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Calculus 2 Notes
1 Integration
Basic Definitions and Theorems
- Let ∆x = b−n a, let xi = a + i∆x, let Ii = [xi− 1 , xi]
- Regular partition of [a, b] into n subintervals: I 1 , I 2 , I 3 ,... In
- Riemann Sum:
Pn i=
f (x∗ i )∆x
Z (^) b
a
f (x) dx = lim n→∞
X^ n
i=
f (x∗ i )∆x
- Theorem: If f (x) is continuous or has finitely many discontinuites on [a, b], then f (x) is integrable on [a, b]
- Ln =
Pn i=
f (xi− 1 )∆x, Rn =
Pn i=
f (xi)∆x
Properties of Integrals
Z (^) b
a
cf (x) dx = c
Z (^) b
a
f (x) dx
Z (^) b
a
(f (x) + g(x)) dx =
Z (^) b
a
f (x) dx +
Z (^) b
a
g(x) dx
- If m ≤ f (x) ≤ M ∀x ∈ [a, b], then m(b − a) ≤
Z (^) b
a
f (x) dx ≤ M (b − a)
- If f (x) ≥ g(x) ∀x ∈ [a, b], then
Z (^) b
a
f (x) dx ≥
Z (^) b
a
g(x) dx
Z (^) b
a
f (x) dx ≤
Z (^) b
a
|f (x)| dx
- Separating the Domain Theorem:
Z (^) b
a
f (x) dx =
Z (^) c
a
f (x) dx +
Z (^) b
c
f (x) dx
- The definite integral represents the signed area under f (x)
- If f (x) is an odd function, then
Z (^) a
−a
f (x) dx = 0
- If f (x) is an even function, then
Z (^) a
−a
f (x) dx = 2
Z (^) a
0
f (x) dx
Average Value
- Average Value of f (x) on [a, b]: (^) b−^1 a
Z (^) b
a
f (x) dx
- Average Value Theorem: If f (x) is continuous on [a, b], then there exists
c ∈ [a, b] such that f (c) = (^) b−^1 a
Z (^) b
a
f (x) dx
Antiderivatives
- Antiderivative: F (x) is an antiderivative of f (x) if F ′(x) = f (x)
- Antiderivative Theorem: If F (x), G(x) are antiderivatives of f (x) on an interval I, then F (x) = G(x) + C for a constant C
- Indefinite Integral:
Z
f (x) dx = F (x) + C
Fundamental Theorem of Calculus
Z (^) x
a
f (x) dx = f (x)
- Extended FTC Part 1: (^) ddx
Z (^) h(x)
g(x)
f (x) dx = f (h(x))h′(x) − f (g(x))g′(x)
Z (^) b
a
f (x) dx = F (b) − F (a)
Evaluating Integrals
- Power Rule: For n ̸= − 1 ,
Z
xn^ dx =
xn+ n + 1
+ C
Z
x−^1 dx = ln |x| + C
Z (^) b
a
f (g(x))g′(x) dx =
Z (^) g(b)
g(a)
f (u) du
Z (^) b 2
b 1
f (x) dx =
Z (^) a
b 1
f (x) dx +
Z (^) b 2
a
f (x) dx
Z 1
0
xp^
dx converges for p < 1
Area Between Curves
- If y 1 (x) ≤ y 2 (x) ∀x ∈ [a, b], then the area between y 1 (x) and y 2 (x) for
x ∈ [a, b] is
Z (^) b
a
(y 2 (x) − y 1 (x)) dx
- If x 1 (y) ≤ x 2 (y) ∀y ∈ [c, d], then the area between x 1 (y) and x 2 (y) for
y ∈ [c, d] is
Z (^) d
c
(x 2 (y) − x 1 (y)) dy
Volumes of Solids of Revolution
Z (^) b
a
πf (x)^2 dx
Z (^) b
a
π(g(x)^2 − f (x)^2 ) dx
Z (^) b
a
2 πxf (x) dx
2 Differential Equations
Basic Definitions
- Differential Equation: Equation relating f (x) and its derivatives
- Order of DE: Highest order derivative that appears in the DE
- General Solution: Complete set of all solutions to a DE
- Particular Solution: A solution to a DE where all arbitrary constants are assigned a value
- Initial Value Problem: A DE with initial conditions that can be used to find the value of constants
- Direction Fields: A plot showing the slope at every point for a DE in the form (^) ddyx = f (x, y)
- Solution Curve: The graph of a particular solution to a DE. Can be visu- alized from the direction field.
Separable DE
- Definition: A DE that can be written in the form d dyx = g(x)h(y)
- Solving a Separable DE
- Write the DE in the form d dyx = g(x)h(y)
- Find any solutions with h(y) = 0
- For h(y) ̸= 0, evaluate
Z
h(y) dy =
Z
g(x) dx and solve for y
- A suitable substitution can be used to transform a non-separable DE into a separable DE
Linear DE
- Defintion: A DE in the form An(x)y(n)^ + An− 1 (x)y(n−1)^ + · · · + A 1 (x)y′^ + A 0 (x)y = B(x)
- Standard Form of First Order Linear DE: y′^ + P (x)y = Q(x)
- Solving a First Order Linear DE
Z
P (x) dx
choosing any value for C
- Multiply by μ(x): μ(x)(y′^ + P (x)y) = μ(x)Q(x) which becomes (yμ(x))′^ = μ(x)Q(x)
- Integrate both sides and solve for y
Applications
- Mixing Problems: d dAt = rin(t) − rout(t)
- Newton’s Law of Heating/Cooling: d dTt = −k(T − Ts) for k ∈ R+
- Solution: T (t) = Ts + Ae−kt, A = T (0) − Ts
- Exponential Growth/Decay: d dPt = kP for k ∈ R
- Solution: P (t) = Aekt, A = P (0)
- Logistic Growth/Decay: d dPt = kP
1 − MP
for k ∈ R
- Solution: P (t) = (^) 1+AeM−kt , A = M P^ − (0)P^ (0)
- p-series
- Can be used for series in the form
P 1
np
- Converges for p > 1
- Diverges for p ≤ 1
- Telescoping Series
- Can be used for series in the form
P
(an − an+k) for any k ∈ Z
- Use cancellations to find a closed-form expression for Sm
- Converges if lim m→∞ Sm converges
- Diverges if lim m→∞ Sm diverges
- Use partial fraction decomposition if needed
- Can be used for find the value of the infinite series if it converges
- Integral Test
- Can be used for series in the form
P
an if f (n) = an ∀n ∈ N and f (x) is positive, continuous, decreasing on [k, ∞) for some k ∈ R
Z ∞
k
f (x) dx converges
Z ∞
k
f (x) dx diverges
- Useful if f (x) is easily integrable
- Direct Comparison Test
- Can be used for two series
P
an,
P
bn if 0 ≤ an ≤ bn eventually
P
bn converges, then
P
an converges
P
an diverges, then
P
bn diverges
- A geometric series or p-series works well as a comparison
- Limit Comparison Test
- Can be used for two series
P
an,
P
bn if an ≥ 0 , bn > 0 eventually
an bn
- If L > 0 and finite, then both series converge or both series diverge
- If L = 0 and
P
bn converges, then
P
an converges
P
bn diverges, then
P
an diverges
- A geometric series or p-series works well as a comparison
- Ratio Test
- Can be used for series in the form
P
an
an+ an
- Converges (absolutely) if L < 1
- Diverges if L > 1
- Inconclusive if L = 1
- Useful if an has exponentials, factorials, nn
- Root Test
- Can be used for series in the form
P
an
p n|a n|
- Converges (absolutely) if L < 1
- Diverges if L > 1
- Inconclusive if L = 1
- Useful if an = (bn)n
- Alternating Series Test
- Can be used for series in the form
P
(−1)nan if an > 0
- Converges if lim n→∞ an = 0 and 0 < an+1 ≤ an
- Diverges if lim n→∞ an ̸= 0
- Inconclusive if ak+1 > ak for some k ∈ N
- Absolute Convergence Test
- Can be used for series in the form
P
an
P
|an| converges
Estimation Theorems
- Integral Test Estimation Theorem: Suppose that an = f (n) for all n ∈ N andZ f (x) is positive, continuous, decreasing on [k, ∞) for some k ∈ Z, with ∞
k
f (x) dx being convergent. Then, for S =
P∞
n=
an,
Z ∞
m+
f (x) dx ≤
S − Sm ≤
Z ∞
m
f (x) dx for m ≥ k
- Alternating Series Estimation Theorem: Suppose that lim n→∞ an = 0 and
0 < an+1 ≤ an. Then, for S =
P∞
n=
(−1)nan, |S − Sm| ≤ am+
- Substitution: For a = 0, f (bxk) =
P∞
n=
cnbnxkn, where b ∈ R, b ̸= 0, k ∈ N
∗ If Rf is finite, then the new radius of convergence is
Rf |b|
1 /k
∗ If Rf is infinite, then the new radius of convergence is also infinite ∗ The new interval of convergence is {x ∈ R | bxk^ ∈ If }
Differentiating and Integrating Power Series
d dx
X^ ∞
n=
cn(x − a)n^ =
X^ ∞
n=
ncn(x − a)n−^1
Z X∞
n=
cn(x − a)n^ dx =
X^ ∞
n=
cn n + 1
(x − a)n+1^ + C
- The radius of convergence will never change, but the interval of conver- gence may change
Taylor Series
- Taylor Series of f (x) at a:
P∞
n=
f (n)(a) n! (x^ −^ a)
n
- Maclaurin Series of f (x):
P∞
n=
f (n)(0) n! x
n
- Order m Taylor Polynomial of f (x) at a: Tm,a(x) =
Pm n=
f (n)(a) n! (x^ −^ a)
n
- Uniqueness of Power Series Theorem: If f (x) =
P∞
n=
cn(x − a)n, then cn is
unique and cn = f^
(n)(a) n!
- Lagrange’s Remainder Formula: If f (m+1)(x) is continuous on an interval I containing a, and x ∈ I, then ∃c between a and x where f (x)−Tm,a(x) = f (m+1)(c) (m+1)! (x^ −^ a)
m+
- Taylor’s Inequality: If f (m+1)(x) is continuous on an interval I containing a, and |f (m+1)(x)| ≤ k for all x ∈ I with k ∈ R, then |f (x) − Tm,a(x)| ≤ k (m+1)! (x^ −^ a)
m+
- Convergence Theorem for Taylor Series: If |f (n)(x)| ≤ k for all n ∈ N and x ∈ I, then f (x) =
P∞
n=
f (n)(a) n! (x^ −^ a)
n (^) for all x ∈ I
Common Power Series
X^ ∞
n=
xn n!
for x ∈ R
X^ ∞
n=
(−1)nx^2 n (2n)!
for x ∈ R
X^ ∞
n=
(−1)nx^2 n+ (2n + 1)!
for x ∈ R
1 − x
X^ ∞
n=
xn^ for |x| < 1
X^ ∞
n=
(−1)n−^1 xn n
for x ∈ (− 1 , 1]
X^ ∞
n=
(−1)nx^2 n+ 2 n + 1
for |x| ≤ 1
X^ ∞
n=
α n
xn^ for |x| < 1 if α /∈ Z≥ 0 and x ∈ R if α ∈ Z≥ 0
α n
nQ− 1 k=
(α − k)
n!
Big-O Notation
- f (x) = O(g(x)) as x → a if ∃M > 0 such that |f (x)| ≤ M |g(x)| for all x in a neighbourhood of a
- Theorems
- If f (x) = O(g(x)) as x → a and g(x) = O(h(x)) as x → a, then f (x) = O(h(x)) as x → a
- If f (x) = O(g(x)) as x → a and lim x→b h(x) = a with h(x) ̸= a for all x in a neighbourhood of b, then f (h(x)) = O(g(h(x))) as x → b
- If f (x) = O(xm) as x → 0 and g(x) = O(xn) as x → 0, then ∗ f (x)g(x) = O(xm+n) as x → 0 ∗ f (x) ± g(x) = O(xmin{m,n}) as x → 0 ∗ xf (x) = O(xm+1) as x → 0 ∗ (^) x^1 f (x) = O(xm−^1 ) as x → 0 if m ≥ 1
