1Slide 4.1 - 1Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-WesleyCopyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-WesleyThe Area under a GraphOBJECTIVES Use the area under a graph to find total cost. Use rectangles to approximate the area under a graph.4.1Slide 4.1 - 3Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-WesleyExample 1: Covington Bakers determines that the cost, in cents per pound, of its sourdough bread iswhere x is the number of pounds of bread baked. Find the total cost of baking 140 lb of bread. Disregard fixed costs.We will use the area under the graph of c(x) to find the total cost of baking 140 lb of bread.4.1 The Area Under a Graphc(x)=−0.2x+125,x<5002Slide 4.1 - 4Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-WesleyExample 1 (continued):Note that the area is a trapezoid.4.1 The Area Under a GraphSlide 4.1 - 5Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-WesleyExample 1 (continued):4.1 The Area Under a Graph=12h b1+ b2( )=12⋅140 c 0( )+ c 140( )()=12⋅140 125 + 97( )= 15,540¢ or $155.40Total cost of baking140 lb. bread = Area of TrapezoidSlide 4.1 - 6Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-WesleyExample 1 (concluded):Check by examining the units involved:Cost per pound · Number of pounds¢/lb · lb = ¢Covington’s total cost of baking 140 lb of sourdough bread is $155.40.4.1 The Area Under a Graph3Slide 4.1 - 7Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-WesleyExample 2: Raggs, Ltd., determines that the marginal cost, in dollars, when suits are manufactured isIgnoring fixed costs, approximate the total cost of producing 400 suits by using four rectangles.4.1 The Area Under a Graph′C (x)=0.0008x2−0.45x+100.Slide 4.1 - 8Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-WesleyExample 2 (continued):SolutionWe divide the interval [0, 600] into four subintervals, each of length ∆x = 400/4=100. To determine the height of each rectangle, we use the left endpoint of each subinterval. 4.1 The Area Under a GraphSlide 4.1 - 9Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-WesleyExample 2 (continued):4.1 The Area Under a Graph4Slide 4.1 - 10Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-WesleyExample 2 (concluded):Then, we have the total cost of producing 400 suits isTotal cost of producing 400 suits is approximately $24,200.4.1 The Area Under a Graph≈Area I + Area II + Area III + Area IV.≈ ′C (0)⋅100 + ′C (100)⋅100+ ′C (200) ⋅100 + ′C (300) ⋅100≈ $100 ⋅100 + $63⋅100 + $42 ⋅100 + $37 ⋅100≈$10,000+$6300+$4200+3700≈$24, 200Slide 4.1 - 11Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-WesleyRiemann Sums:In Example 2, we approximated the total cost of manufacturing 400 suits. How might we make that approximation more exact? We do so by using more rectangles that have a narrower width. This type of calculation is most easily accomplished with the use of summation notation.4.1 The Area Under a GraphSlide 4.1 - 12Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-WesleyRiemann Sums (continued):In the following figure, [a, b] is divided into four subintervals, each having width ∆x = (b – a)/4.The heights of the rectangles are f (x1), f (x2), f (x3) and f (x4).4.1 The Area Under a Graph5Slide 4.1 - 13Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-WesleyRiemann Sums (concluded):The area of the region under the curve is approximately the sum of the areas of the four rectangles:We can denote this sum with summation, or sigma, notation, which uses the Greek capital letter sigma, or Σ:This is read “the sum of the product f (xi)∆x from i = 1 to i = 4.” To recover the original expression, we substitute the numbers 1 through 4 successively for i in f (xi)∆x and write plus signs between the results.4.1 The Area Under a Graphf x1()∆x+f x2()∆x+f x3()∆x+f x4()∆x.f xi( )∆xi=14∑, or i=14f xi( )∆x∑.Slide 4.1 - 14Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-WesleyExample 3: Write summation notation for2 + 4 + 6 + 8 + 10.Note that we are adding consecutive values of 2.4.1 The Area Under a Graph2 + 4 + 6 + 8 + 10 = 2ii=15∑Slide 4.1 - 15Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-WesleyExample 4: Write summation notation for:4.1 The Area Under a Graphg(x1)∆x+g(x2)∆x+⋅⋅⋅+g(x19)∆x.g(x1)∆x + g(x2)∆x + ⋅⋅⋅ + g(x19)∆x = g(xi)∆xi=119∑6Slide 4.1 - 16Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-WesleyExample 5: Express without using summation notation.4.1 The Area Under a Graph3ii=14∑3i= 31+ 32+ 33+ 34= 120i=14∑Slide 4.1 - 17Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-WesleyExample 6: Express without using summation notation.4.1 The Area Under a Graphh(xi)∆xi=130∑h(xi)∆x = h(x1)∆x + h(x2)∆x + ⋅⋅⋅ + h(x30)∆xi=130∑Slide 4.1 - 18Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-WesleyExample 7: Consider the graph of:over the interval [0, 600].a) Approximate the area by dividing the interval into 6 subintervals.b) Approximate the area by dividing the interval into 12 subintervals.4.1 The Area Under a Graphf (x)=600x−x27Slide 4.1 - 19Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-WesleyExample 7 (continued):a) We divide [0, 600] into 6 intervals of sizewith xiranging from x1= 0 to x6= 500.4.1 The Area Under a Graph∆x =600−06= 100,Slide 4.1 - 20Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-WesleyExample 7 (continued):Thus, the area under the curve is approximately4.1 The Area Under a Graphf (xi)∆xi=16∑= f (0)⋅100 + f (100) ⋅100 + f (200)⋅100+ f (300)⋅100 + f (400)⋅100 + f (500) ⋅100= 0 ⋅100 + 50,000 ⋅100 + 80,000 ⋅100+90,000 ⋅100 + 80,000 ⋅100 + 50,000 ⋅100= 35,000,000Slide 4.1 - 21Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-WesleyExample 7 (continued):b) We divide [0, 600] into 12 intervals of sizewith xiranging from x1= 0 to x12= 550.4.1 The Area Under a Graph∆x =600−012= 50,8Slide 4.1 - 22Copyright © 2008