Calculus - UNENE · 2020-03-19 · Calculus — Differentiation ©Wei-Chau Xie Physical Meanings of...
Transcript of Calculus - UNENE · 2020-03-19 · Calculus — Differentiation ©Wei-Chau Xie Physical Meanings of...
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UNENE Math Refresher Course
Calculus
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Calculus — Differentiation © Wei-Chau Xie
Derivatives
Some simple derivatives
d
dxxn = n xn−1
d
dxeax = a eax
d
dxln x = 1
x
d
dxsinax = a cosax
d
dxcosax = −a sinax
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Calculus — Differentiation © Wei-Chau Xie
Important Rules for Evaluating Derivatives
[ f(x) + g(x)]′ = f ′(x) + g ′(x)
[c f(x)]′ = c f ′(x)
[ f(x) g(x)]′ = f ′(x) g(x) + f(x) g ′(x) Product Rule
[
f(x)
g(x)
]′= f ′(x) g(x) − f(x) g ′(x)
g2(x)Quotient Rule
y = y(u), u = u(x) =⇒dy
dx= dy
du· du
dxChain Rule
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Calculus — Differentiation © Wei-Chau Xie
Example
Find the derivative of y = tan x.
y ′ = (tan x)′ =( sin x
cos x
)′
= (sin x)′ cos x − sin x (cos x)′
cos2 xQuotient Rule
( fg
)′= f ′g − f g ′
g2
= cos x · cos x − sin x · (− sin x)
cos2 x
= cos2 x + sin2 x
cos2 xsin2 x + cos2 x = 1
= 1
cos2 x= sec2 x
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Calculus — Differentiation © Wei-Chau Xie
Example
Find the derivative of y = ln cos x.
y = ln cos x can be written as
y = ln u, u = cos x
Apply the Chain Rule
dy
dx= dy
du· du
dx
= 1
u· (− sin x) = 1
cos x· (− sin x)
= − tan x
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Calculus — Differentiation © Wei-Chau Xie
Example
Find the derivative of y = ax.
y = ax can be written as
y = ax = eln ax = ex · ln a
∴ y = eu, u = x · ln a
Apply the Chain Rule
dy
dx= dy
du· du
dx
= eu · ln a = ax ln a
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Calculus — Differentiation © Wei-Chau Xie
Higher-Order Derivatives
d2y
dx2= d
dx
(
dy
dx
)
d3y
dx3= d
dx
(
d2y
dx2
)
...
dny
dxn= d
dx
(
dn−1y
dxn−1
)
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Calculus — Differentiation © Wei-Chau Xie
Physical Meanings of Derivatives
Consider a function y = y(x).
The first-order derivative y ′(x0) is the slope of the curve y(x) at (x0, y0).
The first-order derivative y ′(x0) is the rate of change of the function at x0.
The curvature of the curve is κ = y ′′
(1+ y ′2)3/2.
In many applications of physics and engineering, such as the equation of
bending in beams, the approximations to the fluid flow around surfaces,∣
∣ y ′∣∣≪1, and hence κ≈y ′′ .
Consider the motion of a particle (mass) with displacement x(t).
The first-order derivativedx
dt= v(t) is the velocity of the particle.
The second-order derivatived2x
dt2= a(t) is the acceleration of the particle.
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Calculus — Differentiation © Wei-Chau Xie
Example
Find the second-order derivative of y = ex(sin x + cos x).
y ′ = [ex(sin x + cos x)]′
Product Rule ( f g)′ = f ′g + f g ′
= (ex)′(sin x + cos x) + ex (sin x + cos x)′
= ex (sin x + cos x) + ex (cos x − sin x)
= 2 ex cos x
y ′′ = [2 ex cos x]′
Product Rule ( f g)′ = f ′g + f g ′
= 2 [(ex)′ cos x + ex (cos x)′]
= 2 [ex cos x + ex (− sin x)]
= 2 ex (cos x − sin x)
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Calculus — Differentiation © Wei-Chau Xie
Example
Find the second-order derivative of y = ex2.
Rewrite y = eu, u = x2
∴dy
dx= dy
du· du
dxChain Rule
= eu · 2x = 2x ex2
∴d2y
dx2= [2x ex2
]′
Product Rule ( f g)′ = f ′g + f g ′
= 2 [(x)′ ex2 + x (ex2)′]
= 2(
1 · ex2 + x · 2x ex2)
= 2 ex2(1 + 2x2)
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Calculus — Taylor Series © Wei-Chau Xie
Taylor Series
In many applications, it is desirable to approximate complicated functions by
simple functions, such as polynomials.
Let f(x) and its derivatives f ′(x), f ′′(x), . . . , f (n)(x) exist and be
continuous in a closed interval a6x6b. Then for x0 in [a, b]
f(x) = f(x0) + f ′(x0)
1! (x−x0) + f ′′(x0)
2! (x−x0)2 + · · · + f (n)(x0)
n! (x−x0)n + · · ·
=∞∑
n=0
f (n)(x0)
n! (x−x0)n
Function f(x) can be approximated by a polynomial of degree n:
f(x) = f(x0) + f ′(x0)
1! (x−x0) + f ′′(x0)
2! (x−x0)2 + · · · + f (n)(x0)
n! (x−x0)n + Rn(x)
Rn = f (n+1)(ξ)
(n+1)! (x−x0)n+1, x0 6ξ 6x
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Calculus — Taylor Series © Wei-Chau Xie
MacLaurin Series
When x0 = 0, a Taylor series is also referred to as a MacLaurin series
f(x) = f(0) + f ′(0)
1! x + f ′′(0)
2! x2 + · · · + f (n)(0)
n! xn + · · · =∞∑
n=0
f (n)(0)
n! xn
Approximation of function f(x):
f(x) ≈ f(0) + f ′(0)
1! x + f ′′(0)
2! x2 + · · · + f (n)(0)
n! xn
The error in the approximation is
∣
∣Rn(x)∣
∣ = f (n+1)(ξ)
(n+1)! xn+1, 06ξ 6x
6M
(n+1)!∣
∣x∣
∣
n+1, M = max
∣
∣ f (n+1)(x)∣
∣
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Calculus — Taylor Series © Wei-Chau Xie
Example
Find the Taylor series at x0 = 0 (MacLaurin series) of f(x) = ex.
∵ f(x) = ex, f(0) = e0 = 1
f ′(x) = ex, f ′(0) = e0 = 1
f ′′(x) = ex, f ′′(0) = e0 = 1...
f (n)(x) = ex, f (n)(0) = e0 = 1
∴ ex = f(0) + f ′(0)
1! x + f ′′(0)
2! x2 + · · · + f (n)(0)
n! xn + · · ·
= 1 + x + x2
2! + x3
3! + · · · + xn
n! + · · ·
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Calculus — Taylor Series © Wei-Chau Xie
Example
Find the Taylor series at x0 = 0 (MacLaurin series) of f(x) = sin x.
∵ f(x) = sin x, f(0) = sin 0 = 0
f ′(x) = cos x, f ′(0) = cos 0 = 1
f ′′(x) = − sin x, f ′′(0) = − sin 0 = 0
f ′′′(x) = − cos x, f ′′′(0) = − cos 0 = −1
f (4)(x) = sin x, f (4)(0) = sin 0 = 0...
∴ f (2n)(0) = 0, f (2n+1)(0) = (−1)n
∴ sin x = f(0) + f ′(0)
1! x + f ′′(0)
2! x2 + · · · + f (n)(0)
n! xn + · · ·
= 0 + (−1)0
1! x + 0
2! x2 + (−1)1
3! x3 + · · · + (−1)n
(2n+1)! x2n+1 + · · ·
= x − x3
3! + x5
5! − x7
7! + · · · + (−1)n x2n+1
(2n+1)! + · · ·
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Calculus — Taylor Series © Wei-Chau Xie
sin x = x − x3
3! + x5
5! − x7
7! + · · · + (−1)n x2n+1
(2n+1)! + · · ·
–2
–1
0
1
2
3
0.5 1 321.5 2.5
sinx
x−x
x3
3!−
x7
7!+
x5
5!
−xx3
3!+
x5
5!
x
−xx3
3!
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Calculus — Integration © Wei-Chau Xie
Integrals of Fundamental Functions
∫
xn dx = xn+1
n+1, n 6=−1
∫
1
xdx = ln
∣
∣x∣
∣, x 6=0
∫
ex dx = ex,
∫
eax dx = 1a
eax
∫
sin x dx = − cos x
∫
cos x dx = sin x
∫
tan x dx = − ln∣
∣cos x∣
∣
∫
cot x dx = ln∣
∣sin x∣
∣
∫
sec x dx = ln∣
∣sec x + tan x∣
∣
∫
csc x dx = ln∣
∣csc x − cot x∣
∣
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Calculus — Integration - Change of Variables © Wei-Chau Xie
Techniques of Integration — Change of Variables
Example
Evaluate I =∫
x√
1 − x2 dx.
Let 1 − x2 = u, −2x dx = du =⇒ x dx = − 12
du
I =∫
√
1 − x2 · x dx =∫ √
u ·(
− 12
du)
= − 12
∫
u12 du = − 1
2· u
12+1
12 +1
+ C
∫
un du = un+1
n+1+ C
= − 13
u32 + C = − 1
3(1 − x2)
32 + C
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Calculus — Integration - Change of Variables © Wei-Chau Xie
Example
Evaluate I =∫
√
a2 − x2 dx, a>0.
Let x = a sin u =⇒ u = sin−1 x
a, dx = a cos u du sin2 x + cos2 x = 1
I =∫
√
a2 − a2 sin2 u · a cos u du, cos u =√
1 − sin2 u
= a2
∫
cos2 u du = a2
∫
1 + cos 2u
2du
= a2
2
(
u + 1
2sin 2u
)
+ C = a2
2u + a2
2sin u cos u + C
= a2
2sin−1 x
a+ a2
2· x
a·√
1 −( x
a
)2+ C
= a2
2sin−1 x
a+ 1
2x√
a2 − x2 + C
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Calculus — Integration © Wei-Chau Xie
Some Useful Formulas
d
dx(ax + b) = a =⇒ dx = 1
ad(a x + b)
d
dx(xn+1) = (n+1) xn =⇒ xn dx = 1
n+1d(xn+1)
d
dx(ln x) = 1
x=⇒
1
xdx = d(ln x)
d
dx(eax) = a eax =⇒ ea x dx = 1
ad(ea x)
d
dx(sin x) = cos x =⇒ cos x dx = d(sin x)
d
dx(cos x) = − sin x =⇒ sin x dx = −d(cos x)
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Calculus — Integration - Change of Variables © Wei-Chau Xie
Example
Evaluate I =∫
tan x dx.
I =∫
tan x dx =∫
sin x
cos xdx
= −∫
1
cos xd(cos x), sin x dx = −d(cos x)
= − ln∣
∣cos x∣
∣ + C
∫
1
udu = ln
∣
∣u∣
∣, u = cos x
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Calculus — Integration - Change of Variables © Wei-Chau Xie
Example
Evaluate I =∫
2x ex2dx.
I =∫
ex2 · 2x dx =∫
ex2d(x2), xn dx = 1
n+1d(xn+1)
= ex2 + C
∫
eu du = eu, u = x2
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Calculus — Integration - Change of Variables © Wei-Chau Xie
Example
Evaluate I =∫
3x3
2 − x4dx.
I = 3
∫
1
2 − x4· x3 dx
= 3
∫
1
2 − x4· 1
4d(x3+1), xn dx = 1
n+1d(xn+1)
= 3
4
∫
1
2 − x4· 1
−1d( 2 − x4 ), dx = 1
ad(a x + b)
= − 3
4ln
∣
∣2 − x4∣
∣ + C,
∫
1
udu = ln
∣
∣u∣
∣, u = 2−x4
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Calculus — Integration - Change of Variables © Wei-Chau Xie
Example
Evaluate I =∫
sin3 x dx.
I =∫
sin2 x · sin x dx
= −∫
sin2 x d(cos x), sin x dx = −d(cos x)
= −∫
(1 − cos2 x) d(cos x), sin2 x + cos2 x = 1
= − cos x + 13
cos3 x + C,
∫
un du = un+1
n+1, u = cos x
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Calculus — Integration - Change of Variables © Wei-Chau Xie
Example
Evaluate I =∫
sin2 x cos3 x dx.
I =∫
sin2 x cos2 x · cos x dx
=∫
sin2 x cos2 x d(sin x), cos x dx = d(sin x)
=∫
sin2 x (1 − sin2 x) d(sin x), sin2 x + cos2 x = 1
=∫
(sin2 x − sin4 x) d(sin x)
= 13
sin3 x − 15
sin5 x + C,
∫
un du = un+1
n+1, u = sin x
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Calculus — Integration © Wei-Chau Xie
Techniques of Integration — Integration by Parts
∫
f(x) dx =∫
u dv = u v −∫
v du
The important step is to split function f(x) into the product of two functions∫
f(x) dx =∫
u(x) · g(x) dx =∫
u(x) d[v(x)]
to identify functions u(x) and v(x).
Some rules for selecting functions u(x) and v(x):
1. It is easy to find v(x).
2.
∫
v du is easier to evaluate than
∫
u dv.
3. Functions, such as ln x, sin−1 x, cos−1 x, tan−1 x, . . . , whose derivatives
are of an easier form to integrate, should be left out as u(x).
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Calculus — Integration - Integration by Parts © Wei-Chau Xie
Example
Evaluate I =∫
x2 cos x dx.
I =∫
x2 · cos x dx =∫
x2 d(sin x), cos x dx = d(sin x)
= x2 sin x −∫
sin x · 2x dx, Integration by parts
= x2 sin x − 2
∫
x · sin x dx
= x2 sin x + 2
∫
x d(cos x), sin x dx = −d(cos x)
= x2 sin x + 2(
x cos x −∫
cos x dx)
= x2 sin x + 2 x cos x − 2 sin x + C
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Calculus — Integration - Integration by Parts © Wei-Chau Xie
Example
Evaluate I =∫
x ln x dx.
I =∫
ln x · x dx =∫
ln x · 12
d(x2), xn dx = 1
n+1d(xn+1)
= 12
(
x2 ln x −∫
x2 · 1
xdx
)
, Integration by parts
= 12
(
x2 ln x −∫
x dx)
= 12
(
x2 ln x − 12
x2)
+ C
= 14
x2(2 ln x − 1) + C
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Calculus — Integration - Integration by Parts © Wei-Chau Xie
Example
Evaluate I =∫
ex sin x dx.
I =∫
sin x · ex dx =∫
sin x d(ex), ex dx = d(ex)
= ex sin x −∫
ex · cos x dx, Integration by parts
= ex sin x −∫
cos x d(ex)
= ex sin x −{
ex cos x −∫
ex · (− sin x) dx
}
∫
ex sin x dx = ex sin x − ex cos x −∫
ex sin x dx
∴
∫
ex sin x dx = 12
ex(sin x − cos x) + C
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Calculus — Integration © Wei-Chau Xie
Techniques of Integration — Partial Fractions
Partial fraction decomposition is an essential step in
integrating rational fractions
solving ordinary differential equations using the method of Laplace transform
Consider a rational functionN(x)
D(x), where N(x) and D(x) are polynomials.
If the degree (power) of N(x) is higher or equal to that of Q(x), use long
division to simplify first.
Completely factorize the denominator D(x) into factors of the form
(αx+β)m and (ax2+bx+c)n,
where ax2+bx+c is an unfactorable quadratic.
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Calculus — Integration - Partial Fractions © Wei-Chau Xie
For each factor of the form (αx+β)m, the partial fraction decomposition is
Am
(αx+β)m+
Am−1
(αx+β)m−1+ · · · + A1
(αx+β).
For each factor of the form (ax2+bx+c)n, the partial fraction decomposition
is n terms
Bn x+Cn
(ax2+bx+c)n+
Bn−1 x+Cn−1
(ax2+bx+c)n−1+ · · · + B1 x+C1
(ax2+bx+c).
Example
5x3+2x+7
2x8+7x7−10x5−6x4−x3= 5x3+2x+7
x3(2x−1)(x2+2x−1)2
= A3
x3+ A2
x2+ A1
x+ B
2x−1+ C2 x+D2
(x2+2x−1)2+ C1 x+D1
x2+2x−1
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Calculus — Integration - Partial Fractions © Wei-Chau Xie
The Cover-Up Method
Suppose D(x) has a factor (x−a)m, the partial fraction decomposition is
N(x)
D(x)= N(x)
(x−a)m D1(x)= Am
(x−a)m+
Am−1
(x−a)m−1+ · · · + A1
(x−a)+ N1(x)
D1(x)
in which D1(x) does not have (x−a) as a factor.
Multiplying both sides of the equation by (x−a)m yields
N(x)
D1(x)= Am + Am−1(x−a) + · · · + A1(x−a)m−1 + N1(x)
D1(x)(x−a)m
Setting x = a =⇒ Am = N(x)
D1(x)
∣
∣
∣
∣
x=a
This result can be restated as follows:
To find Am, ‘‘cover-up’’ (remove) the term (x−a)m and set x = a:
Am = N(x)
×(x−a)m D1(x)
∣
∣
∣
∣
x=a
Works only for the highest power of repeated linear factor.
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Calculus — Integration - Partial Fractions © Wei-Chau Xie
Example
Evaluate I =∫
4
(x−1)2(x+1)dx.
Using partial fractions:4
(x−1)2(x+1)= A2
(x−1)2+ A1
x−1+ B
x+1
To find A2, cover-up (x−1)2 and set x = 1 =⇒ A2 = 4
x+1
∣
∣
∣
∣
x=1
= 2
To find B, cover-up (x+1) and set x = −1 =⇒ B = 4
(x−1)2
∣
∣
∣
∣
x=−1
= 1
For A1, set x = 0 : 4
(0−1)2(0+1)= 2
(0−1)2+ A1
0−1+ 1
0+1=⇒ A1=−1
∴ I =∫ {
2
(x−1)2− 1
x−1+ 1
x+1
}
dx = − 2
x−1− ln
∣
∣x−1∣
∣ + ln∣
∣x+1∣
∣ + C
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Calculus — Integration - Partial Fractions © Wei-Chau Xie
Example
Evaluate I =∫
8
(x−1)(x2+2x+5)dx.
Using partial fractions:
8
(x−1)(x2+2x+5)= A
x−1+ Bx+C
x2+2x+5
= A(x2+2x+5) + (Bx+C)(x−1)
(x−1)(x2+2x+5)= (A+B)x2+(2A−B+C)x+(5A−C)
(x−1)(x2+2x+5)
To find A, cover-up (x−1) and set x = 1:
A = 8
x2+2x+5
∣
∣
∣
∣
x=1
= 8
1+2+5= 1
33
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Calculus — Integration - Partial Fractions © Wei-Chau Xie
Compare the coefficients of the numerators
x2 : A + B = 0 =⇒ B = −A = −1
x : 2A − B + C = 0 =⇒ C = B − 2A = −1 − 2·1 = −3
1 : 5A − C = 8 Use this equation as a check: 5·1−(−3)=8
∴8
(x−1)(x2+2x+5)= 1
x−1− x+3
x2+2x+5= 1
x−1− (x+1) + 2
(x+1)2+22
= 1
x−1− (x+1)
(x+1)2+22− 2
(x+1)2+22
Complete the Squares: a2 + 2·a·b + b2 = (a + b)2
3x2 + 2x + 5 = 3(
x2 + 23
x)
+ 5 = 3[
x2 + 2·x·13
+(
13
)2−
(
13
)2]
+ 5
= 3{[
x2 + 2·x·13
+(
13
)2]
+ 53
−(
13
)2}
= 3{(
x + 13
)2+ 14
9
}
= 3{(
x + 13
)2+
(
√143
)2}
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Calculus — Integration - Partial Fractions © Wei-Chau Xie
I =∫ {
1
x−1− (x+1)
(x+1)2+22− 2
(x+1)2+22
}
dx
=∫
1
x−1d(x−1) − 1
2
∫
1
(x+1)2+22d[(x+1)2+22] − 2
∫
1
(x+1)2+22d(x+1)
= ln∣
∣x−1∣
∣ − 12
ln∣
∣(x+1)2 + 22∣
∣ − tan−1 x+1
2+ C
∫
1
xdx = ln
∣
∣x∣
∣,
∫
1
x2 + a2dx = 1
atan−1 x
a
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Calculus — Integration © Wei-Chau Xie
Definite Integrals
∫
f(x) dx = F(x) =⇒∫ b
af(x) dx = F(x)
∣
∣
∣
b
x=a= F(b) − F(a)
The integral A =∫ b
af(x) dx is the area bounded by curve y = f(x) and the
x-axis between a and b.
x
y
y = f(x)
a b
A
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Calculus — Integration - Definite Integrals © Wei-Chau Xie
Example
Find the area bounded by the sine curve y = sin x and the x-axis between 0
and π .
x0 π
y=sin x
A =∫ π
0sin x dx = − cos x
∣
∣
∣
π
0
= −(cos π − cos 0) = −[(−1) − 1] = 2
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Calculus — Integration - Definite Integrals © Wei-Chau Xie
Example
Evaluate I =∫ 4
0
x+2√2x + 1
dx.
Let√
2x + 1 = u, x = u2−1
2, dx = u du
x = 0 =⇒ u =√
2·0+1 = 1; x = 4 =⇒ u =√
2·4+1 = 3
I =∫ 4
x=0
x + 2√2x+1
dx =∫ 3
u=1
u2−1
2+ 2
u· u du
= 12
∫ 3
1(u2 + 3) du = 1
2
[
13
u3 + 3u]3
1
= 12
[
(13·33 + 3·3
)
−(1
3·13 + 3·1
)
]
= 223
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Calculus — Integration - Definite Integrals © Wei-Chau Xie
Example
Determine the area bounded by parabola y2 = 2x and line y = x−4.
x0
y 2=2x
y=x−4y
(8, 4)
(2, −2)
The intersections are given by
{
y2 = 2x
y = x−4
Substitute the second equation into the first
(x−4)2 = 2x =⇒ x2 − 10x + 16 = 0 =⇒ (x−2)(x−8) = 0
x = 2 =⇒ y = x−4 = −2, x = 8 =⇒ y = x−4 = 4
The points of intersection are (2, −2) and (8, 4).
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Calculus — Integration - Definite Integrals © Wei-Chau Xie
x0
x=
y 2=2x
y=x−4
x=y+4
y
ydy
y 2
2(8, 4)
(2, −2)
Consider a horizontal strip of height dy as shown
dA =[
( y+4) − y2
2
]
dy
A =∫ 4
y=−2
(
y + 4 − y2
2
)
dy =[
12
y2 + 4 y − 16
y3]4
y=−2
=(
12·42 + 4·4 − 1
6·43
)
−[
12(−2)2 + 4(−2) − 1
6(−2)3
]
= 18
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Calculus — Partial Derivatives © Wei-Chau Xie
Partial Derivatives
Consider a functiony = f(x1, x2, . . . , xn)
of n independent variables x1, x2, . . . , xn.
The partial derivatives are defined as derivatives when all but the variable of
interest are held fixed during the differentiation.
∂ y
∂xi
is the first-order derivative of y w.r.t. xi, when
x1, x2, . . . , xi−1, xi+1, . . . , xn are held fixed (or regarded as constants).
For ‘‘nice’’ functions (partial derivatives exist and are continuous), mixed
partial derivatives must be equal regardless of the order in which the
differentiation is performed.
For examples, consider f(x, y),
∂2f
∂x ∂ y= ∂2f
∂y ∂x,
∂3f
∂x2∂ y= ∂3f
∂y ∂x2, . . .
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Calculus — Partial Derivatives © Wei-Chau Xie
Example
Given z = x2 − y3 + 2x y2, find∂z
∂x,
∂z
∂ y,
∂2z
∂x2,
∂2z
∂ y2,
∂2z
∂x ∂ y.
∂z
∂x= ∂
∂x(x2 − y3 + 2 x y2) = 2x + 2 y2 y is fixed
∂z
∂ y= ∂
∂ y(x2 − y3 + 2 x y2) = −3 y2 + 4 x y x is fixed
∂2z
∂x2= ∂
∂x
(∂z
∂x
)
= ∂
∂x(2x + 2 y2) = 2 y is fixed
∂2z
∂ y2= ∂
∂ y
( ∂z
∂ y
)
= ∂
∂ y(−3 y2 + 4 x y) = −6 y + 4 x x is fixed
∂2z
∂x ∂y= ∂
∂x
( ∂z
∂ y
)
= ∂
∂x(−3 y2 + 4 x y) = 4 y y is fixed
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Calculus — Partial Derivatives © Wei-Chau Xie
Chain Rules for Partial Derivatives
Suppose z = f(u, v), u = g(x, y), v = h(x, y)
z
u v
y yx x
∂z
∂x= ∂z
∂u
∂u
∂x+ ∂z
∂v
∂v
∂x
∂z
∂ y= ∂z
∂u
∂u
∂ y+ ∂z
∂v
∂v
∂ y
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Calculus — Chain Rules for Partial Derivatives © Wei-Chau Xie
Example
Find∂z
∂sif z = f(x, y), x = g(s, t), y = h(s, t),
in which z = x2 e y + y ln x, x = s2 cos t, y = 4t2 sin 2s.
By the Chain Rule:
∂z
∂s= ∂z
∂x
∂x
∂s+ ∂z
∂ y
∂ y
∂s
z
x y
t ts s
z = x2 e y + y ln x
∂z
∂x= ∂
∂x(x2 e y + y ln x) = 2x e y + y· 1
xy is fixed
∂z
∂ y= ∂
∂ y(x2 e y + y ln x) = x2 e y + ln x x is fixed
x = s2 cos t
∂x
∂s= ∂
∂s(s2 cos t) = 2s cos t t is fixed
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Calculus — Chain Rules for Partial Derivatives © Wei-Chau Xie
y = 4t2 sin 2s
∂ y
∂s= ∂
∂s(4t2sin 2s) = 8t2 cos 2s t is fixed
∴∂z
∂s= ∂z
∂x
∂x
∂s+ ∂z
∂ y
∂ y
∂s
=(
2x e y + y
x
)
(2s cos t) + (x2 e y + ln x) (8t2 cos 2s)
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Calculus — Partial Derivatives © Wei-Chau Xie
Implicit Function
Equation F(x, y) = 0 defines y as an implicit function of x: y = y(x).
Differentiate F[x, y(x)] = F(x, y) = 0 with respect to x:
dF
dx= ∂F
∂x+ ∂F
∂ y
dy
dx= 0
F
y
x
x
∴dy
dx= −
∂F
∂x
∂F
∂y
, provided∂F
∂y6= 0
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Calculus — Partial Derivatives of Implicit Functions © Wei-Chau Xie
Example
F(x, y) = yex + sin(x y) + 4 = 0, finddy
dx.
∂F
∂x= ∂
∂x[ yex + sin(x y) + 4] = y ex + cos(x y)·y y is fixed.
∂F
∂y= ∂
∂y[ yex + sin(x y) + 4] = ex + cos(x y)·x x is fixed.
dy
dx= −
∂F
∂x
∂F
∂y
= −y [ex + cos(x y)]ex + x cos(x y)
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Calculus — Multiple Integrals © Wei-Chau Xie
Double Integrals
Let f(x, y) be defined in a closed region R of the x y-plane.
x
1Ak (xk, yk)
y
R
Divide R into n small regions of area 1Ak, k = 1, 2, . . . , n.
Let (xk, yk) be some point in 1Ak.
∫∫
R
f(x, y) dA = limn→∞
n∑
k=1
f(xk, yk) 1Ak
in which max1Ak → 0 when n → ∞.
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Calculus — Multiple Integrals © Wei-Chau Xie
Double Iterated Integrals
Convert double integrals to double iterated integrals.
Use horizontal lines to divide the region into horizontal strips.
x
x=x1(y) x=x2(y)
y
R
y
c
d
dy
dAy+dy
∫∫
R
f(x, y) dA =∫ d
y=c
{ ∫ x2( y)
x=x1( y)f(x, y) dx
}
dy
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Calculus — Double Iterated Integrals © Wei-Chau Xie
Use vertical lines to divide the region into vertical strips.
y= y2(x)
dx
dA
x
y
xa bx+dx
R
y= y1(x)
∫∫
R
f(x, y) dA =∫ b
x=a
{ ∫ y2(x)
y= y1(x)
f(x, y) dy
}
dx
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Calculus — Double Integrals © Wei-Chau Xie
Some Physical Meanings of I =∫∫
R
f(x, y) dA
If f(x, y)= 1 in R, then
I =∫∫
R
f(x, y) dA =∫∫
R
1 · dA = Area of region R
If f(x, y)=ρ(x, y)= density, I is the mass of a plate with unit thickness,
cross section R, and variable density ρ(x, y).
If z = f(x, y), I is the volume of a cylinder with flat bottom R and height
f(x, y).
x
z
y
z = f(x,y)
R
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Calculus — Double Iterated Integrals © Wei-Chau Xie
Example
Evaluate
∫∫
R
x2 y dA, where R is bounded by y =√
x+4, y = 0, x+ y = 2.
Method 1: Take horizontal strips
x
y
1
y= x+4x
y=0
x+y=2
x=2–yx=y2–4
2
–4 –3 –2 –1 0 1 2
R
∫∫
R
x2 y dA =∫ 2
y=0
[ ∫ 2−y
x=y2−4x2 y dx
]
dy =∫ 2
y=0y[
13
x3]2−y
x=y2−4dy
= 13
∫ 2
y=0(72 y − 12 y2 − 42 y3 − y4 + 12 y5 − y7) dy
= 13
[
36 y2 − 4 y3 − 212
y4 − 15
y5 + 2 y6 − 18
y8]2
y=0= 56
5
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Calculus — Double Iterated Integrals © Wei-Chau Xie
Method 2: Take vertical strips
x
y
1
y= x+4x
y= x+4x
y=0 y=0
x+y=2
y=2–x
2
–4 –3 –2 –1 0 1 2
R1 R2
∫∫
R
x2 y dA =∫ 0
x=−4
[ ∫
√x+4
y=0x2 y dy
]
dx +∫ 2
x=0
[ ∫ 2−x
y=0x2 y dy
]
dx
=∫ 0
x=−4x2
[
12
y2]
√x+4
y=0dx +
∫ 2
x=0x2
[
12
y2]2−x
y=0dx
= 12
[ ∫ 0
x=−4(4x2 + x3) dx +
∫ 2
x=0(4x2 − 4x3 + x4) dx
]
= 12
{
[
43
x3 + 14
x4]0
x=−4+
[
43
x3 − x4 + 15
x5]2
x=0
}
= 565
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Calculus — Multiple Integrals © Wei-Chau Xie
Triple Integrals
Let V be a closed volume bounded by a piecewise smooth surface S.
Divide V into n small volumes 1Vk, k = 1, 2, . . . , n.
Let (xk, yk, zk) be some point in 1Vk.
x
z
S
y
1Vk
V
(xk, yk, zk)
∫∫∫
Vf(x, y, z) dV = lim
n→∞
n∑
k=1
f(xk, yk, zk) 1Vk
in which max1Vk→0 when n→∞.
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Calculus — Multiple Integrals © Wei-Chau Xie
Triple Iterated Integrals
Triple integrals are evaluated using triple iterated integrals.
x
z
yz=z1(x,y)
z=z2(x,y)
Sxy
∫∫∫
Vf(x, y, z) dV =
∫∫
Sxy
{ ∫ z2(x, y)
z=z1(x, y)f(x, y, z) dz
}
dA
A double integral over Sxy is obtained after evaluating the inner integral.
55
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Calculus — Triple Iterated Integrals © Wei-Chau Xie
Example
Find the volume bounded by the surfaces x+ y+z = 4, y = 3z, x = 0, y = 0.
x
z
yy=3z
x+y+z =4
y+z =4
x=4–y–z
x=0
4
4
4
z
Syz
y
y=3z
4
430
V =∫∫∫
VdV =
∫∫
Syz
{ ∫ 4−y−z
x=0dx
}
dA =∫∫
Syz
x∣
∣
∣
4−y−z
x=0dA
=∫∫
Syz
(4 − y − z) dA
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Calculus — Triple Iterated Integrals © Wei-Chau Xie
z
Syz
y
y=3z
z =4–y
4
430z=
y
3
V =∫∫
Syz
(4 − y − z) dA =∫ 3
y=0
{ ∫ 4−y
z= y3
(4 − y − z) dz
}
dy
=∫ 3
y=0
[
4z − yz − 12
z2]4−y
z= y3
dy =∫ 3
y=0
(
89
y2 − 163
y + 8)
dy
=[
827
y3 − 83
y2 + 8 y]3
y=0= 8
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Calculus — Vector Fields © Wei-Chau Xie
Scalar Field
If f(x, y, z) is a scalar function defined at every point (x, y, z) in a region of
space, then the function is a scalar field.
Only one quantity (magnitude) is required to describe a scalar field.
For example, temperature field T(x, y, z) is a scalar field.
Vector Field
If F(x, y, z) = P(x, y, z) i + Q(x, y, z) j + R(x, y, z) k is a vector function de-
fined at every point (x, y, z) in a region of space, then it is a vector field.
Two quantities are required to define a vector field: the scalar field of its
magnitude and the vector field of its direction.
For example, a force field F(x, y, z) and the velocity field V(x, y, z) of a fluid
are vector fields.
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Calculus — Vector Fields © Wei-Chau Xie
Gradient of a Scalar Field
Denote
∇ =( ∂
∂x,
∂
∂y,
∂
∂z
)
= i∂
∂x+ j
∂
∂y+ k
∂
∂z
Gradient of a scalar field f(x, y, z) is defined as the vector field
grad f = ∇f =( ∂ f
∂x,
∂ f
∂y,
∂ f
∂z
)
= i∂ f
∂x+ j
∂ f
∂y+ k
∂ f
∂z
The gradient at a point is a vector pointing in the direction of the steepest
slope or grade at that point, i.e. the direction of the gradient is the direction
of the max rate of change of f.
The steepness of the slope at that point is given by the magnitude of the
gradient, i.e. the magnitude of gradient gives the max rate of change of f.
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Calculus — Vector Fields © Wei-Chau Xie
Normal Vectors of a Surface
If f(x, y, z) = 0 defines a surface S and f has continuous first partial derivatives,
then at any point P of S, the vector ∇f is normal to the surface S.
∴ Normal vector N = ∇f =⇒ Unit normal vector n = ∇f∣
∣∇f∣
∣
x
z
yf(x, y, z)=0
SP
N=∇f
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Calculus — Vector Fields © Wei-Chau Xie
Divergence of a Vector Field
Divergence of a vector field F(x, y, z) = P(x, y, z) i + Q(x, y, z) j + R(x, y, z) k is
defined as the scalar field
div F = ∇·F =( ∂
∂x,
∂
∂y,
∂
∂z
)·(P, Q, R) = ∂P
∂x+ ∂Q
∂y+ ∂R
∂z
Divergence measures the extend to which a vector field flow behaves like a
source or a sink at a given point.
If the divergence is nonzero at some point, then there must be a source (positive
divergence) or a sink (negative divergence) at the point.
A vector field with constant zero divergence is called incompressible or
solenoidal. In this case, no net flow can occur across any closed surface.
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Calculus — Vector Fields © Wei-Chau Xie
Curl of a Vector Field
Curl of a vector field F(x, y, z) = P(x, y, z) i + Q(x, y, z) j + R(x, y, z) k is defined
as the vector field
curl F = ∇×F =
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
i j k
∂
∂x
∂
∂y
∂
∂z
P Q R
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
=(∂R
∂y− ∂Q
∂z
)
i +(∂P
∂z− ∂R
∂x
)
j +(∂Q
∂x− ∂P
∂y
)
k
Curl describes the infinitesimal rotation of a 3-dimensional vector filed.
The direction of the curl is the axis of rotation, as determined by the right-hand
rule. The magnitude of the curl is the magnitude of rotation.
A vector field whose curl is zero is called irrotational.
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Calculus — Vector Fields © Wei-Chau Xie
Laplacian
Laplacian of a scalar function f(x, y, z) is the divergence of the gradient of the
function
∇2f = ∇·∇f =( ∂
∂x,
∂
∂y,
∂
∂z
)·( ∂ f
∂x,
∂ f
∂y,
∂ f
∂z
)
= ∂
∂x
(∂ f
∂x
)
+ ∂
∂y
(∂ f
∂y
)
+ ∂
∂z
(∂ f
∂z
)
∴ ∇2f = ∇·∇f = ∂2f
∂x2+ ∂2f
∂y2+ ∂2f
∂z2
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Calculus — Surface Integrals © Wei-Chau Xie
Surface Integral Involving Vector Fields
Let f(x, y, z) = F(x, y, z)·n, where n is the unit normal vector to surface S.
Surface integral
∫∫
SF(x, y, z)·n dS is called the flux of vector field F(x, y, z).
The Divergence Theorem (Gauss)
Let S be a closed surface and n the unit outward normal to S. V is the region
enclosed by S. Then
∫∫
⊂⊃SF·n dS =
∫∫∫
V∇·F dV
If F(x, y, z) = P(x, y, z) i + Q(x, y, z) j + R(x, y, z) k, then
∫∫
⊂⊃SF·n dS =
∫∫
⊂⊃S(P i + Q j + R k)·n dS =
∫∫∫
V
(∂P
∂x+ ∂Q
∂y+ ∂R
∂z
)
dV
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Calculus — Surface Integrals © Wei-Chau Xie
Example
Evaluate the surface integral∫∫
⊂⊃S(2x yz i − y2z j + xz k)·n dS
where S is the surface enclosing the volume bounded by x= y, x+ y=2, z=0,
z=2, and x=0, and n is the unit inward normal to S.
Apply the Divergence Theorem
I =∫∫
⊂⊃S(2xyz i − y2z j + xz k)·n dS
= −∫∫∫
V
{
∂(2xyz)
∂x+ ∂(−y2z)
∂y+ ∂(xz)
∂z
}
dV
= −∫∫∫
V(2 yz − 2 yz + x) dV
= −∫∫∫
Vx dV
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Calculus — Surface Integrals © Wei-Chau Xie
x
x
z
y
y=x
y=x
x+y=2
y=2–x
z=2
z=0
2
2
2
Sxy
y
2
1
20
z=0Sxy
Intersection of x = y and x+y = 2 =⇒ x = 1, y = 1.
I = −∫∫∫
Vx dV = −
∫∫
Sxy
{ ∫ 2
z=0x dz
}
dA = −∫∫
Sxy
xz∣
∣
∣
2
z=0dA
= −∫ 1
x=0
{ ∫ 2−x
y=x2x dy
}
dx = −∫ 1
x=02xy
∣
∣
∣
2−x
y=xdx
= −∫ 1
x=04(x−x2) dx = −4
[
12
x2 − 13
x3]1
x=0= −2
3
66