Chapter 13 – 2b Integration by Substitution
Example 1. The weekly marginal cost of producing x pairs of tennis shoes is given by
𝐶 ′ (𝑥) = 12 +
500
𝑥+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.
𝐶(𝑥) = ∫ (12 +
500
1
) 𝑑𝑥 = 12 ∫ 𝑑𝑥 + 500 ∫
𝑑𝑥
𝑥+1
𝑥+1
𝐶(𝑥) = 12𝑥 + 500 ln(𝑥 + 1) + 𝐶 = 12𝑥 + 500 ln(𝑥 + 1) + 2,000
𝐶̅ (1,000) =
𝐶(1,000) 12 ∙ 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
𝑆 ′ (𝑡) = 10 − 10𝑒 −0.1𝑡
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.
𝑆(𝑡) = ∫(10 − 10𝑒 −0.1𝑡 )𝑑𝑡 = 10 ∫ 𝑑𝑡 + 100 ∫ 𝑒 −0.1𝑡 (−0.1)𝑑𝑡 = 10𝑡 + 100𝑒 −0.1𝑡 + 𝐶
𝑆(0) = 10 ∙ 0 + 100𝑒 −0.1∙0 + 𝐶 = 0 ⟹ 𝐶 = −100
𝑆(12) = 10 ∙ 12 + 100𝑒
Sales
− 100
−0.1∙12
200
− 100 ≈ 50 million dollars
𝑆(24) = 10 ∙ 24 + 100𝑒 −0.1∙24 − 100 ≈ 149 million dollars
Millions
𝑆(𝑡) = 10𝑡 + 100𝑒
−0.1𝑡
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
𝑅(𝑡) =
5𝑡 + 105
𝑡+1
0 ≤ 𝑡 ≤ 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?
𝑄(𝑡) = ∫
5𝑡 + 105
100
1
𝑑𝑡 = ∫ (5 +
) 𝑑𝑡 = 5 ∫ 𝑑𝑡 + 100 ∫
𝑑𝑡 = 5𝑡 + 100 ln(𝑡 + 1) + 𝐶
𝑡+1
𝑡+1
𝑡+1
𝑄(0) = 0 ⟹ 𝐶 = 0 ⟹ 𝑄(𝑡) = 5𝑡 + 100 ln(𝑡 + 1)
Oil Production
Millions of Barrels
𝑄(10) = 5 ∙ 10 + 100 ln(10 + 1) ≈ 290 million barrels of oil
𝑄(20) = 5 ∙ 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
𝑑𝑁
2,000𝑡
=−
𝑑𝑡
1 + 𝑡2
0 ≤ 𝑡 ≤ 14
What is the bacteria count per milliliter of water one week after the treatment? Two weeks after the treatment?
𝑁(𝑡) = ∫ −
2,000𝑡
1+𝑡 2
𝑑𝑡 = −1,000 ∫
1
1+𝑡 2
2𝑡 𝑑𝑡 = −1000 ln(1 + 𝑡 2 ) + 𝐶
Bacteria Count in Lake Water
𝑁(𝑡) = 5,500 − 1000 ln(1 + 𝑡 2 )
𝑁(7) = 5,500 − 1000 ln(1 + 7
2)
≈ 1,588 bacteria per milliliter
𝑁(14) = 5,500 − 1000 ln(1 + 142 ) ≈ 217 bacteria per milliliter
Bacteria per Milliliter
𝑁(0) = −1000 ln(1 + 02 ) + 𝐶 = 5,500 ⟹ 𝐶 = 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 𝑀′ (𝑡) = 0.2𝑒 0.1𝑡 grams per hour. If the starting culture had a
mass of 2 grams, what will be the mass after 8 hours? After 12 hours?
𝑀(𝑡) = ∫ 0.2𝑒 0.1𝑡 𝑑𝑡 = 2 ∫ 𝑒 0.1𝑡 (0.1𝑑𝑡) = 2𝑒 0.1𝑡 + 𝐶 = 2𝑒 0.1𝑡
Yeast Culture
𝑀(8) = 2𝑒 0.1(8) ≈ 4.45 grams
15
Grams
𝑀(12) = 2𝑒 0.1(12) ≈ 6.64 grams
20
10
5
0
0
5
10
Hours
15
20