Calculus · Units 6–9 · Quick reference

Calculus 2

Grades 11–12+ · College8 pages

Integration techniques and tables, improper integrals, area, volume, arc length, surface area, work and center of mass, sequences and series with every convergence test, power and Taylor series, parametric and polar curves, and differential equations.

1Toolkit2By parts3Trig integrals4Trig substitution5Partial fractions6Numerical7Improper8Area & volume9Length, work, mass10Sequences11Series12Convergence tests13Alternating14Power series15Taylor series16Parametric17Vectors & motion18Polar19Diff. equations20Slips & practice

1Integration Toolkitbasics: Calculus 1 §10–12

Integral table (a > 0; add + C)
∫ f(x) dx	= F(x) + C	∫ f(x) dx	= F(x) + C	∫ f(x) dx	= F(x) + C
xⁿ (n ≠ −1)	xⁿ⁺¹n + 1	tan x	ln\|sec x\|	1a² + x²	1a arctan xa
1x	ln\|x\|	cot x	ln\|sin x\|	1√(a² − x²)	arcsin xa
eˣ; aˣ	eˣ; aˣln a	sec x	ln\|sec x + tan x\|	1x√(x² − a²)	1a arcsec ∣x∣a
sin x; cos x	−cos x; sin x	csc x	−ln\|csc x + cot x\|	1a² − x²	12a ln\|a + xa − x\|
sec² x; csc² x	tan x; −cot x	sin² x	x2 − sin 2x4	1√(x² ± a²)	ln\|x + √(x² ± a²)\|
sec x tan x	sec x	cos² x	x2 + sin 2x4	√(a² − x²)	x2√(a² − x²) + a²2 arcsin xa
csc x cot x	−csc x	tan² x	tan x − x	ln x; x eᵃˣ	x ln x − x; eᵃˣa²(ax − 1)

Choosing a technique
If you see…	Try	If you see…	Try
a function times its derivative	u-substitution	a rational function P(x)/Q(x)	partial fractions (§5)
x eˣ, x sin x, ln x, arctan x	by parts (§2)	x² + bx + c that won’t factor	complete the square
powers of sin, cos, tan, sec	trig integrals (§3)	no antiderivative, or only data	numerical rules (§6)
√(a² − x²), √(a² + x²), √(x² − a²)	trig substitution (§4)	an infinite limit or a blow-up	improper integral (§7)

u-substitution, changing the limits Worked

∫02 x(x² + 1)³ dx: u = x² + 1, du = 2x dx
x = 0 → u = 1; x = 2 → u = 5new limits
12∫15 u³ du = 18(625 − 1) = 78no back-substitution

Recognize a table form Worked

∫ eˣ1 + e²ˣ dx: u = eˣ, du = eˣ dx

= ∫ du1 + u² = arctan(eˣ) + C

Complete the square Worked

x² + 4x + 13 = (x + 2)² + 9

∫ dx(x + 2)² + 9 = 13 arctan x + 23 + Ca = 3

2Integration by Partsthe product rule run backward

Integration by parts

Definite: ∫_a^b u dv = [uv]_a^b − ∫_a^b v du. Pick u so du is simpler; dv must be easy to integrate.

∫ u dv = uv − ∫ v du

LIATE (Traditional)

Choose u from the first type present: Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential. The rest is dv.

One round Worked

∫ x eˣ dx: u = x, dv = eˣ dx → du = dx, v = eˣ

= x eˣ − ∫ eˣ dx = eˣ(x − 1) + C

dv = dx Worked

∫ ln x dx: u = ln x, v = x
= x ln x − ∫ x · 1x dx = x ln x − x + C
∫ arctan x dx = x arctan x − 12 ln(1 + x²) + Csame idea

Cyclic: the integral comes back Worked

I = ∫ eˣ sin x dx = eˣ sin x − ∫ eˣ cos x dxparts once
= eˣ sin x − eˣ cos x − Iparts again
2I = eˣ(sin x − cos x) → I = eˣ(sin x − cos x)2 + Csolve for I

Tabular method: ∫ x² sin x dx
Sign	u and its derivatives	dv and its integrals
+	x²	sin x
−	2x	−cos x
+	2	−sin x
	0	cos x

Multiply down the diagonals with alternating signs: −x² cos x + 2x sin x + 2 cos x + C. Use it when u differentiates to 0.

Definite Worked

∫₁^e ln x dx = [x ln x − x]₁^e

= (e − e) − (0 − 1) = 1

Calculus 2Calculus · Units 6–9 · continued

3Trigonometric Integralsidentities first, then a substitution

Identities you need

Pythagorean: sin² x + cos² x = 1; 1 + tan² x = sec² x; 1 + cot² x = csc² x
Half-angle: sin² x = 1 − cos 2x2; cos² x = 1 + cos 2x2
Double angle: sin 2x = 2 sin x cos x
Product to sum: sin A cos B = 12[sin(A − B) + sin(A + B)]

Odd sine power Worked

∫ sin³x cos²x dx = ∫ (1 − cos²x) cos²x sin x dxu = cos x

= −∫ (u² − u⁴) du = −cos³x3 + cos⁵x5 + C

Sine–cosine powers
∫ sinᵐx cosⁿx dx	Strategy
m odd	save one sin x; rest to cos with sin² = 1 − cos²; u = cos x
n odd	save one cos x; rest to sin; u = sin x
both even	half-angle formulas, repeatedly

Different angles: product to sum, e.g. ∫ sin 3x cos x dx = −cos 2x4 − cos 4x8 + C.

Even secant power Worked

∫ tan²x sec⁴x dx = ∫ tan²x(1 + tan²x) sec²x dxu = tan x

= tan³x3 + tan⁵x5 + C

Tangent–secant powers
∫ tanᵐx secⁿx dx	Strategy
n even	save sec² x; rest to tan with sec² = 1 + tan²; u = tan x
m odd	save sec x tan x; rest to sec; u = sec x
otherwise	identities, or the sec³ x result below

∫ sec³ x dx = 12(sec x tan x + ln|sec x + tan x|) + C (by parts).

Even powers Worked

∫₀^π sin²x dx = ∫₀^π 1 − cos 2x2 dxhalf-angle

= [x2 − sin 2x4]₀^π = π2

4Trigonometric Substitutionsquare roots to trig

The three cases
Expression	Substitute	Identity used	dx
√(a² − x²)	x = a sin θ, −π2 ≤ θ ≤ π2	1 − sin²θ = cos²θ	a cos θ dθ
√(a² + x²)	x = a tan θ, −π2 < θ < π2	1 + tan²θ = sec²θ	a sec²θ dθ
√(x² − a²)	x = a sec θ, 0 ≤ θ < π2 (x ≥ a)	sec²θ − 1 = tan²θ	a sec θ tan θ dθ

Draw the triangle to switch back to x: label the sides from the substitution, then read off any trig ratio.

Sine case Worked

∫ dxx²√(9 − x²): x = 3 sin θ, dx = 3 cos θ dθ, √(9 − x²) = 3 cos θ
= 19∫ csc²θ dθ = −19 cot θ + C
cot θ = √(9 − x²)x → −√(9 − x²)9x + Ctriangle

Tangent case Worked

∫₀² dx√(x² + 4) = [ln(x + √(x² + 4))]₀²x = 2 tan θ

= ln(2 + 2√2) − ln 2 = ln(1 + √2) ≈ 0.8814

Area of a circle, proved Worked

A = 2∫_−r^r √(r² − x²) dx; x = r sin θ

= 2r²∫_−π/2^π/2 cos²θ dθ = πr²r = 3 gives 9π

5Partial Fractionssplit into simple pieces

Steps

If deg(top) ≥ deg(bottom), long-divide first.
Factor the bottom completely.
Write one term per factor (table).
Solve for the constants: cover-up for distinct linear factors, otherwise match coefficients.
Integrate each piece: ln, power or arctan.

Distinct linear: cover-up Worked

5x − 4x² − x − 2 = Ax − 2 + Bx + 1
A = 5(2) − 42 + 1 = 2; B = 5(−1) − 4−1 − 2 = 3cover the factor, plug its root
∫ = 2 ln|x − 2| + 3 ln|x + 1| + C

Setting up
Factor in the bottom	Terms to write
(x − a)	Ax − a
(x − a)ᵏ, repeated	A₁x − a + A₂(x − a)² + … + Aₖ(x − a)ᵏ
x² + bx + c, irreducible	Bx + Cx² + bx + c
(x² + bx + c)ᵏ	one Bᵢx + Cᵢ(x² + bx + c)ⁱ for each i ≤ k

Repeated factor Worked

1x²(x − 1) = −1x − 1x² + 1x − 1

∫ = −ln|x| + 1x + ln|x − 1| + C

Irreducible quadratic Worked

2x² + x + 4x(x² + 4) = 1x + x + 1x² + 4A = 4/4 = 1, then match

∫ = ln|x| + 12 ln(x² + 4) + 12 arctan x2 + C

Cover it, plug it

For a factor (x − a): cover it, plug x = a into the rest. The result is its constant.

Calculus 2Calculus · Units 6–9 · continued

6Numerical Integrationwhen antiderivatives fail

The rules
Rule (Δx = b − an, xᵢ = a + iΔx)	Formula	Error bound
Midpoint Mₙ	Δx[f(m₁) + f(m₂) + … + f(mₙ)], mᵢ = midpoint of strip i	K₂(b − a)³24n²
Trapezoid Tₙ	Δx2[f(x₀) + 2f(x₁) + 2f(x₂) + … + 2f(xₙ₋₁) + f(xₙ)]	K₂(b − a)³12n²
Simpson Sₙ (n even)	Δx3[f(x₀) + 4f(x₁) + 2f(x₂) + 4f(x₃) + … + 4f(xₙ₋₁) + f(xₙ)]	K₄(b − a)⁵180n⁴

Kⱼ ≥ |f⁽ʲ⁾| on [a, b]. Tₙ = average of the left and right sums; S₂ₙ = (2Mₙ + Tₙ)/3. Simpson is exact for cubics. Concave up: Mₙ ≤ exact ≤ Tₙ.

T₄ for ∫₁² dx/x. The curve is concave up, so each trapezoid sits above it: T₄ overestimates.

∫₁² dx/x = ln 2 ≈ 0.693147
n = 4	Value	Error	Bound
T₄	0.697024	+0.003877	0.0104
M₄	0.691220	−0.001927	0.0052
S₄	0.693254	+0.000107	0.00052

|f″| = 2/x³ ≤ 2 = K₂; |f⁽⁴⁾| = 24/x⁵ ≤ 24 = K₄ on [1, 2].

Strips for error < 0.0001 Worked

Trapezoid: 212n² < 0.0001 → n = 41n² > 1666.7

Simpson: 24180n⁴ < 0.0001 → n = 8n⁴ > 1333.3, n even

7Improper Integralsinfinite limits or blow-ups

Two types

If f blows up at c in [a, b], split at c and use one-sided limits. Converges if every limit is finite; otherwise diverges.

∫_a^∞ f dx = lim_b→∞ ∫_a^b f dx

Worked limits Worked

∫₀^∞ e^−2x dx = lim 1 − e^−2b2 = 12

∫₀¹ dx√x = lim_t→0⁺ (2 − 2√t) = 2p = ½

An infinitely long region can have finite area: ∫₁^∞ x⁻² dx = lim (1 − 1/b) = 1.

Comparison test

0 ≤ f ≤ g: ∫g converges ⇒ ∫f converges; ∫f diverges ⇒ ∫g diverges.

EX0 < e^−x² ≤ e^−x on [1, ∞), ∫₁^∞ e^−x dx = 1/e: converges.

Benchmarks
p-integral	Converges when	Value
∫₁^∞ dxxᵖ	p > 1	1p − 1
∫₀¹ dxxᵖ	p < 1	11 − p

p = 1 diverges in both: ∫₁^∞ dx/x = lim ln b = ∞. Two-sided: ∫_−∞^∞ dx/(1 + x²) = π (split at 0).

Hidden blow-up

∫₋₁¹ dxx² is not −2: it blows up at x = 0 and diverges.

8Area & Volumeslice, then add slices

Set-ups (Calculus 1 guide §13 has first examples)
Method	Slice	Formula	Method	Slice	Formula
Area (vertical)	top − bottom	A = ∫_a^b (top − bottom) dx	Washer	⊥ axis, with a hole	V = π∫ (R² − r²) dx
Area (horizontal)	right − left	A = ∫_c^d (right − left) dy	Shell	∥ axis	V = 2π∫ (radius)(height) dx
Disk	⊥ axis, solid	V = π∫ R² dx	Cross sections	⊥ the base	V = ∫ A(x) dx

Region between y = √x and y = x/2 turned about the x-axis: each slice is a washer, outer radius R = √x, inner r = x/2.

Washer Worked

Curves meet where √x = x/2: x = 0, 4

V = π∫₀⁴ (x − x²4) dx = π(8 − 163) = 8π3R² − r²

Region under y = x(3 − x) turned about the y-axis: a shell has radius x, height f(x), thickness dx.

Horizontal slices Worked

Between x = y² and x = y + 2: y² = y + 2 → y = −1, 2

A = ∫₋₁² (y + 2 − y²) dy = 92right − left

Shell Worked

V = 2π∫03 x · x(3 − x) dx = 2π(27 − 814)
= 27π2 ≈ 42.41
About x = −1: radius x + 1 → V = 45π2radius from the axis

Square cross sections Worked

Base between y = x and y = x² on [0, 1]; side x − x²

V = ∫₀¹ (x − x²)² dx = 13 − 12 + 15 = 130

Calculus 2Calculus · Units 6–9 · continued

9Arc Length, Surface Area, Work & Massmore sums of small pieces

Zoom in and a curve is a hypotenuse: ds = √(dx² + dy²) = √(1 + (dy/dx)²) dx.

Surface area Worked

y = √x on [0, 2]: f√(1 + (f′)²) = √(x + ¼)

S = 2π∫₀² √(x + ¼) dx = 13π3 ≈ 13.61

Spring (Hooke: F = kx) Worked

10 N stretches it 0.2 m → k = 50 N/m

W = ∫₀^0.3 50x dx = 2.25 Jstretch 0.3 m

Formulas
Quantity	Integral
Arc length	L = ∫_a^b √(1 + [f′(x)]²) dx
Surface area (about the x-axis)	S = 2π∫_a^b f(x)√(1 + [f′(x)]²) dx
Work (variable force)	W = ∫_a^b F(x) dx
Pumping liquid	W = ∫ (weight density)(area of slice)(distance lifted) dy
Hydrostatic force	F = ∫ (weight density)(depth)(width) dy
Mass of a rod	m = ∫_a^b ρ(x) dx; x̄ = 1m∫_a^b x ρ(x) dx
Centroid of a region	x̄ = 1A∫ x(f − g) dx; ȳ = 1A∫ 12(f² − g²) dx

Arc length Worked

y = x^3/2 on [0, 4]: 1 + (f′)² = 1 + 9x4

L = 827[(1 + 9x4)^3/2]₀⁴ = 827(10√10 − 1) ≈ 9.073

Pumping a tank Worked

Cylinder, radius 2 m, height 5 m, full of water (9,800 N/m³), pumped over the top

W = ∫₀⁵ 9800 · 4π · (5 − y) dy = 490,000π ≈ 1.54 × 10⁶ Jslice at height y

Rod and centroid Worked

ρ = 1 + x on [0, 2]: m = 4, moment = 143 → x̄ = 76

Under y = x², 0 ≤ x ≤ 1: A = 13, centroid (34, 310)

10Sequencesa₁, a₂, a₃, …

Key facts

Converges: lim_n→∞ aₙ = L (finite); otherwise diverges
Function rule: if aₙ = f(n) and f(x) → L, then aₙ → L (so L’Hôpital may help)
Squeeze: bₙ ≤ aₙ ≤ cₙ and bₙ, cₙ → L ⇒ aₙ → L
Monotone & bounded: an increasing sequence bounded above converges (likewise decreasing and bounded below)
Geometric: rⁿ → 0 if |r| < 1; rⁿ → 1 if r = 1; diverges otherwise

Common limits
aₙ	Limit	aₙ	Limit
1nᵖ (p > 0)	0	ln nn	0
n^1/n	1	xⁿn!	0 (any x)
(1 + xn)ⁿ	eˣ	n!nⁿ	0

Recursive sequence Worked

a₁ = 1, aₙ₊₁ = √(2 + aₙ): 1, 1.732, 1.932, 1.983, …
increasing and bounded above by 2, so it convergesmonotone & bounded
limit L = √(2 + L) → L² − L − 2 = 0 → L = 2L > 0

Four quick ones Worked

3n² − n5n² + 2 → 35divide by n²
ln nn → 1/x1 → 0L’Hôpital on f(x)
(1 + 2n)ⁿ → e²
(−1)ⁿ: −1, 1, −1, … divergesno single limit

Growth ladder

For large n: ln n ≪ nᵖ ≪ bⁿ (b > 1) ≪ n! ≪ nⁿ. The faster one on the bottom of a fraction sends it to 0.

11Infinite Serieslimits of partial sums

Partial sums

S_N = a₁ + a₂ + … + a_N. The series converges to S if S_N → S. Changing or dropping finitely many terms never changes convergence (only the sum).

Σ_n=1^∞ aₙ = lim_N→∞ S_N

Geometric Worked

Σn=1∞ 3(25)ⁿ: first term 65, r = 25
sum = 6/51 − 2/5 = 2
0.272727… = 0.271 − 0.01 = 311repeating decimal

Geometric series

Diverges if |r| ≥ 1. a = first term, r = ratio of consecutive terms.

Σ_n=0^∞ a rⁿ = a1 − r if |r| < 1

Telescoping: Σ 1/(n(n + 1)) = Σ (1/n − 1/(n + 1)), so S_N = 1 − 1/(N + 1) → 1. Bars: terms; dots: partial sums.

Basics

nth-term test: lim aₙ ≠ 0 ⇒ the series diverges (Σ n2n + 1: terms → 12). If lim aₙ = 0, the test says nothing.
Harmonic series: Σ 1n diverges, though its terms → 0 (S_N ≈ ln N)
p-series: Σ 1nᵖ converges if p > 1, diverges if p ≤ 1; Σ 1n² = π²6 ≈ 1.6449 (Euler)
Linearity: Σ(c aₙ ± bₙ) = cΣaₙ ± Σbₙ when both converge

Calculus 2Calculus · Units 6–9 · continued

12Convergence Testswhich test, and when

Test summary
Test	Use on	Condition	Conclusion
nth-term	any series	lim aₙ ≠ 0	diverges (never proves convergence)
Geometric	Σ a rⁿ	\|r\| < 1 / \|r\| ≥ 1	converges to a1 − r / diverges
p-series	Σ 1/nᵖ	p > 1 / p ≤ 1	converges / diverges
Integral	aₙ = f(n), f positive, continuous, decreasing	∫₁^∞ f(x) dx finite / infinite	same as the integral; remainder ∫_N+1^∞ f ≤ R_N ≤ ∫_N^∞ f
Direct comparison	0 ≤ aₙ ≤ bₙ	Σbₙ converges / Σaₙ diverges	Σaₙ converges / Σbₙ diverges
Limit comparison	aₙ, bₙ > 0	lim aₙ/bₙ = c, 0 < c < ∞	both converge or both diverge
Alternating series	Σ(−1)ⁿbₙ, bₙ > 0	bₙ decreasing and bₙ → 0	converges; \|error\| ≤ first omitted term
Ratio	factorials, powers like 3ⁿ	L = lim \|aₙ₊₁/aₙ\|	L < 1 converges absolutely; L > 1 diverges; L = 1 no information
Root	aₙ = (…)ⁿ	L = lim \|aₙ\|^1/n	same conclusions as the ratio test

Ratio Worked

Σ n²3ⁿ: aₙ₊₁aₙ = (n + 1)²3n² → 13

L < 1: converges

Ratio with factorials Worked

Σ n!nⁿ: ratio = (nn + 1)ⁿ → 1e
converges
Σn=1∞ 3ⁿn! = e³ − 1converges, ratio → 0

Root Worked

Σ (2n3n + 1)ⁿ: root = 2n3n + 1 → 23

converges

Limit comparison Worked

Σ n + 1n³ + 2 vs Σ 1n²: ratio → 1
p = 2 > 1: converges
Σ 1√(n³ + 1) vs 1n3/2: converges

Integral Worked

Σ_n=2^∞ 1n ln n: ∫₂^b dxx ln x = ln(ln b) − ln(ln 2) → ∞u = ln x

diverges

Never Guess; Always Run It Carefully

Test order: Nth-term, Geometric/p, Alternating, Ratio/root, Integral, Comparison.

13Alternating Series & Absolute Convergencesigns that switch

Alternating series test

Error bound: |S − S_N| ≤ b_N+1, and S lies between consecutive partial sums.

Σ(−1)ⁿ⁺¹ bₙ converges if bₙ ↓ and bₙ → 0

Alternating harmonic series: partial sums zigzag in toward ln 2 ≈ 0.693.

Two kinds of convergence

Absolutely convergent: Σ|aₙ| converges (then Σaₙ converges too)
Conditionally convergent: Σaₙ converges but Σ|aₙ| diverges
Rearrangement: a conditionally convergent series can be reordered to sum to any number (Riemann); absolute sums never change

Classify Worked

Σ (−1)n+1n = ln 2: Σ 1/n divergesconditional
Σ (−1)n+1n²: Σ 1/n² convergesabsolute
Σ (−1)ⁿ nn + 1: terms do not → 0diverges

How many terms? Worked

Σ (−1)n+1n³ with error < 0.001
need 1(N + 1)³ < 0.001 → N + 1 ≥ 111/11³ ≈ 0.00075
N = 10 terms

Calculus 2Calculus · Units 6–9 · continued

14Power Seriesa polynomial that never ends

Power series about x = a

Converges for |x − a| < R (R = radius), diverges for |x − a| > R; test both endpoints separately to get the interval. R may be 0 or ∞.

Σ_n=0^∞ cₙ(x − a)ⁿ

Term by term (same R)

f′(x) = Σ n cₙ(x − a)ⁿ⁻¹
∫ f(x) dx = C + Σ cₙ (x − a)ⁿ⁺¹n + 1
Endpoints may gain or lose convergence: recheck them.

Ratio finds R; endpoints finish it

The ratio test gives the radius but never decides the endpoints: test each one by itself.

Radius and interval Worked

Σ (x − 2)ⁿn · 3ⁿ: ratio → ∣x − 2∣3 < 1
R = 3: open interval (−1, 5)
x = 5: Σ 1/n diverges; x = −1: Σ (−1)ⁿ/n convergesendpoints
interval [−1, 5)

Quick radii (ratio test)
Series	Radius
Σ xⁿn!	∞
Σ n! xⁿ	0
Σ n xⁿ4ⁿ	4

Building from the geometric series
Start	Operation	Result (\|x\| < 1)
11 − x = Σ xⁿ	x → −x²	11 + x² = Σ (−1)ⁿ x²ⁿ
11 + x²	integrate	arctan x = Σ (−1)ⁿ x²ⁿ⁺¹2n + 1
11 + x = Σ (−1)ⁿxⁿ	integrate	ln(1 + x) = Σ (−1)ⁿ⁺¹ xⁿn
11 − x	differentiate, × x	x(1 − x)² = Σ n xⁿ

At x = 1, arctan gives π4 = 1 − 13 + 15 − … (Gregory–Leibniz) and ln(1 + x) gives ln 2 = 1 − 12 + 13 − ….

15Taylor & Maclaurin Seriesmatching every derivative at one point

Series to know
Function	Maclaurin series	Converges for
eˣ	1 + x + x²2! + x³3! + … = Σ xⁿn!	all x
sin x	x − x³3! + x⁵5! − … = Σ (−1)ⁿ x²ⁿ⁺¹(2n + 1)!	all x
cos x	1 − x²2! + x⁴4! − … = Σ (−1)ⁿ x²ⁿ(2n)!	all x
11 − x	1 + x + x² + x³ + … = Σ xⁿ	−1 < x < 1
ln(1 + x)	x − x²2 + x³3 − … = Σ (−1)ⁿ⁺¹ xⁿn	−1 < x ≤ 1
arctan x	x − x³3 + x⁵5 − … = Σ (−1)ⁿ x²ⁿ⁺¹2n + 1	−1 ≤ x ≤ 1
(1 + x)ᵏ	1 + kx + k(k − 1)2!x² + k(k − 1)(k − 2)3!x³ + …	−1 < x < 1 (binomial)

Taylor series of f about x = a

a = 0: Maclaurin series. Stopping at degree n gives the Taylor polynomial Tₙ(x).

f(x) = Σ_n=0^∞ f⁽ⁿ⁾(a)n! (x − a)ⁿ

Lagrange error bound

M ≥ |f⁽ⁿ⁺¹⁾| between a and x. For an alternating series, the first omitted term also bounds the error.

|Rₙ(x)| ≤ M(n + 1)! |x − a|ⁿ⁺¹

Binomial series Worked

√(1 + x) = 1 + x2 − x²8 + x³16 − …k = ½

√1.1 ≈ 1 + 0.05 − 0.00125 = 1.04875true 1.048809

Higher-degree Taylor polynomials hug sin x over a wider interval.

Taylor polynomial and error Worked

ln x about 1: T₃ = (x − 1) − (x − 1)²2 + (x − 1)³3
ln 1.1 ≈ 0.1 − 0.005 + 0.000333 = 0.095333true 0.095310
f⁽⁴⁾ = −6/x⁴, M = 6: |R₃| ≤ 6(0.1)⁴24 = 0.000025actual 0.000023

Integrate a series Worked

e−x² = 1 − x² + x⁴2! − x⁶3! + …x → −x²
∫01 e−x² dx ≈ 1 − 13 + 110 − 142 + 1216 ≈ 0.7475true 0.7468
alternating: error ≤ 11320 ≈ 0.00076

Odd sine, even cosine

Sine: odd powers from x. Cosine: even powers from 1. Both alternate signs; eˣ keeps every power, all plus.

Limits and identities Worked

limx→0 sin x − xx³ = lim −x³/6 + …x³ = −16
e ≈ Σn=05 1n! = 2.716667error < 3/6! ≈ 0.0042
Series show eiθ = cos θ + i sin θ (Euler)

Calculus 2Calculus · Units 6–9 · continued

16Parametric Curves & Motionx and y from t

Formulas
Quantity	Formula
Slope	dydx = dy/dtdx/dt (dx/dt ≠ 0)
Second derivative	d²ydx² = (d/dt)(dy/dx)dx/dt
Horizontal / vertical tangent	dy/dt = 0 / dx/dt = 0 (not both)
Arc length	L = ∫_α^β √((dx/dt)² + (dy/dt)²) dt
Area under the curve	A = ∫ y dx = ∫_α^β y(t) x′(t) dt
Surface area (x-axis)	S = 2π∫ y √((dx/dt)² + (dy/dt)²) dt

x = t², y = t³ − 3t for −2.2 ≤ t ≤ 2.2; arrows show increasing t. The loop closes at (3, 0) (t = ±√3).

Slopes and tangents Worked

dydx = 3t² − 32t; at t = 2: point (4, 2), slope 94
horizontal at t = ±1: (1, −2), (1, 2); vertical at t = 0: (0, 0)
d²ydx² = 3t² + 34t³; at t = 2: 1532 (concave up)

Cycloid (one arch) Worked

x = t − sin t, y = 1 − cos t, 0 ≤ t ≤ 2π
L = ∫02π √(2 − 2 cos t) dt = 8
A = ∫02π (1 − cos t)² dt = 3πRoberval (area), Wren (length)

17Vector-Valued Functions & Motionposition moving with t

Definitions
Quantity	Formula
Position	r(t) = ⟨x(t), y(t)⟩; d/dt and ∫ act on each component (∫ adds a constant vector)
Velocity, acceleration	v = r′; a = v′ = r″
Speed	\|v\| = √(x′² + y′²)
Unit tangent	T = r′/\|r′\| (§16 curve at t = 2: ⟨4, 9⟩/√97)
Displacement	∫_a^b v dt = r(b) − r(a)
Distance	∫_a^b \|v\| dt
Circular motion	r = ⟨R cos ωt, R sin ωt⟩: speed Rω; \|a\| = Rω² toward the center

Projectile: r(t) = ⟨20t, 20t − 4.9t²⟩ m. Velocity arrows (scaled) keep their horizontal part 20 m/s; the vertical part shrinks, hits 0 at the top, then turns negative.

Projectile Worked

a = ⟨0, −9.8⟩, v(0) = ⟨20, 20⟩ m/s, r(0) = ⟨0, 0⟩
v = ⟨20, 20 − 9.8t⟩; r = ⟨20t, 20t − 4.9t²⟩integrate twice
top at t = 209.8 ≈ 2.04 s, height ≈ 20.4 mvertical v = 0
lands at t ≈ 4.08 s, 81.6 m awayy = 0

From velocity to position Worked

v(t) = ⟨t², t⟩, r(0) = ⟨1, 2⟩ → r(t) = ⟨1 + t³3, 2 + t²2⟩
speed at t = √3: √(9 + 3) = 2√3
distance, 0 ≤ t ≤ √3: ∫ t√(t² + 1) dt = 73u = t² + 1

18Polar Curvesradius r at angle θ

Conversions

To rectangular: x = r cos θ, y = r sin θ
To polar: r² = x² + y², tan θ = yx
Slope: dydx = r′ sin θ + r cos θr′ cos θ − r sin θ

Cardioid r = 1 + cos θ with the circle r = 1 (dashed); shaded: inside the cardioid, outside the circle.

Common graphs
Curve	Equation	Shape
Circle	r = a; r = 2a cos θ; r = 2a sin θ	center (0,0); (a, 0); (0, a)
Cardioid	r = a(1 ± cos θ)	heart, through the pole
Limaçon	r = a + b cos θ	inner loop if \|a\| < \|b\|
Rose	r = a cos nθ	n petals (n odd), 2n (n even)
Lemniscate	r² = a² cos 2θ	figure eight
Spiral	r = aθ	Archimedes

Calculus in polar
Quantity	Formula
Area	A = 12∫_α^β r² dθ
Area between	A = 12∫_α^β (R² − r²) dθ
Arc length	L = ∫_α^β √(r² + (dr/dθ)²) dθ

Areas Worked

Whole cardioid: 12∫02π (1 + cos θ)² dθ = 3π2
Shaded: 12∫−π/2π/2 [(1 + cos θ)² − 1] dθ = 2 + π4curves meet at θ = ±π/2
Cardioid length: ∫02π √(2 + 2 cos θ) dθ = 8

Half r squared

A polar region is a fan of thin wedges, each ½r² dθ. Integrate between the angles that bound it.

Slope and conversion Worked

r = 1 + cos θ at θ = π2: r = 1, r′ = −1 → dydx = 1

r = 4 sin θ → x² + y² = 4y → x² + (y − 2)² = 4circle, radius 2

Calculus 2Calculus · Units 6–9 · continued

19Differential Equationsbasics: Calculus 1 §14

Separable: dy/dx = g(x) h(y)

Move y-terms to dy and x-terms to dx: dyh(y) = g(x) dx.
Integrate both sides; one + C.
Use the initial condition; solve for y if you can.

y′ = y, y(0) = 1, h = 0.5: Euler's polygon (solid) reaches 5.06 at x = 2; the true solution eˣ (dashed) reaches 7.39.

Step from where you stand

Use the slope at the point you are standing on, never the point you are stepping to.

Separate Worked

dydx = 2x(1 + y²), y(0) = 0
dy1 + y² = 2x dx → arctan y = x² + C, C = 0
y = tan(x²)for |x| < √(π/2)

Euler's method

Step along the tangent line at the current point; smaller h means smaller error.

yₙ₊₁ = yₙ + h · F(xₙ, yₙ)

Euler for y′ = x + y, y(0) = 1, h = 0.1
n	xₙ	yₙ	slope xₙ + yₙ
0	0	1	1
1	0.1	1.1	1.2
2	0.2	1.22

True y = 2eˣ − x − 1 gives y(0.2) ≈ 1.2428.

Logistic numbers Worked

K = 1000, P₀ = 100, k = 0.5 → A = 9
P(4) = 10001 + 9e−2 ≈ 450.9
P = 500 when 9e−0.5t = 1: t = 2 ln 9 ≈ 4.39P = K/2

Models
Model	Equation	Solution
Exponential growth/decay	dydt = ky	y = y₀eᵏᵗ
Newton’s cooling	dTdt = −k(T − Tₛ)	T = Tₛ + (T₀ − Tₛ)e^−kt
Logistic	dPdt = kP(1 − PK)	P = K1 + Ae^−kt, A = K − P₀P₀

Logistic P = 1000/(1 + 9e^−0.5t): steepest at P = K/2, leveling off at K.

Cooling Worked

Coffee: 90 °C → 70 °C in 10 min, room 20 °Ce^−10k = 5/7

T = 50 °C at t = 10 · ln(3/7)ln(5/7) ≈ 25.2 min

20Common Slips & Practicecheck yourself

Twelve problems (answers checked by the build)
#	Problem	Answer	#	Problem	Answer
1	∫ x cos x dx	x sin x + cos x + C	7	Radius of Σ n xⁿ4ⁿ	R = 4
2	∫ dxx² − 9	16 ln\|x − 3x + 3\| + C	8	Maclaurin series of x eˣ	x + x² + x³2! + x⁴3! + …
3	∫₀^π/2 sin³x dx	23	9	Area inside r = 2 sin θ	π
4	∫₂^∞ dxx³	18	10	Length of y = 23x^3/2, 0 ≤ x ≤ 3	143
5	Σ_n=0^∞ (−25)ⁿ	57	11	Euler: y′ = y − x, y(0) = 2, h = 0.5, two steps	y(1) ≈ 4.25
6	Does Σ n + 1n³ + 2 converge?	yes	12	Volume: y = eˣ, 0 ≤ x ≤ 1, about the x-axis	π(e² − 1)2 ≈ 10.04

Common slips
Slip	Fix
changing variables but keeping the old limits	convert the limits, or switch back to x
missing a blow-up inside [a, b]	look for zeros of the denominator first
“lim aₙ = 0, so Σaₙ converges”	the nth-term test only proves divergence
skipping the endpoints	test each endpoint series separately
shell radius measured from the y-axis for a different axis	radius = distance to the axis of rotation
partial fractions on an improper fraction	long-divide first
polar area without the ½ or with the wrong θ-limits	A = ½∫r² dθ between the right angles
forgetting + C, or the constant vector in motion problems	add it, then use the initial condition
ratio test L = 1 read as “converges”	L = 1 decides nothing: try another test
degrees in a series or derivative	radians always