# Calculus 2 Cheat Sheet/Summary for Exams

Calculus 2, Spring 2020, Practice final exam solutions
Calculus 2, Spring 2020, Practice final exam solutions

Calculus 2 Cheat Sheet/Summary for Exams

University: University of Melbourne

Course: Calculus 2 (MAST10006)

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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 P-SERIES: ∑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 p-series

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

√2-1

√2i)

De Moivre’s Theorem: zn=(reiθ)n= rneinθ

Modulus: |z| = √𝑥2+𝑦2

Argument: Arg(z)

Properties of Modulus and Argument

1. |zw| = |z||w| (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

∫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=y1-n 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

decrease and hence move closer to y = 3 as t inc.

If ½ < y < 3, then 𝑑𝑦

< 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

< 3 increase and hence move further from y = 1

2 as t increases.

If y < ½ , then 𝑑𝑦

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= kp-h h>0

o Logistic model

𝑑𝑝

𝑑𝑡= kp – 𝑘

𝑎p2 = kp(1-𝑝

𝑎) 𝑑𝑝

𝑑𝑡= rate

a = carrying capacity p = population

o Logistic model with harvesting

𝑑𝑝

𝑑𝑡= kp – 𝑘

𝑎p2 = kp(1-𝑝

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)