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Lecture notes for capture one of Calculus MATH1131 at UNSW.
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Calculus Option Notes [RR] IB Math HL Watch this if you need motivation : https:!!!"youtu#e"com!atch$v%&'(N)*y+,-.
"u"re9; 6s pla8list o! Calculus videos made b8 (ro!essor Leonard https:""www#reddit#com"r"IO"comments"07<<8p"lectures9on9the9calculus9option9b89pro!e ssor"
=hat the >uestion asks:
=hat 8ou have to do ? 5otes:
/"0 1 Basics Show that
( x) l i mf
e@ists
( x) l i m f
*ind
( x) l i m f
and
x → a+¿
x →a
the limit
x → a− ¿
as it reaches the point !rom the le!t side and !rom the right sideB and show that the8 are e>ual# Show that !@B is continuous
Show that
( x) l i mf
e@ists and that it is
→a x
e>ual to !aB# Show that !@B is di!!erentiable at a
Show that !@B is continuous at a and: (a+ h)−f(a) f l i m and −¿ h→ a
l i m h→ a +¿
h
( a+ h)−f( a) f h
e@ist and are e>ual#
/"2 3ho! that se4ies 5a n6 is conve47ent dive47ent [Infinite 3e4ies] : 8 comparison test
*ind a series bnD where bnD F anD and anDbnD F - B# I! bnD converges anD is !orced to converge as well# Ghis can also be used to test !or divergence# I! bnD is divergent and anD F bnD anD is divergent as well# =hen to use this:
1#4 se the s>ueee theorem to evaluate a limit lim x → -
g ( x )
4# *ind a !@B and a h@B such that: !@B J g@B J h@B # and
( x)=l =l ( x) l i mf i mh x→ a
x →a
$# I! both o! these conditions appl8
( x)=l l i mg x→ a
8 limit comparison test
*ind a series bnD !or which 8ou know whether it converges or diverges and anDbnD F - B I!
l i m n→∞
an
= c where c is a positive !inite
bn
value both series will either converge or diverge#
8 integral test:
Let ! be a positive decreasing on 4 and let an N !nB# Ghen the series
∑a
∞
∞
K5ote anD is the same as
∑a
n
is convergent i!
n= 4 n
∫ f( x) dx is divergent 1
∞
∑a
1
∞
convergent# I!
∫ f( x) dx is also
n
is also divergent#
n= 4
8 ratio test: Ksed to test i! a series is absolutel8 convergent divergent or inconclusive
*ind an?4# istribute a ? 4 !or all the n6s in anB Calculate L=lim n→∞
A series is a#solutely conve47ent i! ∞ ∑ |a n| is also convergent# Otherwise n= 4
| | an+4 an
I! L4 the series is absolutel8 convergent thus convergent
conditionally conve47ent# I! LP4 the series is divergent I! LN4 the series ma8 be divergent conditionall8 convergent or absolutel8 convergent
5th term test
'
Qemoetric series
'
4
p'series
∑n
p
is convergent i! p P 4 otherwise it is
divergent# I! pN4 it is the harmonic series
5OGE: ('series is use!ul when testing !or convergence using the other tests# E#g# it can be used !or comparison test as it is known that the p series converges " diverges# Absolute value test
'
/"8: 9iffe4ential ;uations <0 Show that a general solution !or di!!erential e>uation
4# *ind derivative o! general solution provided $# (lug in derivative !ound !or di!!erential e>uation given
raw a slope !ield
'
KInitial value problem IRG
Go appro@imate a value !or 8cB: 4# Start with writing down intial condition @- 8$# Obtain @4 b8 adding step sie @ to intial condition @- # @4N @- ? h %# Obtain 84 b8:
Qiven an initial condition use Euler6s method o! numerical integration with a step sie o! h to !ind an appro@imate value !or 8cB
y 4= y - + h (
dy ) where d8 over d@ is !ound dx
using the values @- and 8- and h is step sie# 0# +epeat this process until 8cB is obtained# i#e# 8n?4 N 8n ? hf(xn, yn ) , @n?4 N @n?h o all o! that on calculator KSeperable di!!erential e>uations Integrate a !unction dy f ( x ) = or something such dx g ( y )
4# Qet e>uation in the !orm o!: dy g ( x )⋅ = f ( x ) dx
$# Integrate both sides with respect to @ dy
∫ g ( x )⋅dx dx =∫ f ( x )dx
%# Ghis gives
∫ g ( x )⋅dy ∫ f ( x ) dx 0# Integrate both sides# Sometimes it6s nice to simpli!8 the constant o! integration# E#g# 4
dy
4
4
∫ y dx dx =∫ ( 4 + x−4 − x + 4 )dx 4
∫ y dy = x + ln | x + 4|−ln | x −4| + c let lncB N A ln y = x + ln ( A
| x x−+ |) 4 4
Ghere!ore x
y = Ae (
x −4 ) x + 4
Solve a homogeneous e>uation
4# se substitution 8Nv@
It can be deduced that an e>uation is homogeneous when 8ou cannot simpl8 get the e>uation in the !orm
$# 8 product rule
dy g ( x )⋅ = f ( x ) dx
A homogeneous e>uation is o! the !orm dy y =f ( ) where 8N8@B dx x
where v is a !unction o! @#
dy dv = dx + v dx dx
%# Compare this to di!!!erential e>uation dy y =f ( ) given# dx x
0# Qet this in the !orm o! a seperable di!!erntial e>uation where one side is integrated with respect to v and one side is integrated with respect to @
Solve a di!!erential e>uation in the !orm: dy + P ( x ) y =Q ( x ) dx
4# *ind (@B $# Integrating !actor I@B is !ound b8: ∫ P( x ) dx
I ( x )=e
=hich is not seperable
in !ormula bookletB
%# Calculate this integrating !actor 0# Multipl8 the !ull di!!erential e>uation b8 I@B 3# Simpli!8 the LHS to obtain I ( x ) y =∫ I ( x ) Q ( x ) dx + c
2# Integrate to obtain general solution
/"= Riemann sums> ?*undamental theo4em of calculus?> Imp4ope4 inte74als *ind the area under a curve using riemann sums
4# *ind delta @ given b8:
Δ x =
Identities:
b
integral
n
∑ c =cn ∑ i= n
∑ i= i = 4$
# Lower and upper
values o! interval Kab where b P a#
n (n +4 )
i= 4
∫ f ( x ) dx a
i= 4 n
b− a a and b are given b8 de!inite n
$# *ind @iT given b8:
$
x i∗= a +Δ x ( i ) n ( n + 4 )( $ n+ 4) 2
%# +iemann sum is given b8: b
n
∫ f ( x ) dx =lim ∑ f ( x ∗)Δ x →∞ a
n
i
i= 4
0# Calculate the area b8 using the identities# /ou will want to move an8thing 6n' out o! the sum and keep 6i' in the sum#
Lower and upper values !or +iemann Sum
'
*undamental theorem o! calculus
'
Evaluate an improper integral o! the !orm ∞
4# =rite integral in the !orm o!: b
∫ f ( x ) dx
lim
a
b→∞
O!ten includes techni>ues such as integration b8 parts limit laws l6Hopital6s +ule and integration b8 substitution#
∫ f ( x ) dx a b
$# Integrate
∫ f ( x ) dx a
%# Move all that does not have b in it to the le!t hand side o! the limit as it won6t be a!!ected b8 b tending towards in!init8B 0# Evaluate the limit 3# Solve improper integral etermine whether an imp4ope4 inte74al o! If inte74al can #e found easily: ∞
the !orm
∫ f ( x ) dx
is convergent or
4# Integrate what is given
a
divergent#
b
$# Convergent i!
∫ f ( x ) dx
e@ists !or all b
a b
where a J b and i!
lim b→∞
∫ f ( x ) dx
e@ists#
a
Otherwise divergent#
By compa4ison test: *or convergence: *ind g@B that is greater than or e>ual to !@B !or all @ F a# ∞
∫ g ( x ) dx
I! ∞
is convergent so is
a
∫ f ( x ) dx a
*or divergence: *ind g@B that is smaller or e>ual to !@B !or all @ F a# ∞
I!
∫ g ( x ) dx a
∞
is divergent so is
∫ f ( x ) dx a
O!ten uses p series as convergence and divergence is known
1#2 etermine whether or not Rolle?s theo4em 4# (lug in values a and b in !unction !@B applies to the !unction ! on the given interval Kab $# I! both result in - +olle6s theorem guarantees that there is at least one point c at which !6cBN-# Show that the given !unction ! satis!ies the Mean @alue Aheo4em M@A on the given interval Kab# *ind all values o! c such that !bB'!aB N !6cB @ b'aB
4# Show that !@B is continuous on Kab and di!!erentiable on abK then MRG guarantees !bB'!aB N !6cB @ b'aB !or some value c U abK#
K !bB'!aB N !6cB @ b'aB basicall8 means that $# Ralues !or !6cB can be !ound b8 plugging there must be a value c U abK where the in values in !bB'!aB N !6cB @ b'aB# gradient !6cB is e>ual to the gradient f ( b )− f ( a ) 7oining two outer values o! b− a interval together *ind the nth de74ee Aaylo4 polynomial appro@imation to !@B about @Na
=atch (ro!essor Leonard 6s video about Ga8lor Series and Maclaurin Series Ki!!icult to summarie
se o! error term Ga8lor6s !ormula with remainder
'
Maclaurin series !or e>uations given in !ormula booklet
'
Ga8lor series developed !rom di!!erentials
'
se o! substitution products integration and ' di!!erentiation to obtain other series#
/"D l?Hopital?s 4ule> evaluation of limits of fo4m fE 7E Evaluate limit o! the !orm !@B"g@B using l6Hopital6s
I! given a limit o! the !orm !@B"g@B which results indeterminate !orm either -"- in!init8 over in!init8 or - times in!init8B *ind !6@B and g6@B# Ghe limit o! !6@B " g6@B e>uals the limit o! !@B"g@B provided limit o! !6@B"g6@B e@ists#
3umma4y a#out se4ies and stuff liFe that Concept Alternating series
E@plained Germs in series are alternatel8 positive and negative# ∞
∑ (−4 ) −
I!
n
4
bn=b 4−b$ + b%−## . satis!ies
n= 4
- J bn?4 J bn !or all n U VZ ? and i! lim b n=- then the series is convergent# n→∞
? others (ower series
A series o! the !orm: ∞
n
$
lim cn x = c - + c 4 x + c $ x + ## . n=-
Ghree possibilities !or convergence o! a ∞ $ n lim cn ( x − a) =c -+ c 4 ( x− a )+ c $( x − a) + ## . n=-
' Series converges onl8 when @ N a ' Ghe series converges !or all n U |R+ ' Ghere e@ists a radius o! convergence + U |R+, such that series converges i! V@'aV + and diverges i! V@'aV P + +adius o! convergence
A value + which is the greatest number such that the series converges !or all @ U |R such that V@'aV + and diverges !or all @ U |R such that V@'aV P +
Interval o! convergence
Interval o! convergence I is the set o! all points @ !or which the power series converges#
Gruncating
'
Conditional vs absolute convergence
A series
∞
∑a
n
is absolutel8 convergent i!
n= 4
∞
the series o! absolute values
∑ |a | n
is
n= 4
convergent# I! an F - !or all n absolute convergence is the same as convergence# ∞
A series such as
∑ n= 4
(−4)n− n
4
which is
convergent but not absolutel8 convergent is called conditionall8 convergent# I! a series is absolutel8 convergent then it6s convergent#
inomial series
'
.inds o! continuit8
Continuous i! !aB is de!ined lim o! !@B !rom positive side and lim o! !@B !rom negative side e@ist and are e>ual and i! lim o! @ tending towards a o! !@B e>uals to !aB# I! conditions !ail !@B is discontinuous at @Na# I! lim o! @ tending to a o! !@B e@ists then ! has a 4emova#le discontinuity at @Na# Otherwise essential discontinuity#
*undamental theorem o! calculus
'
Monotone se>uence
'
Monotone convergence
'
(roduct o! series
'
Note on tests: Compa4ison tests can #e done fo4 inte74als> limits> se;uences and se4ies