 332 73b. ( ) ( ) ( ) C x C u duudxxx + + =+==+ ∫∫ 4 7 4 73 434 3332 dxxduxu 23 37Letb. =⇒+=
  23. 23. ©2007 Pearson Education Asia Chapter 14: Integration 14.4 More Integration Formulas Example 3 - Adjusting for du Find Solution: CyC y dyya. +=+=∫ 3/4 33/4 33 4 63 3/4 66 ( ) dx xx xx b. ∫ ++ + 424 3 73 32 ( )dxxxduxxu 6473Let 324 +=⇒++= ( ) C xx C udu u + ++ −=+ − ⋅=      − − ∫ 324 3 4 736 1 32 1 2 dyya. ∫3 6 ( ) dx xx xx b. ∫ ++ + 424 3 73 32
  24. 24. ©2007 Pearson Education Asia Chapter 14: Integration 14.4 More Integration Formulas Example 5 - Integrals Involving Exponential Functions Find Solution: dxxex ∫2a. xdxduxu 2Leta. 2 =⇒= ( ) dxex xx 32 3 1b. + ∫ + [ ] Cedue xdxedxxe xu xx +== = ∫ ∫∫ 2 2 22 ( )dxxduxxu 333Letb. 23 +=⇒+= ( ) Ce Cduedxex xx uxx += +=+ + + ∫∫ 3 32 3 3 3 1 3 1 1
  25. 25. ©2007 Pearson Education Asia Chapter 14: Integration 14.4 More Integration Formulas Example 7 - Integrals Involving Exponential Functions Find Solution: ( ) . 73 32 24 3 dx xx xx ∫ ++ + ( )dxxxduxxu 6473Let 324 +=⇒++= ( ) ( ) Cxx CxxCudx xx xx +++= +++=+= ++ + ∫ 73ln 2 1 73ln 2 1 ln 2 1 73 32 24 24 24 3
  26. 26. ©2007 Pearson Education Asia Chapter 14: Integration 14.5 Techniques of Integration14.5 Techniques of Integration Example 1 - Preliminary Division before Integration Find Cx x dx x xdx x xx ++=      += + ∫∫ ln 2 1 a. 2 2 3 Cx xx dx x xxdx x xxx ++++=       + ++= + +++ ∫∫ 12ln 2 1 23 12 1 12 132 b. 23 2 23
  27. 27. ©2007 Pearson Education Asia Chapter 14: Integration 14.5 Techniques of Integration Example 3 - An Integral Involving bu Find Solution: .23 dxx ∫ − ( )( ) dxduxu 2ln32lnLet −=⇒−= ( )( ) ( )( ) ( ) CCe Ceduedxedx xx uuxx +−=+−= +−=−== −− −− ∫∫∫ 332ln 32ln3 2 2ln 1 2ln 1 2ln 1 2ln 1 2 • General formula for integrating bu is Cb b dub uu +=∫ ln 1
  28. 28. ©2007 Pearson Education Asia Chapter 14: Integration 14.6 The Definite Integral14.6 The Definite Integral Example 1 - Computing an Area by Using Right-Hand Endpoints • For area under the graph from limit a  b, • x is called the variable of integration and f (x) is the integrand. ( )dxxf b a ∫ Find the area of the region in the first quadrant bounded by f(x) = 4 − x2 and the lines x = 0 and y = 0. Solution: Since the length of [0, 2] is 2, ∆x = 2/n.
  29. 29. ©2007 Pearson Education Asia Chapter 14: Integration 14.6 The Definite Integral Example 1 - Computing an Area by Using Right-Hand Endpoints Summing the areas, we get We take the limit of Sn as n→∞: Hence, the area of the region is 16/3. ( )( ) ( )( )       ++ −= ++ −=               −=∆            = ∑∑ == 23 1 2 1 121 3 4 8 6 12188 22 4 2 n nnnnn n n n nn k fx n kfS n k n k n ( )( ) 3 16 3 8 8 121 3 4 8limlim 2 =−=            ++ −= ∞→∞→ n nn S n n n
  30. 30. ©2007 Pearson Education Asia Chapter 14: Integration 14.6 The Definite Integral Example 3 - Integrating a Function over an Interval Integrate f (x) = x − 5 from x = 0 to x = 3. Solution: ( ) 15 1 1 2 9 15 2 1933 1 −      +=− + =            = ∑= nn n nn kfS n k n ( ) 2 21 2 9 15 1 1 2 9 limlim5 3 0 −==      −      +==− ∞→∞→∫ n Sdxx n n n
  31. 31. ©2007 Pearson Education Asia Chapter 14: Integration 14.7 The Fundamental Theorem of14.7 The Fundamental Theorem of Integral CalculusIntegral Calculus Fundamental Theorem of Integral Calculus • If f is continuous on the interval [a, b] and F is any antiderivative of f on [a, b], then Properties of the Definite Integral • If a > b, then • If limits are equal, ( ) ( ) ( )aFbFdxxf b a −=∫ ( ) ( )∫∫ −= a b b a dxxfdxxf ( ) 0=∫ b a dxxf
  32. 32. ©2007 Pearson Education Asia Chapter 14: Integration 14.7 The Fundamental Theorem of Integral Calculus Properties of the Definite Integral 1. is the area bounded by the graph f(x). 2. 3. 4. 5. ( )∫ b a dxxf ( ) ( ) constant.aiswhere kdxxfkdxxkf b a b a ∫∫ = ( ) ( )[ ] ( ) ( )∫∫∫ ±=± b a b a b a dxxgdxxfdxxgxf ( ) ( )∫∫ = b a b a dttfdxxf ( ) ( ) ( )∫∫∫ += c b b a c a dxxfdxxfdxxf
  33. 33. ©2007 Pearson Education Asia Chapter 14: Integration 14.7 The Fundamental Theorem of Integral Calculus Example 1 - Applying the Fundamental Theorem Find Solution: ( ) .63 3 1 2 ∫− +− dxxx ( ) ( ) ( ) ( ) ( ) 48 16 2 1 136 2 3 3 6 2 63 2 3 2 3 3 1 2 3 3 1 2 =       −+ − −−−      +−=       +−=+− −− ∫ x x xdxxx
  34. 34. ©2007 Pearson Education Asia Chapter 14: Integration 14.7 The Fundamental Theorem of Integral Calculus Example 3 - Evaluating Definite Integrals Find Solution: ( )[ ]∫ ++ 2 1 323/1 14a. dxttt ( )[ ] ( ) ( ) ( ) ( ) 8 585 2625 8 1 123 4 1 2 1 414a. 3443/4 2 1 2/12 3 4 3/42 1 323/1 +=−+−=         +       +=++∫ tt dxttt [ ] ( ) ( )1 3 1 3 1 3 1 b. 3031 0 3 1 0 3 −=−=      =∫ eeeedte tt ∫ 1 0 3 b. dte t
  35. 35. ©2007 Pearson Education Asia Chapter 14: Integration 14.7 The Fundamental Theorem of Integral Calculus Example 5 - Finding a Change in Function Values by Definite Integration The Definite Integral of a Derivative • The Fundamental Theorem states that ( ) ( ) ( )afbfdxxf a b −=∫ ' A manufacturer’s marginal-cost function is . If production is presently set at q = 80 units per week, how much more would it cost to increase production to 100 units per week? Solution: The rate of change of c is dc/dq is 26.0 += q dq dc ( ) ( ) ( ) [ ] 112020803200 23.026.080100 100 80 2 100 80 =−= +=+=− ∫ qqdqqcc
  36. 36. ©2007 Pearson Education Asia Chapter 14: Integration 14.8 Approximate Integration14.8 Approximate Integration Trapezoidal Rule • To find the area of a trapezoidal area, we have ( ) ( ) ( ) ( ) ( )( ) ( )[ ] ( ) ./where 12222 2 nb-ah bfhnafhafhafaf h dxxf b a = +−+++++++≈∫ 
  37. 37. ©2007 Pearson Education Asia Chapter 14: Integration 14.8 Approximate Integration Example 1 - Trapezoidal Rule Use the trapezoidal rule to estimate the value of for n = 5. Compute each term to four decimal places, and round the answer to three decimal places. Solution: With n = 5, a = 0, and b = 1, dx x∫ + 1 0 2 1 1 2.0 5 01 = − = − = n ab h
  38. 38. ©2007 Pearson Education Asia Chapter 14: Integration 14.8 Approximate Integration Example 1 - Trapezoidal Rule Solution: The terms to be added are Estimate for the integral is ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) sum fbf fhaf fhaf fhaf fhaf faf = == ==+ ==+ ==+ ==+ == 8373.7 0.50001 2195.18.0242 4706.16.0232 7241.14.0222 9231.12.022 0000.10 ( ) 784.08373.7 2 2.0 1 1 1 0 2 ≈≈ +∫ dx x
  39. 39. ©2007 Pearson Education Asia Chapter 14: Integration 14.8 Approximate Integration Simpson’s Rule • Approximating the graph of f by parabolic segments gives ( ) ( ) ( ) ( ) ( ) ( )( )[ ] ( ) even.isand/where 14224 3 nnabh bfhnafhafhafaf h dxxf b a −= +−+++++++≈∫ 
  40. 40. ©2007 Pearson Education Asia Chapter 14: Integration 14.8 Approximate Integration Example 3 - Demography A function often used in demography (the study of births, marriages, mortality, etc., in a population) is the life-table function, denoted l. In a population having 100,000 births in any year of time, l(x) represents the number of persons who reach the age of x in any year of time. For example, if l(20) = 98,857, then the number of persons who attain age 20 in any year of time is 98,857.
  41. 41. ©2007 Pearson Education Asia Chapter 14: Integration 14.8 Approximate Integration Example 3 - Demography Suppose that the function l applies to all people born over an extended period of time. It can be shown that, at any time, the expected number of persons in the population between the exact ages of x and x + m, inclusive, is given by The following table gives values of l(x) for males and females in the United States. Approximate the number of women in the 20–35 age group by using the trapezoidal rule with n = 3. ( )dttl mx x ∫ +1
  42. 42. ©2007 Pearson Education Asia Chapter 14: Integration 14.8 Approximate Integration Example 3 - Demography Life table:
  43. 43. ©2007 Pearson Education Asia Chapter 14: Integration 14.8 Approximate Integration Example 3 - Demography Solution: We want to estimate Thus The terms to be added are By the trapezoidal rule, ( ) . 35 20 dttl∫ 5 3 2035 = − = − = n ab h ( ) ( ) ( ) ( ) ( ) ( ) sum l l l l = = == == = 90,7755 964,9735 700,1966230,982302 254,197627,982252 857,9820 ( ) ( ) 5.937,476,1775,590 3 5 35 20 =≈∫ dttl
  44. 44. ©2007 Pearson Education Asia Chapter 14: Integration 14.9 Area14.9 Area Example 1 - Using the Definite Integral to Find Area • The width of the vertical element is ∆x. The height is the y-value of the curve. • The area is defined as ( ) ( ) areadxxfxxf b a =→∆ ∫∑ Find the area of the region bounded by the curve and the x-axis. 2 6 xxy −−=
  45. 45. ©2007 Pearson Education Asia Chapter 14: Integration 14.9 Area Example 1 - Using the Definite Integral to Find Area Solution: Summing the areas of all such elements from x = −3 to x = 2, ( )( )326 2 +−−=−−= xxxxy areadxyxy =→∆ ∫∑ − 2 3 ( ) 6 125 3 37 2 9 18 3 8 3 4 12 32 66 2 3 322 3 2 =      − −−−−      −−=       −−=−−= −− ∫ xx xdxxxarea
  46. 46. ©2007 Pearson Education Asia Chapter 14: Integration 14.9 Area Example 3 - Finding the Area of a Region Find the area of the region between the curve y = ex and the x-axis from x = 1 to x = 2. Solution: We have [ ] ( )1 2 1 2 1 −=== ∫ eeedxearea xx
  47. 47. ©2007 Pearson Education Asia Chapter 14: Integration 14.9 Area Example 5 - Statistics Application In statistics, a (probability) density function f of a variable x, where x assumes all values in the interval [a, b], has the following properties: The probability that x assumes a value between c and d, which is written P(c ≤ x ≤ d), where a ≤ c ≤ d ≤ b, is represented by the area of the region bounded by the graph of f and the x-axis between x = c and x = d. ( ) ( ) 1(ii) 0(i) b a = ≥ ∫ dxxf xf
  48. 48. ©2007 Pearson Education Asia Chapter 14: Integration 14.9 Area Example 5 - Statistics Application Hence For the density function f(x) = 6(x − x2), where 0 ≤ x ≤ 1, find each of the following probabilities. ( ) ( )dxxfdxcP d c ∫=≤≤ ( )4 1 0. ≤≤ xPa ( )2 1 . ≥xPb
  49. 49. ©2007 Pearson Education Asia Chapter 14: Integration 14.9 Area Example 5 - Statistics Application Solution: a. b. ( ) ( ) 32 5 32 6 60 4/1 0 32 4/1 0 2 4 1 =      −= −=≤≤ ∫ xx dxxxxP ( ) ( ) 2 1 32 6 6 1 2/1 32 1 2/1 2 2 1 =      −= −=≥ ∫ xx dxxxxP
  50. 50. ©2007 Pearson Education Asia Chapter 14: Integration 14.10 Area between Curves14.10 Area between Curves Example 1 - Finding an Area between Two Curves Vertical Elements • The area of the element is Find the area of the region bounded by the curves y = √x and y = x. Solution: Eliminating y by substitution, [ ] .xyy lowerupper ∆− 1or0 == xx ( ) 6 1 22/3 1 0 22/31 0 =            −=−= ∫ xx dxxxarea
  51. 51. ©2007 Pearson Education Asia Chapter 14: Integration 14.10 Area between Curves Example 3 - Area of a Region Having Two Different Upper Curves Find the area of the region between the curves y = 9 − x2 and y = x2 + 1 from x = 0 to x = 3. Solution: The curves intersect when 2 19 22 ±= +=− x xx ( ) ( ) 22 22 9and1,2,5ofrightFor 1and9,2,5ofleftFor xyxy xyxy lowerupper lowerupper −=+= +=−= [ ] ( ) ( )[ ] ( ) [ ] ( ) ( )[ ] ( ) xx xxxyyxx xx xxxyyxx lowerupper lowerupper ∆−= −−+=∆−== ∆−= +−−=∆−== 82 91,3to2From 28 19,2to0From 2 22 2 22 ( ) ( ) 3 46 8228 3 2 2 2 0 2 =−+−= ∫∫ dxxdxxarea
  52. 52. ©2007 Pearson Education Asia Chapter 14: Integration 14.10 Area between Curves Example 5 - Advantage of Horizontal Elements Find the area of the region bounded by the graphs of y2 = x and x − y = 2. Solution: The intersection points are (1,−1) and (4, 2). The total area is ( ) 2 9 2 2 1 2 =−+= ∫− dyyyarea
  53. 53. ©2007 Pearson Education Asia Chapter 14: Integration 14.11 Consumers’ and Producers’ Surplus14.11 Consumers’ and Producers’ Surplus Example 1 - Finding Consumers’ Surplus and Producers’ Surplus • Consumers’ surplus, CS, is defined as • Producers’ surplus, PS, is defined as The demand function for a product is where p is the price per unit (in dollars) for q units. The supply function is . Determine consumers’ surplus and producers’ surplus under market equilibrium. ( )[ ]dqpqfCS q ∫ −= 0 0 0 ( )[ ]dqqgpPS q ∫ −−= 0 0 0 ( ) qqfp 05.0100 −== ( ) qqgp 1.010 +==
  54. 54. ©2007 Pearson Education Asia Chapter 14: Integration 14.11 Consumers’ and Producers’ Surplus Example 1 - Finding Consumers’ Surplus and Producers’ Surplus Solution: Find the equilibrium point (p0, q0), Consumers’ surplus is Producers’ surplus is ( ) 706001.010Thus 600 05.01001.010 0 0 =+= == −=+ p qq qq ( )[ ] 000,18$ 2 1.060 600 0 2 0 0 0 =      −=−= ∫ q qdqpgpPS q ( )[ ] 9000$ 2 05.030 600 0 2 0 0 0 =      −=−= ∫ q qdqpqfCS q

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