Chapter 6
Integration
Section 1
Antiderivatives and
Indefinite Integrals
Learning Objectives for Section 6.1
Antiderivatives and Indefinite Integrals
 The student will be able to
• formulate problems involving
antiderivatives
• use the formulas and
properties of antiderivatives
and indefinite integrals
• solve applications using
antiderivatives and indefinite
integrals
Barnett/Ziegler/Byleen Business Calculus 12e
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The Antiderivative
Many operations in mathematics have inverses.
• Addition & subtraction
• Multiplication & division
• Powers and roots
• In Calculus, we have inverse functions too!
• Derivative & antiderivative
𝑓′(𝑥) ↔ 𝑓(𝑥)
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The Antiderivative
A function F is an antiderivative of a function f if
F (x) = f (x).
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Example 1
Find a function that has a derivative of 2x.
Answer: x2, since
𝑑 2
(x )
𝑑𝑥
= 2x.
Would these also be correct answers?
x2 + 2 or x2 – 6 or 𝑥 2 + 𝜋
Note that the antiderivative is not unique. To account for all
antiderivatives, we write our answer like this:
x2 + C
(where C represents a constant)
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Example 2
Find all functions that have a derivative of 3𝑥 2 .
Answer: 𝑥 3 + C
Find all functions that have a derivative of 𝑒 𝑥 .
Answer: 𝑒 𝑥 + C
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Theorem 1: Antiderivatives
Conceptual Interpretation: If F(x) and G(x) are both
antiderivatives of f(x), then the graphs of F(x) and G(x)
are vertical translations of each other.
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Example 3
 Let 𝑓 𝑥 = 2𝑥
 Then two antiderivatives of f(x) are:
• 𝐹 𝑥 = 𝑥2 + 1
• 𝐺 𝑥 = 𝑥2 − 2
 The graphs of F(x) and G(x) are vertical translations of
each other.
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Indefinite Integrals
Let f (x) be a function. The family of all functions that are
antiderivatives of f (x) is called the indefinite integral and has
the symbol  f ( x) dx
The symbol  is called an integral sign, and the function f (x) is
called the integrand. The symbol dx indicates that antidifferentiation is performed with respect to the variable x.
𝑓 𝑥 𝑑𝑥 = 𝐹 𝑥 + 𝐶
𝐹 ′ 𝑥 = 𝑓(𝑥)
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Example 4
Evaluate each indefinite integral:
𝑥 2 𝑑𝑥
3
x
2
x
 dx  3  C
Barnett/Ziegler/Byleen Business Calculus 12e
𝑥 4 𝑑𝑥
5
𝑥
𝑥 4 𝑑𝑥 =
+𝐶
5
10
Indefinite Integral
Formulas and Properties
n 1
x
1.  x n dx 
 C , n  1
n 1
2.
(power rule)
x
x
e
dx

e
C

1
3.  dx  ln | x | C
x
4.
5.
 k f ( x) dx  k  f ( x) dx
  f ( x)  g ( x) dx   f ( x) dx   g ( x) dx
It is important to note that property 4 states that a constant
factor can be moved across an integral sign. A variable
factor cannot be moved across an integral sign.
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Examples using the Power Rule
  444 dx = 444𝑥 + 𝐶

𝑑𝑥 = 𝑥 + 𝐶
4
𝑥
  x3 dx =
+𝐶
−3+1
4
𝑥
 5𝑥 −3 𝑑𝑥 = 5
+𝐶
−3 + 1
=
𝑥 −2
5
−2
=
5 −2
− 𝑥
2
+𝐶
+C
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Using the Power Rule
 
x2/3
5
𝑥
dx =
5
3
+𝐶
3
3 5
= 𝑥
5
3
+𝐶
  (x4 + x + x1/2 + 1 + x –1/2) dx =
=
=
𝑥5
5
𝑥5
5
𝑥2
+
2
+
𝑥2
2
Barnett/Ziegler/Byleen Business Calculus 12e
+
𝑥3 2
2
3
3
2
+ 𝑥3
+𝑥+
2
𝑥1 2
1
2
+𝐶
+ 𝑥 + 2𝑥 1
2
+𝐶
13
More Examples
  x –1 dx
not
𝑥0
0
+ C (which is undefined).
Recall Property #3:
  2 x-1 dx = 2
1
𝑑𝑥
𝑥
= ln 𝑥 + 𝐶
1
𝑑𝑥
𝑥
= 2 ln |x| + C
  4 ex dx = 4 𝑒 𝑥 𝑑𝑥
= 4 ex + C
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More Examples
 Sometimes you need to rewrite the problem first.

3
5 𝑢2 𝑑𝑢 = 5 𝑢2 3 𝑑𝑢
=5
𝑢5 3
5
3
= 3𝑢5

𝑧 3 −𝑧 2
𝑑𝑧
𝑧
=
=
=
3
+𝐶
+𝐶
𝑧3
𝑧2
− dz
𝑧
𝑧
𝑧 2 − 𝑧 𝑑𝑧
𝑧3
3
𝑧2
−
2
+𝐶
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Homework
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Homework
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Chapter 6
Integration
Section 1
Antiderivatives and
Indefinite Integrals
(continued)
Learning Objectives for Section 6.1
Antiderivatives and Indefinite Integrals
 The student will be able to
• formulate problems involving
antiderivatives
• use the formulas and
properties of antiderivatives
and indefinite integrals
• solve applications using
antiderivatives and indefinite
integrals
Barnett/Ziegler/Byleen Business Calculus 12e
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Example 1
 Find 𝑓 𝑥 given that
𝑑𝑦
𝑑𝑥 =
𝑑𝑥
𝑑𝑦
𝑑𝑥
= 4𝑒 𝑥 − 2𝑥 3
(4𝑒 𝑥 − 2𝑥 3 ) 𝑑𝑥
4
𝑥
𝑦 = 4𝑒 𝑥 − 2 + 𝐶
4
4
𝑥
𝑦 = 4𝑒 𝑥 − + 𝐶
2
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Application 1
 Find the equation of the curve that passes through (2, 6) if
𝑑𝑦
the slope of the curve is given by = 3𝑥 2 at any point x.
𝑑𝑥
First find f(x):
𝑑𝑦
𝑑𝑥 = 3𝑥 2 𝑑𝑥
𝑑𝑥
Substitute (2, 6) :
𝑥3
𝑦 =3 +𝐶
3
6 = 23 + 𝐶
𝐶 = −2
𝑦 = 𝑥3 − 2
𝑦 = 𝑥3 + 𝐶
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Application 2
 Find the particular antiderivative of the derivative that
satisfies the given condition.
𝑑𝑥 10
=
; x 4 = 10
𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒 (4,10)
𝑑𝑡
𝑡
1
2 +𝐶
10
=
20(4)
𝑑𝑥
−1 2
𝑑𝑡 = 10𝑡
𝑑𝑡
𝑑𝑡
10 = 20(2) + 𝐶
1
𝑡
𝑥 = 10
1
𝑥 = 20𝑡
1
2
+𝐶
2
2
+𝐶
Barnett/Ziegler/Byleen Business Calculus 12e
10 = 40 + 𝐶
−30 = 𝐶
𝑥 = 20𝑡
1
2
− 30
22
Review




𝐶(𝑥)  Cost function
𝐶(𝑥)  Average Cost function
𝐶′(𝑥)  Marginal cost function
𝐶′(𝑥)  Marginal average cost function
 Similar notation for revenue and profit.
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Application 3
 The marginal cost of producing x widgets is given by
𝐶 ′ (𝑥) = 0.3𝑥 2 + 2𝑥
• Find the cost function when the fixed cost is $2,000.
• Find the cost of producing 20 widgets.
𝐶 𝑥 =
3
2
𝑥
𝑥
(0.3𝑥 2 + 2𝑥)𝑑𝑥 = 0.3 + 2 + C
3
2
𝐶(𝑥) = 0.1𝑥 3 + 𝑥 2 + 𝐶
Since the fixed cost is $2,000 this means C(0)=2000
2000 = 0.1(0)3 +03 + 𝐶
𝐶 = 2000
𝐶(𝑥) = 0.1𝑥 3 + 𝑥 2 + 2000
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Application 3 (cont.)
 Find the cost of producing 20 widgets.
𝐶 𝑥 = 0.1𝑥 3 + 𝑥 2 + 2000
𝐶 20 = 0.1(20)3 +(20)2 +2000
= 3,200
𝑇ℎ𝑒 𝑐𝑜𝑠𝑡 𝑜𝑓 𝑝𝑟𝑜𝑑𝑢𝑐𝑖𝑛𝑔 20 𝑤𝑖𝑑𝑔𝑒𝑡𝑠 𝑖𝑠 $3,200.
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Application 4
 There are 64,000 subscribers to the online fashion magazine
Vogue. Due to competition of a new magazine TO Trends,
the number of Vogue subscribers is expected to decrease at
the rate of 𝑁 ′ 𝑥 = −600𝑥 1 3 per month where x is the
time in months since TO Trends came out. How long will it
take until the number of subscribers to Vogue drops to
46,000?
Find N(x) = number of Vogue subscribers after x months.
𝑁 𝑥 =
4 3
𝑥
+𝐶
−600𝑥 1 3 𝑑𝑥 = −600
4 3
𝑁(𝑥) = −450𝑥 4 3 + 𝐶
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Application 4 (continued)
There were 64,000 subscribers before the new
magazine came out (N=64000, x=0):
64000 = −450(0)4 3 +𝐶
64000 = 𝐶
𝑁(𝑥) = −450𝑥 4 3 + 64000
How long until the number of subscribers drops to
46,000? (N=46000, x=?):
3 4
4 3
4
3
40
=
𝑥
46000 = −450𝑥
+ 64000
−18000 = −450𝑥 4
40 = 𝑥 4
3
3
3 4
403 4 = 𝑥
𝑥 ≈ 15.9
It will take about 15.9 months for the number of Vogue
subscribers to drop to 46,000.
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𝐻𝑜𝑚𝑒𝑤𝑜𝑟𝑘
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