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Calculus 2 Cheat Sheet/Summary for Exams
University: University of Melbourne
Course: Calculus 2 (MAST10006)
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NOTES
Rejecting a value for quadratics: as it is not in the range of __
Integration of fractions: partial fractions or completing
square then integration by substitution or just integration by
substitution
When lim
𝑥→𝑛 1
(𝑥−𝑛)2, it becomes unbounded as x n, hence
cannot be arbitrarily close to any real number as xn,
therefore limit does not exist.
Derive of cosx = sinx, coshx = sinhx
LIMIT LAWS
1. lim
𝑥→𝑎[𝑓(𝑥)+𝑔(𝑥)] = lim
𝑥→𝑎𝑓(𝑥) + lim
𝑥→𝑎𝑔(𝑥)
2. lim
𝑥→𝑎[𝑐𝑓(𝑥)] = 𝑐 lim
𝑥→𝑎𝑓(𝑥)
3. lim
𝑥→𝑎[𝑓(𝑥)𝑔(𝑥)] = lim
𝑥→𝑎 𝑓(𝑥)∙lim
𝑥→𝑎𝑔(𝑥)
4. lim
𝑥→𝑎[𝑓(𝑥)
𝑔(𝑥)]= lim
𝑥→𝑎𝑓(𝑥)
lim
𝑥→𝑎𝑔(𝑥) provided that lim
𝑥→𝑎𝑔(𝑥)≠0
5. lim
𝑥→𝑎𝑐 = 𝑐 6. lim
𝑥→𝑎𝑥 = 𝑎
SANDWICH THEOREM
Now lim
𝑥→∞−𝑥=lim
𝑥→∞𝑥= n by standard limits
Hence by the sandwich rule, lim
𝑥→∞𝑓(𝑥)= n
GEOMETRIC SERIES: ∑𝑎𝑟𝑛∞
𝑛=0 SN=a(1−𝑟𝑛
1−𝑟 ) S∞=𝑎
1−𝑟
HARMONIC PSERIES: ∑1
𝑛𝑝
∞
𝑛=1 , p>1 converges, p<1 diverges
TESTS
Divergence: lim
𝑛→∞𝑎𝑛≠0, diverges
Comparison
1. an ≤ bn, converges 2. an ≥ bn, diverges
Process:
1. Check if f(n)>0 (positive)
2. Find for large n + expectation (div/conv)
𝑛+2
4𝑛3−2 ≈ 𝑛
4𝑛3 = 1
4𝑛2 (conv.)
3. For all n≥1
𝑛+2
4𝑛3−2 ≤𝑛+2𝑛
4𝑛3−2𝑛3 = 3
2𝑛2
4. Check (3) conv/div by harmonic pseries
5. Therefore f(n) conv/div by comparison test
Ratio (has to be f(n)>0)
L = lim
𝑛→∞𝑎𝑛+1
𝑎𝑛, L<1 conv, L>1 div, L=1 inconclusive
Example:
6𝑛
𝑛! is of form ∑𝑎𝑛∞
𝑛=1 , where an>0
𝑎𝑛+1
𝑎𝑛=6𝑛+1
(𝑛+1)! X 𝑛!
6𝑛 = 6
𝑛+1
CONTINUITY LAWS
Theorem 2
Combination of continuous functions
Theorem 3
nth root functions, trig (state the domain,
and if it is being derived for a definite limit,
state that continuity at x = n). poly
(anything that is not just x), ex (x ∈ 𝑅),
log(x) (x>0)
Definition of continuity
lim
𝑥→𝑛𝑓(𝑥)= f(n)
For lim
𝑥→𝑛𝑓(𝑥) to exist, we require
lim
𝑥→𝑛−𝑓(𝑥) = lim
𝑥→𝑛+𝑓(𝑥)
When c= a, f(n)= value = lim
𝑥→𝑛𝑓(𝑥), so f
continuous at x = n
Hence f continuous on (∞, n]U[n, ∞)
COMPLEX NUMBERS
Cartesian form: z= x +iy
Complex conjugate: z= x – iy (Im changes sign)
Properties:
1. z + z real
2. z – z imaginary
3. zz real
4. z + w = z + w
5. zw = zw
Division of complex numbers
𝑎+𝑖𝑏
𝑐+𝑖𝑑 multiply with conjugate to find Cartesian form
Finding Re: only use the real values
Finding Im: only use the imaginary values
Complex number: z=r(cosθ + isinθ)
Principle argument: π < θ < π
Polar form: z = reiθ
Polar to Cartesian:
r( 1
√21
√2i)
De Moivre’s Theorem: zn=(reiθ)n= rneinθ
Modulus: z = √𝑥2+𝑦2
Argument: Arg(z)
Properties of Modulus and Argument
1. zw = zw (for Im numbers, don’t square i)
2. 𝑧
𝑤 = 𝑧
𝑤
3. arg(zw) = arg(z) + arg(w)
4. arg(𝑧
𝑤) = arg(z) – arg(w)
INTEGRALS
Integration of square roots
Example: ∫√𝑥2−4 dx
Partial fractions
1. 𝑝𝑥+𝑞
(𝑥−𝑎)(𝑥−𝑏)=𝐴
𝑥−𝑎+𝐵
𝑥−𝑏
2. 𝑝𝑥+𝑞
(𝑥−𝑎)2=𝐴
𝑥−𝑎+𝐵
(𝑥−𝑎)2
3. 𝑝𝑥2+𝑞𝑥+𝑟
(𝑥−𝑎)(𝑥−𝑏)(𝑥−𝑐)=𝐴
𝑥−𝑎+𝐵
𝑥−𝑏 +𝐶
𝑥−𝑐
4. 𝑝𝑥2+𝑞𝑥+𝑟
(𝑥−𝑎)2(𝑥−𝑏)=𝐴
𝑥−𝑎+𝐵
(𝑥−𝑎)2+𝐶
𝑥−𝑏
5. 𝑝𝑥2+𝑞𝑥+𝑟
(𝑥−𝑎)(𝑥2+𝑏𝑥+𝑐)=𝐴
𝑥−𝑎+𝐵𝑥+𝐶
𝑥2+𝑏𝑥+𝑐
6. 𝑞𝑥+𝑟
(𝑥2−𝑎)(𝑥2−𝑏𝑥)=𝐴+𝐵𝑥
𝑥2−𝑎+𝐶+𝐷𝑥
𝑥2−𝑏
Integrand
Substitution
√𝑎2−𝑥2, 1
√𝑎2−𝑥2, (𝑎2−𝑥2)3
2, etc.
x = asinθ
x = acosθ
√𝑎2+𝑥2, 1
√𝑎2+𝑥2, (𝑎2+𝑥2)−3
2, etc.
x = asinhθ
√𝑥2−𝑎2, 1
√𝑥2−𝑎2, (𝑥2−𝑎2)5
2, etc.
x = acoshθ
1
𝑎2+𝑥2, 1
(𝑎2+𝑥2)2, etc.
x = atanθ
Long division
Numerator/denominator = result of fraction +
remainder/denominator
Integrating quadratics
∫𝑥2
1+𝑥2 dx = ∫𝑥2+1
1+𝑥2−1
1+𝑥2 dx = ∫1− 1
1+𝑥2 dx
= x – arctan(x) + C
Integrating exy
∫𝑒3𝑥 sin(2𝑥)𝑑𝑥
I = 𝑒3𝑥 sin(2𝑥)
Let u = e3x 𝑑𝑢
𝑑𝑥 = 3e3x
𝑑𝑣
𝑑𝑥 = sin(2x) v = ½ cos(2x)
By int. by parts,
I = e3x(½cos(2x))+½ ∫cos(2𝑥)3𝑒3𝑥 dx
I = −𝑒3𝑥
2cos(2x) + 3
2∫cos(2𝑥)3𝑒3𝑥dx
Let u = e3x 𝑑𝑢
𝑑𝑥 = 3e3x
𝑑𝑣
𝑑𝑥 = cos(2x) v = ½ sin(2x)
∫cos(2𝑥)3𝑒3𝑥dx = e3x½ sin(2x) – 3
2∫sin(2𝑥)𝑒3𝑥dx
Substitute into the first equation,
I = −𝑒3𝑥
2cos(2x) + 3
2(e3x½ sin(2x) – 3
2𝐼)
I = −2𝑒3𝑥
13 cos(2x) + 3
13e3xsin(2x) + C
FIRST ODE
If the arbitrary constant is c ∈ R, the solution will give a more general
solution than when A = ±ed ∈ R\{0}
Separable
𝑑𝑦
𝑑𝑥 = M(x)N(y) (M(x)≠0, N(y)≠0)
1
𝑁(𝑦)𝑑𝑦
𝑑𝑥=M(x)
∫1
𝑁(𝑦)dy = ∫𝑀(𝑥)dx
Linear
𝑑𝑦
𝑑𝑥 + P(x)y= Q(x)
𝐼(𝑥)= 𝑒∫𝑃𝑑𝑥
E.g: differential of cosh(u) is double angle of cosh(𝑢
2)
Homogeneous type ODE
𝑑𝑦
𝑑𝑥=f(𝑦
𝑥), substituting u=𝑦
𝑥 to make a separable
Bernoulli’s eqn
𝑑𝑦
𝑑𝑥+P(x)y = Q(x)yn, substituting u=y1n to make linear
Graphs
o Phase plot (𝒅𝒑
𝒅𝒕,p): looks like a curve, gen solns 𝑑𝑝
𝑑𝑥=0
o Family of solutions (p,t): lines curve in, p(0)=x
o Stability
In terms of whether or not it moves closer/further from the
point as t inc. Unrealistic: k>0 (unbounded exponential), k<0
(dies), k=0 (constant)
Example of explanations:
y = 3:
If y > 3, then 𝑑𝑦
𝑑𝑡 < 0, so solutions which start with y(0) > 3
decrease and hence move closer to y = 3 as t inc.
If ½ < y < 3, then 𝑑𝑦
𝑑𝑡> 0, so solutions which start with ½ < y(0)
< 3 increase and hence move closer to y = 3 as t increases. So
y = 3 is a stable equilibrium.
y = ½
If ½ < y < 3, then 𝑑𝑦
𝑑𝑡> 0, so solutions which start with ½ < y(0)
< 3 increase and hence move further from y = 1
2 as t increases.
If y < ½ , then 𝑑𝑦
𝑑𝑡 < 0, so solutions which start with y(0) < ½
decrease and hence move further from y = ½ as t increase. So
y = ½ is an unstable equilibrium.
Population models
o Malthus’s Doomsday model
p(t) = p0ekt p0 (initial) k= constant of proportionality (net
births/time)
o Doomsday model with harvesting
𝑑𝑝
𝑑𝑡= kph h>0
o Logistic model
𝑑𝑝
𝑑𝑡= kp – 𝑘
𝑎p2 = kp(1𝑝
𝑎) 𝑑𝑝
𝑑𝑡= rate
a = carrying capacity p = population
o Logistic model with harvesting
𝑑𝑝
𝑑𝑡= kp – 𝑘
𝑎p2 = kp(1𝑝
𝑎)h
𝑑𝑝
𝑑𝑡= rate a = carrying capacity p = population
h = harvesting rate
Example:
Finding the value of h for a sustainable ecosystem
Transient: decaying to 0 as t∞ Steady state: constants
(state that as lim
𝑡→∞𝑥 =0, thus at t∞ it will be at a fixed value)
FIRST ODE CONTINUED
Mixing problems
1. Find volume
V(t) = initial amount of liquid + volume added per hour x t –
volume removed per hour x t
2. Find conc.
C = 𝑥
𝑉
3. Find 𝒅𝒙
𝒅𝒕
𝑑𝑥
𝑑𝑡 = rate thing added in – rate thing removed
= flow rate in x conc (from Q) – flow rate out x conc (from prev
step)
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