Calculus 2: cram pack

Formula sheets, the key method steps for every topic and 8 practice problems, with full solutions at the back.

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1Formulas and methods

Integrals

Power rule
∫xⁿ dx = xⁿ⁺¹/(n + 1) + C
Definite integral
∫ab f dx = F(b) − F(a)
Common integrals
∫eˣ dx = eˣ, ∫1/x dx = ln|x|, ∫sin x dx = −cos x
Area between curves
∫ab (f − g) dx

Method

  1. Rewrite the integrand as a sum of simple terms where possible.
  2. Apply the power rule: ∫xⁿ dx = xⁿ⁺¹/(n + 1), and standard integrals for ex, 1/x, sin and cos.
  3. Use substitution when you see a function and its derivative together.
  4. For a definite integral, evaluate F(b) − F(a).

Sequences & series

Arithmetic sequence
aₙ = a₁ + (n − 1)d, Sₙ = n(a₁ + aₙ)/2
Geometric sequence
aₙ = a₁rⁿ⁻¹, Sₙ = a₁(1 − rⁿ)/(1 − r)
Infinite geometric series
S = a₁/(1 − r), |r| < 1
p-series
Σ1/kᵖ converges only for p > 1
Ratio test
L = lim |ak+1/ak|: L < 1 converges, L > 1 diverges
Radius of convergence
R = lim |ck/ck+1|
Taylor polynomial
Pₙ(x) = Σ f⁽ᵏ⁾(a)(x − a)ᵏ/k!
Maclaurin series
eˣ = Σxᵏ/k!, sin x = x − x³/3! + …, cos x = 1 − x²/2! + …, ln(1 + x) = x − x²/2 + …
Lagrange error bound
|Rₙ| ≤ M·|x − a|ⁿ⁺¹/(n + 1)!

Method

  1. Identify the type: arithmetic (constant difference), geometric (constant ratio), a standard series, a power series or a Taylor polynomial.
  2. For sums use Sₙ = n(a₁ + aₙ)/2, Sₙ = a₁(1 − rⁿ)/(1 − r), or S = a₁/(1 − r) when |r| < 1, with a₁ the first term of the sum.
  3. For convergence: n-th term test first, then compare with Σ1/kᵖ or a geometric series, and use the ratio or root test for factorials and k-th powers.
  4. For Taylor polynomials list f(a), f′(a), f″(a), …, divide the k-th derivative by k!, and bound the error with M·|x − a|ⁿ⁺¹/(n + 1)!.

Multivariable calculus

Partial derivatives
fx = ∂f/∂x (y constant), fxy = fyx
Gradient
∇f = (fx, fy), largest slope |∇f|
Directional derivative
Duf = ∇f·u, u = v/|v|
Tangent plane
z = f(a, b) + fx(x − a) + fy(y − b)
Chain rule
dz/dt = fx·dx/dt + fy·dy/dt
Second-derivative test
D = fxxfyy − fxy²: D > 0, fxx > 0 minimum; D > 0, fxx < 0 maximum; D < 0 saddle
Double integral
∬R f dA = ∫ab∫g(x)h(x) f dy dx
Polar coordinates
x = r cos θ, y = r sin θ, dA = r dr dθ

Method

  1. For a partial derivative, treat the other variable as a constant and use the one-variable rules.
  2. For gradient questions compute ∇f = (fx, fy) at the point; use |∇f| for the steepest slope and ∇f·u with a unit vector u for a directional derivative.
  3. For extrema solve fx = 0 and fy = 0 together, then classify every point with D = fxx·fyy − fxy².
  4. For a double integral sketch the region, choose the order (or polar coordinates for discs) so the inner limits are the bounding curves, and integrate inside first.

Differential equations

Exponential growth/decay
dy/dt = ky → y = y0ekt
Half-life
k = ln 2 / t½
Newton's law of cooling
T(t) = Ta + (T0 − Ta)e−kt
Integrating factor
y′ + p(x)y = q(x): μ = e∫p dx, (μy)′ = μq
Mixing tank
dm/dt = r·c − (r/V)·m

2Practice problems

1.IntegralsMedium
xy13

Evaluate I = ∫13 4/x dx.

f(x) = 4/x[1, 3]
I =
2.IntegralsHard
xy−11

Find the area A of the region between the parabola y = 1 − x² and the x-axis.

y = 1 − x²
A =
3.Sequences & seriesMedium
AΣ (k + 3)/k²k = 1 … ∞BΣ k/2ᵏk = 1 … ∞CΣ (2k + 1)/(k + 2)k = 1 … ∞DΣ (4/3)ᵏk = 1 … ∞

Exactly one of the four series converges. Click it.

A: Σ (k + 3)/k²B: Σ k/2ᵏC: Σ (2k + 1)/(k + 2)D: Σ (4/3)ᵏ
  1. 1A: Σ (k + 3)/k²
  2. 2B: Σ k/2ᵏ
  3. 3C: Σ (2k + 1)/(k + 2)
  4. 4D: Σ (4/3)ᵏ
4.Sequences & seriesHard

Recognise the series and find its sum: S = Σk=0∞ (−1)ᵏ/k!.

ak = (−1)ᵏ/k!
S =
5.Multivariable calculusMedium
xy−4−20024vP

For f(x, y) = 3xy + 3x² − 4y³, find the directional derivative Duf at (−2, 2) in the direction of v = (3, 1).

f(x, y) = 3xy + 3x² − 4y³P = (−2, 2)v = (3, 1)
Duf =
6.Multivariable calculusHard
xyy = 0.8y = x0.80.8

The integral I = ∫00.8 ∫x0.8 cos(y²) dy dx cannot be done in this order. Reverse the order of integration and evaluate it.

f = cos(y²)0 ≤ x ≤ y ≤ 0.8
I =
7.Differential equationsMedium
tyy(0) = 2y(1.5) = ?

Solve y″ + 6y′ + 9y = 0 with y(0) = 2 and y′(0) = 2. Find y(1.5).

y″ + 6y′ + 9y = 0y(0) = 2y′(0) = 2
y(1.5) =
8.Differential equationsHard
tyy(0) = 10y(0.75) = ?

Solve y″ + 4y′ + 20y = 0 with y(0) = 10 and y′(0) = 0 (released from rest). Find y(0.75); angles in radians.

y″ + 4y′ + 20y = 0y(0) = 10y′(0) = 0
y(0.75) =

3Solutions

1.Integrals

Answer: I = 4.394

Definite integrals: ∫ₐᵇ f(x) dx is the signed area between the graph and the x-axis from a to b. Find an antiderivative F (with F′ = f) and evaluate F(b) − F(a). Power rule: ∫xⁿ dx = xⁿ⁺¹/(n + 1). The area between two curves is ∫(upper − lower) dx.

Antiderivative

F(x) = 4·ln x

Evaluate at the limits

I = 4·(ln 3 − ln 1) = 4·ln 3 = 4.394

Answer

I = 4.394
2.Integrals

Answer: A = 1.333

Definite integrals: ∫ₐᵇ f(x) dx is the signed area between the graph and the x-axis from a to b. Find an antiderivative F (with F′ = f) and evaluate F(b) − F(a). Power rule: ∫xⁿ dx = xⁿ⁺¹/(n + 1). The area between two curves is ∫(upper − lower) dx.

Zeros

1 − x² = 0 → x = ±1

Integrate

A = [1x − x³/3]−11 = 0.6667 − (−0.6667) = 1.333

Answer

A = 1.333
3.Sequences & series

Answer: B: Σ k/2ᵏ

Convergence: if the terms do not tend to 0 the series diverges (n-th term test). Σ1/kᵖ converges only for p > 1, and Σrᵏ only for |r| < 1. Compare a series with one of these when its terms behave like them for large k. Ratio test: L = lim |ak+1/ak|; root test: L = lim |ak|1/k. L < 1 converges, L > 1 diverges, L = 1 decides nothing. A power series Σck(x − x₀)ᵏ converges for |x − x₀| < R with R = lim |ck/ck+1|; an alternating series with terms decreasing to 0 converges.

Test each series

A: Σ (k + 3)/k² → Behaves like 1/k for large k: diverges by comparison with the harmonic series.
B: Σ k/2ᵏ → Ratio test: L = 1/2 < 1, converges.
C: Σ (2k + 1)/(k + 2) → The terms tend to 2, not 0: diverges by the n-th term test.
D: Σ (4/3)ᵏ → Geometric with r = 4/3 > 1: the terms grow, so it diverges.

The answer

B: Σ k/2ᵏ

Answer

B: Σ k/2ᵏ
4.Sequences & series

Answer: S = 0.3679

Taylor polynomials: near x = a a smooth function is approximated by Pₙ(x) = f(a) + f′(a)(x − a) + f″(a)(x − a)²/2! + … + f⁽ⁿ⁾(a)(x − a)ⁿ/n!; for a = 0 this is the Maclaurin polynomial. Standard series: eˣ = Σxᵏ/k!, sin x = x − x³/3! + x⁵/5! − …, cos x = 1 − x²/2! + x⁴/4! − …, ln(1 + x) = x − x²/2 + x³/3 − …, 1/(1 − x) = Σxᵏ. Lagrange error bound: |f(x) − Pₙ(x)| ≤ M·|x − a|ⁿ⁺¹/(n + 1)!, where M is the largest value of |f⁽ⁿ⁺¹⁾| between a and x.

Standard series

Σ xᵏ/k! = eˣ

Substitute

x = −1
S = e−1 = 0.3679

Answer

S = 0.3679
5.Multivariable calculus

Answer: Duf = −22.77

Gradient: ∇f = (fx, fy) points in the direction of steepest ascent, and |∇f| is the largest rate of change. The directional derivative along a direction v is Duf = ∇f·u with the unit vector u = v/|v|. The tangent plane at (a, b) is z = f(a, b) + fx(a, b)(x − a) + fy(a, b)(y − b); it gives the linear approximation of f near that point. Chain rule: dz/dt = fx·dx/dt + fy·dy/dt.

Gradient at the point

fx = 3y + 6x → fx(−2, 2) = −6
fy = 3x − 12y² → fy(−2, 2) = −54

Unit vector

|v| = √(3² + 1²) = 3.162
u = (0.9487, 0.3162)

Dot product

Duf = (−6)·0.9487 + (−54)·0.3162 = −22.77

Answer

Duf = −22.77
6.Multivariable calculus

Answer: I = 0.2986

Critical points and double integrals: a critical point has fx = fy = 0. Second-derivative test with D = fxxfyy − fxy²: D > 0 and fxx > 0 is a local minimum, D > 0 and fxx < 0 a local maximum, D < 0 a saddle point, and D = 0 decides nothing. For a double integral over a general region, let the inner limits be the curves that bound the region; for discs and sectors use polar coordinates, x = r cos θ, y = r sin θ and dA = r dr dθ (do not forget the factor r).

Region

0 ≤ x ≤ y, 0 ≤ y ≤ 0.8

Reversed order

I = ∫00.8 ∫0y cos(y²) dx dy = ∫00.8 y·cos(y²) dy

Substitution u = y²

I = [sin(y²)/2]00.8 = 0.2986

Answer

I = 0.2986
7.Differential equations

Answer: y(1.5) = 0.1555

Second-order linear equations: for y″ + b·y′ + c·y = 0 try y = ert: this gives the characteristic equation r² + b·r + c = 0. Two real roots r₁ ≠ r₂ give y = A·er₁t + B·er₂t (overdamped); a double root r gives y = (A + Bt)·ert (critically damped); complex roots −α ± jω give y = e−αt(A cos ωt + B sin ωt) (underdamped, oscillating). The initial values y(0) and y′(0) fix A and B. Written as y″ + 2ζωny′ + ωn²y = 0, ωn is the natural frequency and ζ the damping ratio: ζ < 1 oscillates. With a constant input, y″ + by′ + cy = c·K, the solution is the steady value K plus the free response.

Double root

(r + 3)² = 0 → r = −3

Constants

A = y(0) = 2
y′(0) = B − 3A → B = 2 + 6 = 8

Evaluate

y(1.5) = (2 + 8·1.5)·e−4.5 = 0.1555

Answer

y(1.5) = 0.1555
8.Differential equations

Answer: y(0.75) = −2.052

Second-order linear equations: for y″ + b·y′ + c·y = 0 try y = ert: this gives the characteristic equation r² + b·r + c = 0. Two real roots r₁ ≠ r₂ give y = A·er₁t + B·er₂t (overdamped); a double root r gives y = (A + Bt)·ert (critically damped); complex roots −α ± jω give y = e−αt(A cos ωt + B sin ωt) (underdamped, oscillating). The initial values y(0) and y′(0) fix A and B. Written as y″ + 2ζωny′ + ωn²y = 0, ωn is the natural frequency and ζ the damping ratio: ζ < 1 oscillates. With a constant input, y″ + by′ + cy = c·K, the solution is the steady value K plus the free response.

Complex roots

r = −2 ± 4j

Constants

A = 10, B = 2·10/4 = 5

Evaluate

y(0.75) = e−1.5(10·cos 3 + 5·sin 3) = −2.052

Answer

y(0.75) = −2.052

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