Chapter 13 – 2b Integration by Substitution Example 1. The weekly marginal cost of producing x pairs of tennis shoes is given by πΆ ′ (π₯) = 12 + 500 π₯+1 Find the cost function assuming fixed costs of $2,000 per week. Find the average cost per pair of shoes if 1,000 pairs of shoes are produced per week. πΆ(π₯) = ∫ (12 + 500 1 ) ππ₯ = 12 ∫ ππ₯ + 500 ∫ ππ₯ π₯+1 π₯+1 πΆ(π₯) = 12π₯ + 500 ln(π₯ + 1) + πΆ = 12π₯ + 500 ln(π₯ + 1) + 2,000 πΆΜ (1,000) = πΆ(1,000) 12 β 1,000 + 500 ln(1,001) + 2,000 = ≈ $17.45 a pair 1,000 1,000 $150 $20,000 $120 $15,000 $90 $10,000 $60 $5,000 $30 $0 Average Cost per Pair Total Cost of Production Production Cost $25,000 Cost Avg Cost $0 0 250 500 750 1000 1250 1500 Number of Pairs of Tennis Shoes Example 2. An automobile company is ready to introduce a new line of cars through a national sales campaign. After test marketing the line in a carefully selected city, the marketing research department estimates that, for the next two years, total sales (in millions of dollars) will increase at the monthly rate of π ′ (π‘) = 10 − 10π −0.1π‘ where t is the time in months. Find the total sales function assuming that there are no sales at the beginning of the campaign. Find the estimated sales for the first and second years of the campaign. π(π‘) = ∫(10 − 10π −0.1π‘ )ππ‘ = 10 ∫ ππ‘ + 100 ∫ π −0.1π‘ (−0.1)ππ‘ = 10π‘ + 100π −0.1π‘ + πΆ π(0) = 10 β 0 + 100π −0.1β0 + πΆ = 0 βΉ πΆ = −100 π(12) = 10 β 12 + 100π Sales − 100 −0.1β12 200 − 100 ≈ 50 million dollars π(24) = 10 β 24 + 100π −0.1β24 − 100 ≈ 149 million dollars Millions π(π‘) = 10π‘ + 100π −0.1π‘ 150 100 50 0 0 During the second year, the total sales will be about 99 million dollars. 3 6 9 12 15 18 21 24 Months Example 3. Using data from the first 3 years of production and data from geological studies, the management of an oil company estimates that the rate of production for a given oil field is π (π‘) = 5π‘ + 105 π‘+1 0 ≤ π‘ ≤ 20 where t is the time in years since pumping began and R(t) is the instantaneous rate of production in millions of barrels per year. Find the equation for the total production Q(t). How much oil will be produced the first 10 years? The last 10 years? π(π‘) = ∫ 5π‘ + 105 100 1 ππ‘ = ∫ (5 + ) ππ‘ = 5 ∫ ππ‘ + 100 ∫ ππ‘ = 5π‘ + 100 ln(π‘ + 1) + πΆ π‘+1 π‘+1 π‘+1 π(0) = 0 βΉ πΆ = 0 βΉ π(π‘) = 5π‘ + 100 ln(π‘ + 1) Oil Production Millions of Barrels π(10) = 5 β 10 + 100 ln(10 + 1) ≈ 290 million barrels of oil π(20) = 5 β 20 + 100 ln(20 + 1) ≈ 404 million barrels of oil In the last 10 years, the oil field will produce about 115 million barrels of oil. 500 400 300 200 100 0 0 4 8 12 16 20 Years Example 4. A contaminated lake is found to contain 5,500 harmful bacteria per milliliter of water. The Parks and Recreation Department decides to treat the lake water with bactericide. The rate of change in the number of harmful bacteria per milliliter of water t days after the treatment is given by ππ 2,000π‘ =− ππ‘ 1 + π‘2 0 ≤ π‘ ≤ 14 What is the bacteria count per milliliter of water one week after the treatment? Two weeks after the treatment? π(π‘) = ∫ − 2,000π‘ 1+π‘ 2 ππ‘ = −1,000 ∫ 1 1+π‘ 2 2π‘ ππ‘ = −1000 ln(1 + π‘ 2 ) + πΆ Bacteria Count in Lake Water π(π‘) = 5,500 − 1000 ln(1 + π‘ 2 ) π(7) = 5,500 − 1000 ln(1 + 7 2) ≈ 1,588 bacteria per milliliter π(14) = 5,500 − 1000 ln(1 + 142 ) ≈ 217 bacteria per milliliter Bacteria per Milliliter π(0) = −1000 ln(1 + 02 ) + πΆ = 5,500 βΉ πΆ = 5,500 6000 5000 4000 3000 2000 1000 0 0 2 4 6 8 10 Days after Treatment 12 14 Example 5. A yeast culture is growing at the rate of π′ (π‘) = 0.2π 0.1π‘ grams per hour. If the starting culture had a mass of 2 grams, what will be the mass after 8 hours? After 12 hours? π(π‘) = ∫ 0.2π 0.1π‘ ππ‘ = 2 ∫ π 0.1π‘ (0.1ππ‘) = 2π 0.1π‘ + πΆ = 2π 0.1π‘ Yeast Culture π(8) = 2π 0.1(8) ≈ 4.45 grams 15 Grams π(12) = 2π 0.1(12) ≈ 6.64 grams 20 10 5 0 0 5 10 Hours 15 20