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6.1 Matrices
• Definition: A Matrix A is a rectangular array of the formA11 A12 · · · A1n
A21 A22 · · · A2n... ... ...
Am1 Am2 · · · Amn
• The size of A is m× n, where m is the number of rows and n is the number of
columns.
• The entry Aij locates in the ith row and the jth column.
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6.1 Matrices
• Equality of two matrices: A = B if and only if
1. A and B have the same size.2. The corresponding entries are equal: Aij = Bij.
• Transpose of a matrix:
1. If A has the size m× n, then AT has the size n×m.2. The ith row of AT is the ith column of A.3. (AT )T = A.
• Special matrices:
1. zero matrix.2. square matrix: diagonal matrix; upper/lower triangular matrix.
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6.2 Matrix Addition and Scalar Multiplication
• Matrix addition: (A+B)ij = Aij +Bij.
1. commutative property: A+B = B +A.2. associative property: A+ (B + C) = (A+B) + C.3. identity property: A+ 0 = 0 +A = A.
• Scalar multiplication: (kA)ij = kAij.
• Some properties:
1. k(A+B) = kA+ kB.2. (k + l)A = kA+ lA.3. k(lA) = (kl)A.4. 0A = 0 and k0 = 0.5. (A+B)T = AT +BT and (kA)T = kAT .
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6.3 Matrix Multiplication
• Definition. Let A be an m× n matrix, and B be an n× p matrix. Then
1. AB is an m× p matrix.
2. (AB)ik =n∑j=1
AijBjk = Ai1B1k +Ai2B2k + · · ·+AinBnk.
• The number of columns of A should be equal to the number of rows of B.
• (AB)ik is the product of ith row of A and kth column of B.
• Matrix addition and matrix multiplication.
1. A has size m× n and B has size m× n, then A+B has size m× n.2. A has size m× n and B has size n× p, then AB has size m× p.
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6.3 Matrix Multiplication
• Properties:
1. associative: A(BC) = (AB)C.2. distributive: A(B + C) = AB +AC and (A+B)C = AC +BC.3. k(AB) = (kA)B = A(kB).4. (AB)T = BTAT .5. In general, AB 6= BA and (AB)T 6= ATBT . (Matrix multiplication is
non-commutative!)
• If A is a square matrix of order n, I is an identity matrix of order n and 0 is azero matrix of size n× n, then AI = IA = A and A0 = 0A = 0.
• Power of a square matrix: A2 = A×A, A3 = A×A×A, · · ·
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6.3 Matrix Multiplication
• System of linear equationsA11X1 +A12X2 + · · ·+A1nXn = B1
A21X1 +A22X2 + · · ·+A2nXn = B2
...
Am1X1 +Am2X2 + · · ·+AmnXn = Bm
• Matrix equation AX = B
1. Coefficient matrix A of size m× n is known.2. The m× 1 column vector B is also given.3. The n× 1 column vector X is the variable to be solved.4. Solution C is a column vector of size n× 1 such that AC = B.
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6.4 Solving Systems by Reducing Matrices
• Three elementary row operations:
1. Ri ↔ Rj: interchange rows Ri and Rj.2. kRi: multiply row Ri by a nonzero constant k.3. kRi +Rj: add k times row Ri to row Rj (but leave Ri unchanged).
• Reduced matrix:
1. All zero-rows are at the bottom of the matrix.2. The leading entry in each nonzero-row is 1, and all others entries in the
column of the leading entry is 0.3. The leading entry in each nonzero-row is to the right of the leading entry in
any row above it.
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6.5 Solving Systems by Reducing Matrices (continued)
• Homogeneous system: AX = 0, or equivalently,A11X1 +A12X2 + · · ·+A1nXn = 0
A21X1 +A22X2 + · · ·+A2nXn = 0...
Am1X1 +Am2X2 + · · ·+AmnXn = 0
• X = 0 is always a trivial solution.
• To solve a homogeneous system, we only need to reduce the coefficient matrixA, instead of the augmented coefficient matrix [A | 0].
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6.5 Solving Systems by Reducing Matrices (continued)
• Homogeneous system: AX = 0, or equivalently,A11X1 +A12X2 + · · ·+A1nXn = 0
A21X1 +A22X2 + · · ·+A2nXn = 0...
Am1X1 +Am2X2 + · · ·+AmnXn = 0
• The number of equations m is equal to the number of rows of A.
• The number of unknowns n is equal to the number of columns of A.
• If A is a reduced matrix with k nonzero-rows, then k ≤ m.
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6.5 Solving Systems by Reducing Matrices (continued)
• Theorem. Let A be a reduced coefficient matrix of size m× n for ahomogeneous system AX = 0. (Note that X is an n× 1 column vector and 0is an m× 1 column vector.)
1. If A has exactly k nonzero-rows, then k ≤ n.2. If A has exactly k nonzero-rows and k < n, then the system has infinitely
many solutions.3. If A has exactly k nonzero-rows and k = n, then the system has a unique
solution X = 0 (the trivial solution).
• Corollary. If m < n, (the number of equations is less than the number ofunknowns), then k ≤ m < n, which by the above theorem implies that thehomogeneous system has infinitely many solutions.
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6.6 Inverses
• A square matrix A of order n is invertible if there exists a square matrix C oforder n such that CA = AC = I, where I is the identity matrix of order n. Weuse A−1 to denote the inverse of A.
• A should be a square matrix.
• Recall that AI = IA = A for any square matrix A having the same order of I.
• The identity matrix I is invertible and I−1 = I.
• If A is invertible, then A−1 is unique and AA−1 = A−1A = I, namely,multiplication of A and A−1 is commutative. Recall that in general, matrixmultiplication is NOT commutative.
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6.6 Inverses
• Find the inverse of
A =
[1 23 7
]• Let C be the inverse of A such that AC = I, where
C =
[x zy w
]
• It follows that [1 23 7
] [x zy w
]=
[x+ 2y z + 2w3x+ 7y 3z + 7w
]=
[1 00 1
]
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6.6 Inverses
• We have two linear systems{x+ 2y = 1
3x+ 7y = 0and
{z + 2w = 0
3z + 7w = 1
• Reducing augmented coefficient matrix[1 23 7
10
]−3R1+R2−−−−−−→
[1 20 1
1−3
]−2R2+R1−−−−−−→
[1 00 1
7−3
]⇒
{x = 7
y = −3
[1 23 7
01
]−3R1+R2−−−−−−→
[1 20 1
01
]−2R2+R1−−−−−−→
[1 00 1
−21
]⇒
{z = −2
w = 1
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6.6 Inverses
• Reducing augmented coefficient matrix[1 23 7
10
]−3R1+R2−−−−−−→
[1 20 1
1−3
]−2R2+R1−−−−−−→
[1 00 1
7−3
]⇒
{x = 7
y = −3
[1 23 7
01
]−3R1+R2−−−−−−→
[1 20 1
01
]−2R2+R1−−−−−−→
[1 00 1
−21
]⇒
{z = −2
w = 1
• Combining the above two reducing procedures[1 23 7
1 00 1
]−3R1+R2−−−−−−→
[1 20 1
1 0−3 1
]−2R2+R1−−−−−−→
[1 00 1
7 −2−3 1
]
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6.6 Inverses
• Find the inverse of
A =
[1 23 7
]• Reducing the matrix [A | I] to [I | C][
1 23 7
1 00 1
]→ · · · →
[1 00 1
7 −2−3 1
]
• The inverse matrix
A−1 = C =
[7 −2−3 1
]
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6.6 Inverses
• If A is square and [A | I] can be reduced to the form [I | C], then A isinvertible and A−1 = C.
• If A is square and [A | I] is reduced to [R | C] but R 6= I, then A is notinvertible.
• If A is not square then A is not invertible.
• If the square matrix A is invertible, then the matrix equation AX = B has aunique solution X = A−1B.
• If the coefficient matrix A is square but not invertible, then the homogeneousequation AX = 0 has infinitely many solutions.
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6.6 Inverses
Matrix Algebra AlgebraA is a square matrix of order n a is a real numberI is the identity matrix of order n 1 is the real identity
AI = IA = A a · 1 = 1 · a = aI−1 = I 1−1 = 1
If A is invertible, then A−1 is unique If a is invertible, then a−1 is uniqueAA−1 = A−1A = I aa−1 = a−1a = 1
If A 6= 0, A may still not be invertible If a 6= 0, then a is invertibleIn general AB 6= BA ab = ba
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6.7 Leontief’s Input-Output Analysis
• Input-output matrix
InputSector 1 Sector 2 External Demand
Output TotalsSector 1
[240 500 460
]1200
Sector 2 360 200 940 1500Other 600 800 -Totals 1200 1500
• Leontief matrix
A =
2401200
5001500
3601200
2001500
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6.7 Leontief’s Input-Output Analysis
• Leontief matrix
A =
2401200
5001500
3601200
2001500
• To produce one unit of sector 1, it spends 240
1200 on sector 1, and 3601200 on sector
2.
• To produce one unit of sector 2, it spends 5001500 on sector 1, and 200
1500 on sector2.
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6.7 Leontief’s Input-Output Analysis
• Leontief matrix
A =
2401200
5001500
3601200
2001500
• External Demand matrix
D =
[460940
]• Production matrix
X =
[12001500
]
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6.7 Leontief’s Input-Output Analysis
• Production=Internal Demand+External Demand:
X = AX +D
1200
1500
=
2401200
5001500
3601200
2001500
1200
1500
+
460
940
• Solving production matrix X from external demand matrix D:
X = AX +D ⇒ X −AX = D ⇒ (I −A)X = D ⇒ X = (I −A)−1D
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6.7 Leontief’s Input-Output Analysis
1. Determine the Leontief matrix A.
2. Find the external demand matrix D.
3. Solve the production matrix X by X = (I −A)−1D.
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6.7 Leontief’s Input-Output Analysis
A very simple economy consists of two sectors: agriculture and milling. Toproduce one unit of agricultural products requires 1
3 of a unit of agriculturalproducts and 1
4 of a unit of milled products. To produce one unit of milledproducts requires 3
4 of a unit of agricultural products and no units of milledproducts. Determine the production levels needed to satisfy an external demandfor 300 units of agriculture and 500 units of milled products.
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6.7 Leontief’s Input-Output Analysis
A very simple economy consists of two sectors: agriculture and milling. Toproduce one unit of agricultural products requires 1
3 of a unit of agriculturalproducts and 1
4 of a unit of milled products. To produce one unit of milledproducts requires 3
4 of a unit of agricultural products and no units of milledproducts. Determine the production levels needed to satisfy an external demandfor 300 units of agriculture and 500 units of milled products.
1. Leontief matrix
A =
13
34
14 0
2. External demand matrix
D =
[300500
]3. Production matrix X = (I −A)−1D
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6.7 Leontief’s Input-Output Analysis
Find the inverse of I −A: 23 −3
4
−14 1
1 0
0 1
32R1−−→
1 −98
−14 1
32 0
0 1
14R1+R2−−−−−→
1 −98
0 2332
32 0
38 1
3223R2−−−→
1 −98
0 1
32 0
1223
3223
98R2+R1−−−−−→
1 0
0 1
4823
3623
1223
3223
Production matrix
X = (I −A)−1D =
4823
3623
1223
3223
300
500
=
3240023
1960023
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6.7 Leontief’s Input-Output Analysis
Input-output matrix 40 120 40120 90 9040 90 −
Production matrix
X =
[40 + 120 + 40120 + 90 + 90
]=
[200300
]Leontief matrix
A =
40200
120300
120200
90300
=
15
25
35
310
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6.7 Leontief’s Input-Output Analysis
Leontief matrix
A =
40200
120300
120200
90300
=
15
25
35
310
Question: determine the corresponding production matrix if the external demandmatrix changes to
D =
64
64
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6.7 Leontief’s Input-Output Analysis
Find the inverse of I −A: 45 −2
5
−35
710
1 0
0 1
54R1−−→
1 −12
−35
710
54 0
0 1
35R1+R2−−−−−→
1 −12
0 25
54 0
34 1
52R2−−→
1 −12
0 1
54 0
158
52
12R2+R1−−−−−→
1 0
0 1
3516
54
158
52
Production matrix
X = (I −A)−1D =
3516
54
158
52
64
64
=
220
280
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7.1 Linear Inequalities in Two Variables
• Linear inequality in variables x and y (either a or b is nonzero):ax+ by + c < 0 ax+ by + c ≤ 0 ax+ by + c > 0 ax+ by + c ≥ 0
• The nonvertical line y = mx+ b separates the plane into three distinct parts:
1. the line itself, y = mx+ b2. the region (open half-plane) above the line, y > mx+ b3. the region (open half-plane) below the line, y < mx+ b
• The vertical line x = a separates the plane into three distinct parts:
1. the line itself, x = a2. the region (open half-plane) to the left of the line, x < a3. the region (open half-plane) to the right of the line, x > a
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7.2 Linear Programming
• If the feasible region is nonempty, closed and bounded, then the objectivefunction has a maximum (minimum) value which is achieved at a corner point.
• If the feasible region is empty, then the objective function has no maximum(minimum).
• Suppose the feasible region is nonempty, closed and unbounded.
1. If there exists an isovalue (isocost) line below the feasible region, then theobjective function has a minimum which is achieved at a corner point.Otherwise, the objective function has no minimum.
2. If there exists an isovalue (isoprofit) line above the feasible region, then theobjective function has a maximum which is achieved at a corner point.Otherwise, the objective function has no maximum.
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7.3 Multiple Optimum Solutions
If (x1, y1) and (x2, y2) are two corner points at which an objective function isoptimum, then the objective function will also be optimum at all points (x, y),where
x = (1− t)x1 + tx2
y = (1− t)y1 + ty2
with 0 ≤ t ≤ 1. The above two expressions can be also written in terms of matrixform: [
xy
]= (1− t)
[x1y1
]+ t
[x2y2
]
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7.4 The Simplex Method
Standard Linear Programming Problem:
Maximize the linear function Z = c1x1 + c2x2 + · · ·+ cnxn subject toa11x1 + a12x2 + · · ·+ a1nxn≤b1a21x1 + a22x2 + · · ·+ a2nxn≤b2...
am1x1 + am2x2 + · · ·+ amnxn≤bm
where x1, x2, · · · , xn and b1, b2, · · · , bm are nonnegative.
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7.4 The Simplex Method
Add slack variables s1, s2, · · · , sm:a11x1 + a12x2 + · · ·+ a1nxn + s1 = b1
a21x1 + a22x2 + · · ·+ a2nxn + s2 = b2...
am1x1 + am2x2 + · · ·+ amnxn + sm = bm
where x1, x2, · · · , xn, s1, s2, · · · , sm, and b1, b2, · · · , bm are nonnegative.
From Z = c1x1 + c2x2 + · · ·+ cnxn we have
−c1x1 − c2x2 − · · · − cnxn + Z = 0
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7.4 The Simplex Method
• Initial Simplex Table:
x1 x2 · · · xn s1 s2 · · · sm Zs1
a11 a12 · · · a1n 1 0 · · · 0 0a21 a22 · · · a2n 0 1 · · · 0 0
... ... ... ... ... . . . ... ...am1 am2 · · · amn 0 0 · · · 1 0−c1 −c2 · · · −cn 0 0 · · · 0 1
b1b2...bm0
s2...smZ
• Basic feasible solution (BFS):
1. Basic variables: s1 = b1, s2 = b2, · · · , sm = bm.2. Nonbasic variables: x1 = x2 = · · · = xn = 0.3. Objective function: Z = 0.
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7.4 The Simplex Method
• Each basic feasible solution (BFS) corresponds to a corner point in feasibleregion.
• The simplex method provides an algorithm to search the BFS (corner point)where the objective function achieves its maximum.
• Two principles of simplex method:
1. Eliminate negative indicators (start from the most negative indicator first).2. Keep the last column nonnegative (normalize the row with smallest
quotient).
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7.5 Degeneracy, Unbounded Solutions, and Multiple Solutions
• Degeneracy occurs when there are two rows have the smallest quotient, orwhen the smallest quotient is zero.
• If there exists no quotient, then the linear programming problem has anunbounded solution.
• If the indicators are all nonnegative and the number of zero indicators is largerthan the number of basic variables, then the linear programming problem hasmultiple solutions.
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7.6 Artificial Variables & 7.7 Minimization
• We need to add an artificial variable t when the feasible region does notcontain the origin (i.e., 0 is not a solution to the constraints). The objectivefunction becomes W = Z −Mt = c1x1 + c2x2 + · · ·+ cnxn −Mt with Mbeing a large parameter.
1. The case ai1x1 + ai2x2 + · · ·+ ainxn ≥ bi with bi > 0. We add a slackvariable si ≥ 0 and an artificial variable t such thatai1x1 + ai2x2 + · · ·+ ainxn − si + t = bi.
2. The case ai1x1 + ai2x2 + · · ·+ ainxn = bi with bi > 0. Since x1 = 0,x2 = 0, · · · , xn = 0 is not a solution to this constraint, we need to add anartificial variable t such that ai1x1 + ai2x2 + · · ·+ ainxn + t = bi.
• Minimizing the objective function Z = c1x1 + c2x2 + · · ·+ cnxn is the same asmaximizing the objective function −Z = −c1x1 − c2x2 − · · · − cnxn.
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7.8 The Dual
Maximize MinimizeZ = c1x1 + c2x2 + · · ·+ cnxn W = b1y2 + b2y2 + · · ·+ bmymsubject to subject toa11x1 + a12x2 + · · ·+ a1nxn≤b1a21x1 + a22x2 + · · ·+ a2nxn≤b2...
am1x1 + am2x2 + · · ·+ amnxn≤bm
a11y1 + a21y2 + · · ·+ am1ym≥c1a12y1 + a22y2 + · · ·+ am2ym≥c2...
a1ny1 + a2ny2 + · · ·+ amnym≥cnand andx1, x2, · · · , xn ≥ 0 y1, y2, · · · , ym ≥ 0
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7.8 The Dual
Maximize Minimize
Z = [c1, c2, · · · , cn]
x1x2...xn
W = [b1, b2, · · · , bm]
y1y2...ym
subject to subject to
a11 a12 · · · a1na21 a22 · · · a2n
... ... ...am1 am2 · · · amn
x1x2...xn
≤b1b2...bm
a11 a21 · · · am1
a12 a22 · · · am2... ... ...a1n a2n · · · amn
y1y2...ym
≥c1c2...cn
x1x2...xn
≥ 0
y1y2...ym
≥ 0
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7.8 The Dual
Maximize Z = CX Minimize W = BTYsubject to subject to
AX≤B and X ≥ 0 ATY≥CT and Y ≥ 0
where
C = [c1, c2, · · · , cn], X = [x1, x2, · · · , xn]T
Y = [y1, y2, · · · , ym]T , B = [b1, b2, · · · , bm]T
and
A =
a11 a12 · · · a1na21 a22 · · · a2n
... ... ...am1 am2 · · · amn
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Differentials
• The derivative of a function f is the function denoted f ′ and defined by
f ′(x) = limz→x
f(z)− f(x)
z − x= limh→0
f(x+ h)− f(x)
h
provided that the limit exists.
• Let y = f(x). The differential of y is denoted dy and defined by
dy = f ′(x)dx.
• Chain rule. Let y = f(x) and x = g(t), then
dy
dt=dy
dx
dx
dt= f ′(x)g′(t).
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Integrals
• An antiderivative of a function f(x) is a function F (x) such that F ′(x) = f(x).
• The indefinite integral of f(x) with respect to x is defined by∫f(x)dx = F (x) + C,
where∫
is called the integral sign, f(x) is the integrand, x is the variable ofintegration, and C is the constant of integration.
• Any two antiderivatives of a function differ only by a constant.
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Differentiation and Integration Formulas
Differentiation Integration(1)′ = 0
∫0dx = C
(kx)′ = k∫kdx = kx+ C
(xa+1)′ = (a+ 1)xa, a 6= −1∫xadx = xa+1
a+1 + C, a 6= −1
(ln |x|)′ = 1x, x 6= 0
∫1xdx = ln |x|+ C, x 6= 0
(ex)′ = ex∫exdx = ex + C
(bx)′ = bx(ln b), b > 0∫bxdx = bx
ln b + C, b > 0(kf(x))′ = kf ′(x)
∫kf(x)dx = k
∫f(x)dx
(f(x) + g(x))′ = f ′(x) + g′(x)∫f(x) + g(x)dx =
∫f(x)dx+
∫g(x)dx
Any two antiderivatives of a function differ only by a constant.
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The Definite Integrals
Let f(x) be continuous on [a, b], then∫ b
a
f(x)dx = limn→∞
n∑k=1
f(a+b− an
k).
The definite integral is the limit of an infinite sum.
Fundamental theorem: Let F (x) be an antiderivative of a continuous functionf(x) on [a, b], then ∫ b
a
f(x)dx = F (x)
∣∣∣∣ba
= F (b)− F (a).
The average of a continuous function f(x) over [a, b] is f̄ = 1b−a
∫ baf(x)dx.
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The Definite Integrals
1.∫ ba
(f(x) + g(x))dx =∫ baf(x)dx+
∫ bag(x)dx.
2.∫ bakf(x)dx = k
∫ baf(x)dx.
3.∫ baf(x)dx+
∫ cbf(x)dx =
∫ caf(x)dx.
4.∫ baf(x)dx = −
∫ abf(x)dx.
5.∫ baF ′(x)dx = F (x)
∣∣∣∣ba
= F (b)− F (a).
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Areas between Curves
• The area of the region (vertical strip) bounded by the two curves y = f(x) andy = g(x) from x = a to x = b is
∫ b
a
|f(x)− g(x)|dx.
• The area of the region (horizontal strip) bounded by the two curves x = u(y)and x = v(y) from y = c to y = d is
∫ d
c
|u(y)− v(y)|dy.
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Consumers’ and Producers’ Surplus
• Demand function: p = f(q).
• Supply function: p = g(q).
• Equilibrium: p0 = f(q0) = g(q0).
• Consumers’ surplus: CS =∫ q00
(f(q)− p0)dq.
• Producers’ surplus: PS =∫ q00
(p0 − g(q))dq.
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Methods of Integration
• Change of variable (let u = u(x)):
∫f(x)dx =
∫g(u)
dx
dudu and
∫ b
a
f(x)dx =
∫ u(b)
u(a)
g(u)dx
dudu
where g(u) = g(u(x)) = f(x) and dxdu = (dudx)−1 = 1
u′(x).
• Integration by parts:
∫uv′dx = uv −
∫u′vdx and
∫ b
a
uv′dx = uv
∣∣∣∣ba
−∫ b
a
u′vdx.
Lecture Notes for Math 1540 First Previous Next Last 48
Integration of Rational Functions
Step 1: Long division.
Step 2: Factorize denominator.
Step 3: Partial fractions (four cases).
Case 1) distinct linear factors:Pn−1(x)
(x−x1)···(x−xn)= A1
x−x1+ · · ·+ An
x−xn.
Case 2) repeated linear factors:Pn−1(x)(x−a)n = A1
x−a + · · ·+ An(x−a)n.
Case 3) distinct irreducible quadratic factors:P2n−1(x)
(x2+a1x+b1)···(x2+anx+bn)= A1x+B1
x2+a1x+b1+ · · ·+ Anx+Bn
x2+anx+bn.
Case 4) repeated irreducible quadratic factors:P2n−1(x)
(x2+ax+b)n= A1x+B1
x2+ax+b+ · · ·+ Anx+Bn
(x2+ax+b)n.
Lecture Notes for Math 1540 First Previous Next Last 49
Partial Derivatives
• The partial derivative of z = f(x, y) with respect to x is defined by
∂z
∂x=∂f
∂x= ∂xf(x, y) = fx(x, y) = lim
h→0
f(x+ h, y)− f(x, y)
h
provided that the limit exists. We treat y as a constant.
• The partial derivative of z = f(x, y) with respect to y is defined by
∂z
∂y=∂f
∂y= ∂yf(x, y) = fy(x, y) = lim
h→0
f(x, y + h)− f(x, y)
h
provided that the limit exists. We treat x as a constant.
Lecture Notes for Math 1540 First Previous Next Last 50
Competitive and Complementary Products
• Demand functions: qA = f(pA, pB) and qB = g(pA, pB) are given.
• Competitive products (Apple and Blackberry): ∂qA∂pB
> 0 and ∂qB∂pA
> 0.
• Complementary products (cars and gasoline): ∂qA∂pB
< 0 and ∂qB∂pA
< 0.
• Remark: it is always true that ∂qA∂pA
< 0 and ∂qB∂pB
< 0.
Lecture Notes for Math 1540 First Previous Next Last 51
Higher-Order Partial Derivatives
• Let z = f(x, y).
• fxx = (fx)x = ∂xxf = ∂2f∂x2
= ∂∂x
(∂f∂x
).
• fyy = (fy)y = ∂yyf = ∂2f∂y2
= ∂∂y
(∂f∂y
).
• Mixed partial derivative: fxy = fyx = ∂2f∂x∂y = ∂2f
∂y∂x = ∂∂y
(∂f∂x
)= ∂
∂x
(∂f∂y
).
Lecture Notes for Math 1540 First Previous Next Last 52
Chain Rule and Implicit Partial Differentiation
• Let z = f(x, y) with x = x(r, s) and y = y(r, x). Then
∂z
∂r=
∂z
∂x
∂x
∂r+∂z
∂y
∂y
∂r
∂z
∂s=
∂z
∂x
∂x
∂s+∂z
∂y
∂y
∂s
• Let F (x, y, z) = 0. Then∂F
∂x+∂F
∂z
∂z
∂x= 0.
By solving the above equation, we obtain
∂z
∂x= −∂xF
∂zF.
Lecture Notes for Math 1540 First Previous Next Last 53
Relative Extrema for two-variable functions
• The function f(x, y) has a relative maximum at the point (a, b) if
f(x, y) ≤ f(a, b)
for any (x, y) close to (a, b).
• The function f(x, y) has a relative minimum at the point (a, b) if
f(x, y) ≥ f(a, b)
for any (x, y) close to (a, b).
Lecture Notes for Math 1540 First Previous Next Last 54
Relative Extrema for two-variable functions
• The critical points of a function f(x, y) are the solutions to the followingsystem {
fx(x, y) = 0
fy(x, y) = 0
• Second-derivative test for a critical point (a, b): D(x, y) = fxxfyy − f2xy.
1) If D(a, b) > 0 and fxx(a, b) < 0, then f has a relative maximum at (a, b).2) If D(a, b) > 0 and fxx(a, b) > 0, then f has a relative minimum at (a, b).3) If D(a, b) < 0, then f has a saddle point at (a, b).4) If D(a, b) = 0, then NO conclusion.
Lecture Notes for Math 1540 First Previous Next Last 55
Relative Maximum
Case 1: D = fxxfyy − f2xy > 0 and fxx < 0.
Lecture Notes for Math 1540 First Previous Next Last 56
Relative Minimum
Case 2: D = fxxfyy − f2xy > 0 and fxx > 0.
Lecture Notes for Math 1540 First Previous Next Last 57
Saddle
Case 3: D = fxxfyy − f2xy < 0.
Lecture Notes for Math 1540 First Previous Next Last 58
Lagrange Multipliers
• Find the critical points of w = f(x1, · · · , xn) subject to g(x1, · · · , xn) = 0.
• Introduce a lagrange multiplier λ:F (x1, · · · , xn, λ) = f(x1, · · · , xn)− λg(x1, · · · , xn).
• Find the critical points of F (x1, · · · , xn, λ), namely, solve the system∂F∂x1
(x1, · · · , xn, λ) = ∂f∂x1
(x1, · · · , xn)− λ ∂g∂x1
(x1, · · · , xn) = 0...
∂F∂xn
(x1, · · · , xn, λ) = ∂f∂xn
(x1, · · · , xn)− λ ∂g∂xn
(x1, · · · , xn) = 0
∂F∂λ (x1, · · · , xn, λ) = −g(x1, · · · , xn) = 0
Lecture Notes for Math 1540 First Previous Next Last 59
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