soa exam P Prerequisite
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Transcript of soa exam P Prerequisite
Prerequisites for Exam P/1
PREREQUISITES FOR EXAM P/1 Much of probability and statistics is applied mathematics. The concepts that you need to know for this exam are built on a strong foundation in a select area in mathematics. This document will review many of the mathematical skills that you should be equipped with to ensure unobstructed progress when studying the material.
CONTENTS
1 Algebra ........................................................................................................................ 1
1.1 Quadratic Formula ......................................................................................................................................... 1 1.2 Laws of Exponents ......................................................................................................................................... 1 1.3 Laws of Logarithms ....................................................................................................................................... 2 1.4 Sum of Arithmetic Series ............................................................................................................................. 3 1.5 Sum of Geometric Series .............................................................................................................................. 3 1.6 Matrix Algebra ................................................................................................................................................. 5
2 Geometry .................................................................................................................... 6
2.1 Pythagorean Theorem .................................................................................................................................. 6 2.2 Areas .................................................................................................................................................................... 7
3 Differential Calculus ............................................................................................... 9
3.1 Basic Formulas and Rules ........................................................................................................................... 9 3.2 Chain Rule ........................................................................................................................................................ 10 3.3 Product Rule ................................................................................................................................................... 11 3.4 Quotient Rule ................................................................................................................................................. 11 3.5 Derivative of Absolute Value Functions............................................................................................... 12 3.6 Second Derivative ......................................................................................................................................... 13 3.7 Optimization ................................................................................................................................................... 14
4 Integral Calculus .................................................................................................... 16
4.1 Fundamental Theorem of Calculus ........................................................................................................ 16 4.2 Basic Formulas and Rules ......................................................................................................................... 17 4.3 Integration by Substitution ...................................................................................................................... 18 4.4 Integration by Parts ..................................................................................................................................... 20 4.5 Tabular Integration...................................................................................................................................... 22
5 Multivariate Calculus ........................................................................................... 25
5.1 Partial Derivatives ........................................................................................................................................ 25 5.2 Double Integration ....................................................................................................................................... 27
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1 Algebra 1.1 Quadratic Formula If ππ₯2 + ππ₯ + π = 0, then
π₯ =βπ Β± βπ2 β 4ππ
2π
Example
π₯2 β π₯ β 1 = 0
π = 1, π = β1, π = β1
π₯ =1 Β± β(β1)2 β 4(1)(β1)
2(1) β ππ =π β βπ
π, ππ =
π + βππ
1.2 Laws of Exponents
ποΏ½ β ππ = ποΏ½+π
ποΏ½
ππ= ποΏ½βπ
(ποΏ½)π = ποΏ½π
ποΏ½π = βποΏ½π
π0 = 1, π β 0
ποΏ½ β ποΏ½ = (ππ)οΏ½
ποΏ½
ποΏ½= (
ππ)οΏ½
1ππ
= πβπ
(ππ)βπ
= (ππ)π
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1.3 Laws of Logarithms
π¦ = ππ₯ β π₯ = ln(π¦)
ln(ππ₯) = πln(π₯) = π₯
ln(π₯π) = π ln(π₯)
ln(π₯π¦) = ln(π₯) + ln(π¦)
ln(π₯ + π¦) β ln(π₯) + ln(π¦)
ln (π₯π¦) = ln(π₯) β ln(π¦)
ln(π₯ β π¦) β ln(π₯) β ln(π¦)
Examples
πln(1
1βπ‘) =π
π β π
πβπ₯ β π₯ = πβπ₯ β πln(π₯) = πβπ+π₯π§(π)
π₯βπ₯
+πβ2π₯
π₯4= π₯1β0.5 +
(πβπ₯)2
(π₯2)2 = ππ.π + (πβπ
ππ)π
[π‘π
π‘πβ2β 2 (
1π‘)β1
+ π‘0]
12
= [π‘πβ(πβ2) β 2π‘ + 1]12
= (π‘2 β 2π‘ + 1)12
= [(π‘ β 1)2]12
= π β π ln(πβ(π₯2+π¦2)) = β(ππ + ππ) ln(π₯2 + 4π₯ + 4) = ln(π₯ + 2)2 = π π₯π§(π + π)
ln(π₯2π¦3
6π§) = ln(π₯2π¦3) β ln(6π§)
= [ln(π₯2) + ln(π¦3)] β [ln(6) + ln(π§)] = π π₯π§(π) + π π₯π§(π) β π₯π§(π) β π₯π§(π)
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1.4 Sum of Arithmetic Series The general formula is
ππ =π2
(π1 + ππ), where
π is the number of terms π1 is the first term of the series ππ is the last term of the series
An expression for π is
π =ππ β π1
π+ 1
where π is the common difference, defined as the difference between a term and the term prior to it. This is useful when we have the first and last terms but have to find the number of terms. Example
ππ = 1 + 5 + 9 + β―+ 41
π1 = 1, ππ = 41 , π =41 β 1
4+ 1 = 11
ππ =112
(1 + 41) = πππ
1.5 Sum of Geometric Series Finite The standard formula is
ππ =π(ππ β 1)π β 1
=π(1 β ππ)
1 β π, where
π is the number of terms π is the first term of the series π is the common ratio, i.e., the ratio between a term and the term prior to it
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The next formula is derived from the previous formula but is easier to remember.
ππ =first term β first omitted term
1 β common ratio
The first omitted term is the term that would come after the last term if the series continued. Example
ππ = 1.02 + (1.02)2 + β―+ (1.02)20
π = 1.02, π = 1.02, π = 20 Using the standard formula,
ππ =1.02[(1.02)20 β 1]
1.02 β 1β ππ.ππππ
Using the alternative formula, the first omitted term is the term after (1.02)20, which is (1.02)21. Thus,
ππ =1.02 β (1.02)21
1 β 1.02β ππ.ππππ
Infinite If the common ratio is between β1 and 1 (non-inclusive), then an infinite geometric series converges to zero and its sum is finite, which equals
πβ =π
1 β π
Example
πβ = 0.9 + 0.92 + 0.93 + β―
π = 0.9, π = 0.9
πβ =0.9
1 β 0.9= π
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1.6 Matrix Algebra Suppose we have a two-by-two matrix, π΄, defined as follows:
π΄ = [π ππ π]
The determinant of π΄ is
|π΄| = |π ππ π| = ππ β ππ
Example
The determinant of π΄ = [2 05 β2] is
|π΄| = 2(β2) β 0(5) = βπ
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2 Geometry 2.1 Pythagorean Theorem
Figure 2.1
If π, π, and π are the sides of a right triangle as shown in Figure 2.1, then
π2 + π2 = π2 Example Find π in Figure 2.2.
Figure 2.2
π2 = π2 β π2
= (1 + π₯)2 β (1 β π₯)2 = 1 + 2π₯ β π₯2 β 1 + 2π₯ β π₯2 = 4π₯
β π = β4π₯ = 2βπ₯
c
a
b
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2.2 Areas
Diagram Formula for Area
Rectangle & square
Figure 2.3
π΄ = πβ
Triangle
Figure 2.4
π΄ =12πβ
Parallelogram
Figure 2.5
π΄ = πβ
Trapezoid
Figure 2.6
π΄ = (average of π and π) Γ β
=12
(π + π)β
b
h
b
h
b
h
b
h
a
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Circle
Figure 2.7
π΄ = ππ2
Examples
Figure 2.8
Area of rectangle π΄ = 2 β 5 = ππ
Area of triangle π΄ =12β 2 β 5 = π
Area of parallelogram π΄ = 1 β 3 = π
Area of trapezoid π΄ =12β (1 + 3) β 2 = π
Area of circle π΄ = π β 12 = π
r
80 1 2 3 4 5 6 7
6
1
2
3
4
5
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3 Differential Calculus 3.1 Basic Formulas and Rules
πππ₯
π = 0
πππ₯
ππ₯ = π
πππ₯
π₯π = π β π₯πβ1
πππ₯
ππ₯ = ππ₯
πππ₯
ππ₯ = ππ₯ β ln(π) , π > 0
πππ₯
ln(π₯) =1π₯
, π₯ > 0
πππ₯
[ππ(π₯) + ππ(π₯)] = πποΏ½(π₯) + πποΏ½ (π₯)
πππ₯
π β ππ(π₯) = π β πππ₯
ππ(π₯)
πππ₯
sin(π₯) = cos(π₯)
πππ₯
cos(π₯) = β sin(π₯)
Note: π is a constant. Examples
πππ₯
(2π₯ + 35) = 2 + 0 = π
πππ‘
(π‘ + 2)(2π‘ β 3) =πππ‘
(2π‘2 + π‘ β 6) = ππ + π
πππ₯
(1π₯3) =
πππ₯
(π₯β3) = β3π₯β4 = βπππ
πππ₯
(2βπ₯ β1βπ₯
) =πππ₯
(2π₯0.5 β π₯β0.5) = π₯β0.5 + 0.5π₯β1.5 =πβπ
+π
ππβπ
πππ‘
ln(π‘10) =πππ‘
10 ln(t) = 10 β 1π‘
=πππ
πππ¦
(ππ¦ β 4π¦) = ππ β ππ β π₯π§(π)
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3.2 Chain Rule If π and π are differentiable functions where π¦ = ππ(ππ(π₯)), then
ππ¦ππ₯
= πποΏ½(ππ(π₯)) β πποΏ½ (π₯) This formula can be extended to include more than two differentiable functions. For three functions,
π¦ = ππ (ππ(βπ(π₯))) ππ¦ππ₯
= πποΏ½ (ππ(βπ(π₯))) β πποΏ½ (βπ(π₯)) β βποΏ½ (π₯)
Examples
πππ‘
(4π‘2 + 5π‘)7 = 7(4π‘2 + 5π‘)6 β πππ‘
(4π‘2 + 5π‘)
= π(πππ + ππ)π β (ππ + π) πππ₯
(1
1 β π₯2) =
πππ₯
(1 β π₯2)β1
= β(1 β π₯2)β2 β πππ‘
(1 β π₯2)
= β(1 β π₯2)β2 β (0 β 2π₯)
=ππ
(π β ππ)π
πππ₯
(1 β πβπ₯10) = 0 β
πππ₯
(πβπ₯10)
= β(πβπ₯10) β
πππ₯
(βπ₯
10)
= β(πβπ₯10) β (β
110)
=πππ
πβπππ
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πππ₯
ln(1 + π₯2) =1
1 + π₯2β πππ₯
(1 + π₯2)
=1
1 + π₯2β (0 + 2π₯)
=ππ
π + ππ
πππ
(1 + sinπ)4 = 4(1 + sinπ)3 β πππ
(1 + sinπ)
= 4(1 + sin π)3 β (0 + cos π) = π ππ¨π¬ π½ (π + π¬π’π§ π½)π
3.3 Product Rule Letβs define π’ = ππ(π₯) and π£ = ππ(π₯), where π and π are differentiable functions. If π¦ = π’ β π£, then
ππ¦ππ₯
= π’οΏ½ β π£ + π’ β π£οΏ½ 3.4 Quotient Rule Letβs define π’ = ππ(π₯) and π£ = ππ(π₯), where π and π are differentiable functions. If π¦ = π’
π£,
then
ππ¦ππ₯
=π£ β π’οΏ½ β π’ β π£οΏ½
π£2
Tip: It is not necessary to memorize the quotient rule, because π’
π£ can be written as π’ β π£β1, which
is a product. Then, we can use the product rule instead.
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Example 1
πππ‘
(3π‘ + 1)2(π‘2 + 2) = (π‘2 + 2) β πππ‘
(3π‘ + 1)2 + (3π‘ + 1)2 β πππ‘
(π‘2 + 2)
= (π‘2 + 2) β 2(3π‘ + 1) β πππ‘
(3π‘ + 1) + (3π‘ + 1)2 β 2π‘
= (π‘2 + 2) β 2(3π‘ + 1) β 3 + (3π‘ + 1)2 β 2π‘ = π(ππ + π)(ππ + π) + ππ(ππ + π)π
πππ₯
[(4π₯2 β 3)3
ππ₯] =
πππ₯
[(4π₯2 β 3)3 β πβπ₯]
= πβπ₯ β πππ₯
(4π₯2 β 3)3 + (4π₯2 β 3)3 β πππ₯
πβπ₯
= πβπ₯ β 3(4π₯2 β 3)2 β πππ₯
(4π₯2 β 3) + (4π₯2 β 3)3 β πβπ₯ β πππ₯
(βπ₯)
= πβπ₯ β 3(4π₯2 β 3)2 β 8π₯ + (4π₯2 β 3)3 β πβπ₯ β (β1) = πππ(πππ β π)ππβπ β (πππ β π)ππβπ
3.5 Derivative of Absolute Value Functions We can write any absolute value function as follows:
|ππ(π₯)| = { ππ(π₯), ππ(π₯) β₯ 0
βππ(π₯), ππ(π₯) < 0
Then, the derivative of the absolute value function is the derivative of its individual components. πππ₯
|ππ(π₯)| = { πποΏ½(π₯), ππ(π₯) > 0
βπποΏ½(π₯), ππ(π₯) < 0
The derivative may or may not be defined at ππ(π₯) = 0.
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Example 1
|π₯| = { π₯, π₯ β₯ 0βπ₯, π₯ < 0
πππ₯
|π₯| = { πππ₯
(π₯), π₯ > 0
πππ₯
(βπ₯), π₯ < 0= { π, π > 0
βπ, π < 0
The derivative is undefined at π₯ = 0.
Example 2
|π₯2| = { π₯2, π₯2 β₯ 0
βπ₯2, π₯2 < 0
Because π₯2 cannot be negative, |π₯2| = π₯2 for all π₯. Therefore, πππ₯
|π₯2| = ππ, ββ < π₯ < β 3.6 Second Derivative The second derivative of a function π is the derivative of the first derivative.
π2
ππ₯2ππ(π₯) =
πππ₯
(πππ₯
ππ(π₯))
Example
π¦ =1
1 β π₯
ππ¦ππ₯
=πππ₯
(1
1 β π₯)
=πππ₯
(1 β π₯)β1
= β(1 β π₯)β2 β πππ₯
(1 β π₯)
=1
(1 β π₯)2
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π2π¦ππ₯2
=πππ₯
(ππ¦ππ₯
)
=πππ₯
[1
(1 β π₯)2]
=πππ₯
(1 β π₯)β2
= β2(1 β π₯)β3 β πππ₯
(1 β π₯)
=2
(1 β π₯)3
3.7 Optimization Derivatives can be used to find the maximum or minimum of a function, specifically the local max or min. We will illustrate this application using the following examples. Example 1 Find the local minimum of the function ππ(π₯) = π₯2 β 2π₯ β 5. Step 1: Take derivative of function.
πποΏ½(π₯) = 2π₯ β 2 Step 2: Set derivative equal to zero, and solve for π₯.
πποΏ½(π₯) = 2π₯ β 2 = 0 β π₯ = 1
If the question asks for the value of π₯ that minimizes ππ(π₯), then the answer is π₯ = 1. To find the minimum value of ππ(π₯), we continue with Step 3. Step 3: Substitute π₯ into function.
ππ(1) = (1)2 β 2(1) β 5 = β6 The minimum value of ππ(π₯) is βπ .
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Step 4: Perform second derivative test (optional). If we know for sure that β6 is the minimum and not the maximum point, then we can skip this step. Otherwise, the second derivative test will tell us whether it is a minimum or maximum point. The second derivative of ππ(π₯) is
πποΏ½οΏ½(π₯) = 2 If πποΏ½οΏ½(π₯) is positive at the value of π₯ that we calculated, then the point is a minimum point. If πποΏ½οΏ½(π₯) is negative at the value of π₯ that we calculated, then the point is a maximum point. Because πποΏ½οΏ½(1) = 2 > 0, (1,β6) is a minimum point, as shown in Figure 3.1.
Figure 3.1
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4 Integral Calculus 4.1 Fundamental Theorem of Calculus First Theorem The integral of a continuous function π from π to π is
β« ππ(π₯) ππ₯π
π= πΉπ(π) β πΉπ(π), where πΉπ(π₯) = β«ππ(π₯) ππ₯
Second Theorem For a continuous function π,
πππ₯
β« ππ(π‘) ππ‘π₯
π= π(π₯), π = constant
This is because
πππ₯
β« ππ(π‘) ππ‘π₯
π=
πππ₯
[πΉπ(π₯) β πΉπ(π)]
= ππ(π₯) β 0 = ππ(π₯)
A more general version of the previous equation with functions π’ = ππ(π₯) and π£ = βπ(π₯) in place of π and π₯ is
πππ₯
β« ππ(π‘) ππ‘π£
π’= ππ(π£) β π£οΏ½ β ππ(π’) β π’οΏ½
Because the lower limit of integration is not a constant, its derivative is not zero. Furthermore, the derivatives of π’ and π£ are the result of the chain rule.
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4.2 Basic Formulas and Rules
β«π ππ₯ = ππ₯ + πΆ
β«π₯π ππ₯ =π₯π+1
π + 1+ πΆ, π β β1
β«π₯β1 ππ₯ = ln(π₯) + πΆ
β«πππ₯ ππ₯ =1
π ln(π)πππ₯ + πΆ
β«πππ₯ ππ₯ =1ππππ₯ + πΆ
β«[ππ(π₯) + ππ(π₯)] ππ₯ = β«ππ(π₯) ππ₯ + β«ππ(π₯) ππ₯
β«π β ππ(π₯) ππ₯ = π β β«ππ(π₯) ππ₯
β« sin(π₯)ππ₯ = β cos(π₯) + πΆ
β« cos(π₯)ππ₯ = sin(π₯) + πΆ
Note: π, π, and πΆ are constants. Examples
β« (5π₯3 + 4) ππ₯1
0= [
5π₯3+1
3 + 1+ 4π₯]
0
1
= [54π₯4 + 4π₯]
0
1
= (54
+ 4) β (0 + 0)
=πππ
β« (1
2βπ₯β πβ2.3π₯) ππ₯
3
2= [
π₯β0.5+1
2(β0.5 + 1) +1
2.3πβ2.3π₯]
2
3
= [βπ₯ +1
2.3πβ2.3π₯]
2
3
= (β3 +1
2.3πβ2.3(3)) β (β2 +
12.3
πβ2.3(2))
β π.ππππ
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β«25π‘ππ‘
10
1= [
25
ln(π‘)]1
10
=25
[ln(10) β ln(1)]
β π.ππππ
β« (πβ3π₯ + 35π₯) ππ₯0.1
0= [
πβ3π₯
β3+
35π₯
5 ln(3)]0
0.1
= (πβ3(0.1)
β3+
35(0.1)
5 ln(3)) β (1β3
+1
5 ln(3))
β π.ππππ 4.3 Integration by Substitution Integration by substitution is the reverse of the chain rule. We use integration by substitution if the integrand looks like it is the result of the chain rule. Example 1
β« 2π₯πβπ₯2 ππ₯2
1
Notice that 2π₯ is the derivative of π₯2. So, we make the substitution π’ = π₯2. Then, we convert the variable of integration from π to π. The relationship between ππ₯ and ππ’ comes from the derivative of π’ = π₯2, i.e., ππ’ = 2π₯ ππ₯. We convert the limits of integration by substituting them into π’ = π₯2:
Lower limit: π’ = 11 = 1 Upper limit: π’ = 22 = 4
Therefore,
β« 2π₯πβπ₯2 ππ₯2
1= β« πβπ’ ππ’
4
1
= [βπβπ’]14 = πβ1 β πβ4 β π.ππππ
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Example 2
β«2π‘
(π‘ + 3)4 ππ‘8
1
Identify the substitution π’ = π‘ + 3, where π‘ = π’ β 3. Then, we have ππ’ = ππ‘. The new limits of integration are:
Lower limit: π’ = 1 + 3 = 4 Upper limit: π’ = 8 + 3 = 11
Therefore,
β«2π‘
(π‘ + 3)4 ππ‘8
1= β«
2(π’ β 3)π’4
ππ’11
4
= β« (2π’β3 β 6π’β4)ππ’11
4
= [2π’β2
β2+
6π’β3
3]4
11
= [β1π’2
+2π’3]4
11
= β1
112+
2113
+1
42β
243
β π.ππππ Example 3
β«βπ₯
(1 + βπ₯)3 ππ₯
9
4
Identify the substitution π’ = 1 + βπ₯, where βπ₯ = π’ β 1. Then, we have ππ’ = 1
2βπ₯ππ₯, so that ππ₯ = 2βπ₯ ππ’ = 2(π’ β 1) ππ’.
The new limits of integration are:
Lower limit: π’ = 1 + β4 = 3
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Upper limit: π’ = 1 + β9 = 4 Therefore,
β«βπ₯
(1 + βπ₯)3 ππ₯
9
4= β«
π’ β 1π’3
β 2(π’ β 1) ππ’4
3
= 2β«π’2 β 2π’ + 1
π’3ππ’
4
3
= 2β« (π’β1 β 2π’β2 + π’β3)ππ’4
3
= 2 [ln(π’) + 2π’β1 βπ’β2
2]3
4
= 2 (ln(4) +24β
12 β 42
) β 2 (ln(3) +23β
12 β 32
)
β π.ππππ 4.4 Integration by Parts Integration by parts is the reverse of the product rule.
πππ₯
π’π£ = π’ππ£ππ₯
+ π£ππ’ππ₯
Integrating both sides of the equation,
β«π(π’π£) = β«π’ ππ£ + β«π£ ππ’
π’π£ = β«π’ ππ£ + β«π£ ππ’
Then, the integration by parts formula is
β«π’ ππ£ = π’π£ β β«π£ ππ’
In general, integration by parts is applicable if the integrand is a product of two terms, one of which becomes zero when differentiated repeatedly.
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Example 1
β« π₯πβπ₯ ππ₯β
0
The first step is to identify π’ and ππ£, where π’ is the term to be differentiated, and ππ£ is the term to be integrated. In this example
π’ = π₯, ππ£ = πβπ₯ ππ₯ Then, we find ππ’ and π£.
ππ’ = ππ₯, π£ = βπβπ₯ Next, we apply the integration by parts formula as follows:
β« π₯πβπ₯ ππ₯β
0= [π₯ β (βπβπ₯)]0β β β« βπβπ₯ ππ₯
β
0
= [βπ₯πβπ₯]0β β [πβπ₯]0β = [βπ₯πβπ₯ β πβπ₯]0β = (0 β 0) β (0 β 1) = π
Example 2
β« (3π₯2 + 1)πβπ₯10 ππ₯
β
0
The first step is to identify π’ and ππ£, where π’ is the term to be differentiated, and ππ£ is the term to be integrated. In this example
π’ = (3π₯2 + 1), ππ£ = πβπ₯10 ππ₯
Then, we find ππ’ and π£.
ππ’ = 6π₯ ππ₯, π£ = β10πβπ₯10
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Next, we apply the integration by parts formula as follows:
β« π₯πβπ₯ ππ₯β
0= [(3π₯2 + 1) β (β10πβ
π₯10)]
0
ββ β« 6π₯ β (β10πβ
π₯10) ππ₯
β
0
= [β10πβπ₯10(3π₯2 + 1)]
0
β+ 60β« π₯πβ
π₯10 ππ₯
β
0
To evaluate the integral in blue, we use integration by parts a second time.
π’ = π₯, ππ£ = πβπ₯10 ππ₯
ππ’ = ππ₯, π£ = β10πβπ₯10
β« π₯πβπ₯10 ππ₯
β
0= [π₯ β (β10πβ
π₯10)]
0
ββ β« β10πβ
π₯10 ππ₯
β
0
= [β10π₯πβπ₯10]
0
ββ [100πβ
π₯10]
0
β
= [β10π₯πβπ₯10 β 100πβ
π₯10]
0
β
= (0 β 0) β (0 β 100) = 100
In our final step,
β« π₯πβπ₯ ππ₯β
0= [(3π₯2 + 1) β (β10πβ
π₯10)]
0
β+ 60(100)
= 0 β 1 β (β10) + 6,000 = π,πππ
4.5 Tabular Integration Integration by parts can sometimes be messy to evaluate, which makes it prone to errors, especially when it needs to be done more than once. Tabular integration is an alternative to integration by parts that keeps our calculations organized. Besides, it takes less time and is less prone to errors. Tabular integration is best explained with examples. We will rework the two previous examples to show that integration by parts and tabular integration are equivalent.
Prerequisites for Exam P/1
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Example 1
β« π₯πβπ₯ ππ₯β
0
The first step is to choose the term that would be differentiated if were to do integration by parts, i.e., π’. We write this term in the first row of the first column of the table (see Table 4.1). In this example, this term is π₯. Then, we differentiate π₯ repeatedly until we get zero, while writing down each derivative in a column. The derivative of π₯ is 1, and the derivative of 1 is 0. Next, we write the term that would be integrated if we were to do integration by parts in the first row of the second column. This term is πβπ₯. Then, we integrate πβπ₯ repeatedly until we have the same number of rows in both columns. Then, we draw diagonal lines from the first row of the first column to the second row of the second column, and from the second row of the first column to the third row of the second column. If we had more rows, we would continue drawing lines in this matter until we get to the last row of the second column. The next step is to label the lines with alternating signs, starting with β+β for the first line. We multiply the terms that are connected by the lines and then multiply by 1 if the line has a plus sign, or β1 if the line has a minus sign. We do the same for each pair of terms and add them to get the indefinite integral.
π₯
πβπ₯ + 1 βπβπ₯ β 0 πβπ₯
Table 4.1
Therefore,
β« π₯πβπ₯ ππ₯β
0= [+1 β π₯ β (βπβπ₯) + (β1) β 1 β πβπ₯]0β
= [βπ₯πβπ₯ β πβπ₯]0β = (0 β 0) β (0 β 1) = π
Prerequisites for Exam P/1
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Example 2
β« (3π₯2 + 1)πβπ₯10 ππ₯
β
0
The term to be differentiated is 3π₯2 + 1. The term to be integrated is πβπ₯10.
3π₯2+ 1
πβπ₯10 +
6π₯ β10πβπ₯10 β
6 100πβπ₯10 +
0 β1000πβπ₯10
Table 4.2
Using Table 4.2,
β« (3π₯2 + 1)πβπ₯10 ππ₯
β
0
= [+1 β (3π₯2 + 1) β (β10πβπ₯10) + (β1) β 6π₯ β 100πβ
π₯10 + (+1) β 6 β (β1000πβ
π₯10)]
0
β
= [β10(3π₯2 + 1)πβπ₯10 β 600π₯πβ
π₯10 β 6000πβ
π₯10]
0
β
= β0 β 0 β 0 + 10 + 0 + 6,000 = π,πππ
Prerequisites for Exam P/1
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5 Multivariate Calculus 5.1 Partial Derivatives When taking the partial derivative of a multivariate function with respect to a variable, π₯, we treat all other variables as constants.
Partial derivative of ππ,οΏ½(π₯,π¦) with respect to π₯ πππ₯
ππ,οΏ½(π₯, π¦)
Partial derivative of ππ,οΏ½(π₯,π¦) with respect to π¦ πππ¦
ππ,οΏ½(π₯,π¦)
Second order partial derivative of ππ,οΏ½(π₯,π¦) with respect to π₯
π2
ππ₯2ππ,οΏ½(π₯,π¦) =
πππ₯
(πππ₯
ππ,οΏ½(π₯, π¦))
Second order partial derivative of ππ,οΏ½(π₯,π¦) with respect to π¦
π2
ππ¦2ππ,οΏ½(π₯,π¦)
Second order partial derivative of ππ,οΏ½(π₯,π¦) with respect to π₯ and π¦
π2
ππ₯ ππ¦ππ,οΏ½(π₯,π¦) =
πππ₯
(πππ¦
ππ,οΏ½(π₯,π¦))
π2
ππ¦ ππ₯ππ,οΏ½(π₯,π¦) =
πππ¦
(πππ₯
ππ,οΏ½(π₯,π¦))
π2
ππ₯ ππ¦ππ,οΏ½(π₯,π¦) =
π2
ππ¦ ππ₯ππ,οΏ½(π₯, π¦)
Example 1
ππ,οΏ½(π₯,π¦) = π₯2π¦
πππ₯
(π₯2π¦) = πππ Treat π¦ as constant.
πππ¦
(π₯2π¦) = ππ Treat π₯ as constant.
π2
ππ₯2(π₯2π¦) =
πππ₯
(2π₯π¦) = ππ Differentiate π
ππ₯ with respect to π₯.
Treat π¦ as constant.
Prerequisites for Exam P/1
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π2
ππ¦2(π₯2π¦) =
πππ¦
(π₯2) = π Differentiate π
ππ¦ with respect to π¦.
Treat π₯ as constant.
π2
ππ₯ ππ¦(π₯2π¦) =
πππ₯
(π₯2) = ππ Differentiate π
ππ¦ with respect to π₯.
Treat π¦ as constant. π2
ππ¦ ππ₯(π₯2π¦) =
πππ¦
(2π₯π¦) = ππ Differentiate π
ππ₯ with respect to π¦.
Treat π₯ as constant. π2
ππ₯ ππ¦(π₯2π¦) =
π2
ππ¦ ππ₯(π₯2π¦) Order of differentiation is reversible.
Example 2
ππ,π(π , π‘) = π2π 2+π π‘+3π‘2
πππ
(π2π 2+π π‘+3π‘2) = π2π 2+π π‘+3π‘2 β πππ
(2π 2 + π π‘ + 3π‘2)
= π2π 2+π π‘+3π‘2 β (4π + π‘ + 0)
= (ππ + π)ππππ+ππ+πππ πππ‘(π2π 2+π π‘+3π‘2) = π2π 2+π π‘+3π‘2 β
πππ‘
(2π 2 + π π‘ + 3π‘2)
= π2π 2+π π‘+3π‘2 β (0 + π + 6π‘)
= (π + ππ)ππππ+ππ+πππ π2
ππ 2(π2π 2+π π‘+3π‘2) =
πππ
[(4π + π‘)π2π 2+π π‘+3π‘2]
= π2π 2+π π‘+3π‘2 β πππ
(4π + π‘) + (4π + π‘) β πππ
(π2π 2+π π‘+3π‘2)
= π2π 2+π π‘+3π‘2 β (4 + 0) + (4π + π‘) β π2π 2+π π‘+3π‘2 β (4π + π‘ + 0)
= πππππ+ππ+πππ + (ππ + π)πππππ+ππ+πππ π2
ππ‘2(π2π 2+π π‘+3π‘2) =
πππ‘[(π + 6π‘)π2π 2+π π‘+3π‘2]
= π2π 2+π π‘+3π‘2 β πππ‘
(π + 6π‘) + (π + 6π‘) β πππ‘(π2π 2+π π‘+3π‘2)
= π2π 2+π π‘+3π‘2 β (0 + 6) + (π + 6π‘) β π2π 2+π π‘+3π‘2 β (0 + π + 6π‘)
= πππππ+ππ+πππ + (π + ππ)πππππ+ππ+πππ
Prerequisites for Exam P/1
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5.2 Double Integration We often use double integrals to calculate probabilities and moments. When evaluating double integrals, we always start with the inner integral and work our way out to the outer integral. There are two cases to consider: Case 1: Region of integration is rectangular. Case 2: Region of integration is not rectangular. Case 1 In the first case, double integrals are easier to evaluate. Besides, the order of integration is reversible without the need to redefine the limits of integration. In mathematical notation, this means
β« β« ππ,οΏ½(π₯,π¦) ππ¦π
πππ₯
π
π= β« β« ππ,οΏ½(π₯,π¦) ππ₯
π
πππ¦
π
π
Figure 5.1 shows a rectangular region of integration.
Figure 5.1
x0 c d
y
a
b
Prerequisites for Exam P/1
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Example Integrate the following function over the shaded region in Figure 5.2.
ππ,οΏ½(π₯,π¦) =π₯π¦
2500, 0 < π₯ < 10, 0 < π¦ < 10
Figure 5.2
If we chose π¦ as the variable for the inner integral, then the double integral is
β« β«π₯π¦
2500ππ¦
6
2ππ₯
7
1
Starting from the inside out, the inner integral is with respect to π¦, so we treat π₯ as a constant when performing the inner integral. Also, after performing the inner integral, we substitute the limits into π¦, not π₯, because the inner integral is with respect to π¦. Once we evaluate the inner integral, the double integral becomes a single integral that we evaluate in the usual way.
β« β«π₯π¦
2500ππ¦
6
2ππ₯
7
1= β« [
π₯π¦2
5000]π¦=2
π¦=6
ππ₯7
1
= β«π₯
5000(62 β 22) ππ₯
7
1
x0 1 7
y
2
6
Prerequisites for Exam P/1
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=32
5000β« π₯ ππ₯7
1
=16
5000[π₯2]17
=16
5000(72 β 12)
= π.ππππ If we evaluate the same double integral in the reverse order of integration, we will get the same answer.
β« β«π₯π¦
2500ππ₯
7
1ππ¦
6
2= β« [
π₯2π¦5000
]π₯=1
π₯=7
ππ¦6
2
= β«π¦
5000(72 β 12) ππ¦
7
1
=48
5000β« π¦ ππ¦6
2
=24
5000[π¦2]26
=24
5000(62 β 22)
= π.ππππ Case 2 In the second case, double integrals are trickier. They are reversible, but the limits of integration have to be redefined. We will demonstrate this with the following example. Example 1 Integrate the following function over the shaded region in Figure 5.3.
ππ,οΏ½(π₯,π¦) = πβ(π₯+π¦), π₯ > 0, π¦ > 0
Prerequisites for Exam P/1
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Figure 5.3
Step 1: Draw a diagram that satisfies all the constraints. In this example, the diagram has been given. However, on the exam, you often have to draw your own diagram. The constraints of the region are π₯ > 0, π¦ > 0, and π₯ + π¦ < 2. The region that satisfies these constraints is the shaded region in Figure 5.3. Note that Step 2 and Step 3 can be interchanged.
Step 2: Choose one variable for the outer integral. If we choose π₯, then the limits of the outer integration would be the minimum and maximum value of π₯ in the region. From the diagram, we see that the minimum value of π₯ is 0 and the maximum value of π₯ is 2. Thus, the lower limit is 0 and the upper limit is 2.
β« (β«ππ,οΏ½(π₯,π¦)ππ¦)ππ₯2
0
Likewise, if we choose π¦, then the limits of the outer integration would be the minimum and maximum value of π¦ in the region, which are 0 and 2, respectively.
β« (β«ππ,οΏ½(π₯,π¦)ππ₯) ππ¦2
0
x0 1 2
y
1
2
y
xοΌy= 2
Prerequisites for Exam P/1
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Both integrals will produce the same answer, but the variable that you choose for the outer integration will determine the order of integration; so you should choose the variable that will not require separating the region of integration, if possible. Step 3: Determine the limits of the inner integral. The third step is to define the rest of the integral. The variable for the inner integral will automatically be the variable not chosen for the outer integral. For example, if we chose π₯ for the outer integral, then the inner integral will be with respect to π¦. To determine the limits of integration when the integration is on π¦, draw a line parallel to the π¦-axis that cuts across the region of integration, as shown in Figure 5.4. The limits are the π¦-coordinates of the intersection points of the vertical line that was just drawn and the top and bottom bounds of the region.
Figure 5.4
In this example, the lower limit is π¦ = 0 because the vertical line intersects the lower bound at 0. Similarly, the upper limit is π¦ = 2 β π₯ because the vertical line intersects the upper bound at 2 β π₯. Since the vertical line can be drawn at any arbitrary location within the range of values of π₯, we always express the π¦-coordinates in terms of π₯, or the variable chosen for the outer integral. Therefore, our double integral is
x0 1 2
y
1
2
y
xοΌy= 2
y = 2 β x
y = 0
Prerequisites for Exam P/1
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β« (β«ππ,οΏ½(π₯,π¦)ππ¦)ππ₯2
0= β« (β« ππ,οΏ½(π₯, π¦) ππ¦
2βπ₯
0)ππ₯
2
0
On the other hand, if we chose π¦ for the outer integral, then the inner integral will be for the variable π₯. In this case, we draw a line parallel to the π₯-axis that cuts across the region, shown below in Figure 5.5.
Figure 5.5
We determine the limits of integration in the same way as before except the limits are now the π₯-coordinates of the intersection points of the horizontal line and the left and right bounds of the region, expressed in terms of π¦. In this example, the horizontal line intersects the left bound at π₯ = 0 and the right bound at π₯ = 2 β π¦. Therefore, our double integral is
β« (β«ππ,οΏ½(π₯,π¦)ππ₯) ππ¦2
0= β« (β« ππ,οΏ½(π₯,π¦) ππ₯
2βπ¦
0)ππ¦
2
0
Step 4: Evaluate the double integral. The last step is to evaluate the double integral. There are 2 possible orders of integration, ππ¦ ππ₯ or ππ₯ ππ¦. Regardless, we always start with the inner integral and work our way out.
x0 1 2
y
1
2
y
xοΌy= 2
x = 2 β yx = 0
Prerequisites for Exam P/1
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We will start with the ππ¦ ππ₯ case.
β« (β« ππ,οΏ½(π₯,π¦) ππ¦2βπ₯
0)ππ₯
2
0= β« (β« πβ(π₯+π¦) ππ¦
2βπ₯
0)ππ₯
2
0
= β« [βπβ(π₯+π¦)]π¦=0π¦=2βπ₯
ππ₯2
0
= β« [βπβ(π₯+2βπ₯) β πβ(π₯+0)]ππ₯2
0
= β« (βπβ2 + πβπ₯)ππ₯2
0
= [βπβ2π₯ β πβπ₯]02 = (β2πβ2 β πβ2) β (0 β πβ0) = π β ππβπ
Next, we will show that the ππ₯ ππ¦ case produces the same answer.
β« (β« ππ,οΏ½(π₯,π¦) ππ₯2βπ¦
0) ππ¦
2
0= β« (β« πβ(π₯+π¦) ππ₯
2βπ¦
0) ππ¦
2
0
= β« [βπβ(π₯+π¦)]π₯=0π₯=2βπ¦
ππ¦2
0
= β« [βπβ(2βπ¦+π¦) β πβ(0+π¦)]ππ¦2
0
= β« (βπβ2 + πβπ¦)ππ¦2
0
= [βπβ2π¦ + πβπ¦]02 = (β2πβ2 β πβ2) β (0 β πβ0) = π β ππβπ