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
xy01

Evaluate I = ∫01 2x·(x² + 1)3 dx.

f(x) = 2x·(x² + 1)3[0, 1]
I =
2.IntegralsHard
xy1.333y = 4xy = 3x²

The region between the line y = 4x and the parabola y = 3x² is rotated about the x-axis. Find the volume V of the solid.

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

Exactly one of the four series converges. Click it.

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

Find the coefficient c₃ of (x − 3)³ in the Taylor series of f(x) = 1/x about a = 3.

f(x) = 1/xa = 3k = 3
c₃ =
5.Multivariable calculusMedium

z = 5x²y + x² − 4y³ + 3xy with x = −3t − 1 and y = −t² + 2. Use the chain rule to find dz/dt at t = 0.

z = 5x²y + x² − 4y³ + 3xyx = −3t − 1y = −t² + 2t = 0
dz/dt =
6.Multivariable calculusHard
xy−505−202(3, 0)(−3, 0)

f(x, y) = x³ − 27x + y² has a critical point at (−3, 0). Compute the discriminant D = fxxfyy − fxy² there.

f(x, y) = x³ − 27x + y²P = (−3, 0)
D =
7.Differential equationsMedium
tyy(0) = 1y(1.5) = ?

Solve y″ + 4y′ + 4y = 0 with y(0) = 1 and y′(0) = 2. Find y(1.5).

y″ + 4y′ + 4y = 0y(0) = 1y′(0) = 2
y(1.5) =
8.Differential equationsHard
xy12(1, 5)y(2) = ?

Solve the initial value problem y′ + 3y/x = 2 for x > 0, with y(1) = 5. Find y(2).

y′ + 3y/x = 2y(1) = 5
y(2) =

3Solutions

1.Integrals

Answer: I = 3.75

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.

Substitution

u = x² + 1, du = 2x dx
I = ∫12 u3 du

Integrate and evaluate

I = [u4/4]12 = (16 − 1)/4 = 3.75

Answer

I = 3.75
2.Integrals

Answer: V = 15.89

Volumes of revolution: rotating the region under y = f(x) from x = a to x = b about the x-axis makes a solid whose cross-section at x is a disc of radius f(x). Adding the discs gives V = π∫ₐᵇ f(x)² dx. When the region lies between an outer curve R(x) and an inner curve r(x), each slice is a washer: V = π∫ₐᵇ (R(x)² − r(x)²) dx. Square the radii before you subtract them.

Intersection points

4x = 3x² → x = 0, x = 1.333

Washer method

V = π∫₀1.333 ((4x)² − (3x²)²) dx

Integrate

V = π(16·1.333³/3 − 9·1.333⁵/5) = π·5.057 = 15.89

Answer

V = 15.89
3.Sequences & series

Answer: C: Σ 1/(k(k + 2))

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 (3k + 1)/(k + 2) → The terms tend to 3, not 0: diverges by the n-th term test.
B: Σ 1/(4k + 3) → Behaves like 1/(4k) for large k; compared with the harmonic series it diverges.
C: Σ 1/(k(k + 2)) → Smaller than 1/k²: converges by comparison.
D: Σ (k + 4)/k² → Behaves like 1/k for large k: diverges by comparison with the harmonic series.

The answer

C: Σ 1/(k(k + 2))

Answer

C: Σ 1/(k(k + 2))
4.Sequences & series

Answer: c₃ = −0.01235

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.

Derivative f⁽³⁾

f⁽³⁾(x) = −6/x⁴

Coefficient

c₃ = −6/(3⁴·6) = −0.01235
c₃ = −0.01235

Answer

c₃ = −0.01235
5.Multivariable calculus

Answer: dz/dt = 48

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.

Point and rates

x(0) = −1, y(0) = 2
dx/dt = −3, dy/dt = −2t = 0

Partial derivatives

fx = 10xy + 2x + 3y = −16
fy = 5x² − 12y² + 3x = −46

Chain rule

dz/dt = (−16)·(−3) + (−46)·0 = 48

Answer

dz/dt = 48
6.Multivariable calculus

Answer: D = −36

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).

Second derivatives

fxx = −18, fyy = 2, fxy = 0

Discriminant

D = (−18)·2 − 0² = −36
→ saddle point

Answer

D = −36
7.Differential equations

Answer: y(1.5) = 0.3485

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 + 2)² = 0 → r = −2

Constants

A = y(0) = 1
y′(0) = B − 2A → B = 2 + 2 = 4

Evaluate

y(1.5) = (1 + 4·1.5)·e−3 = 0.3485

Answer

y(1.5) = 0.3485
8.Differential equations

Answer: y(2) = 1.563

First-order differential equations: dy/dt = k·y has the solution y = y₀·ekt (growth for k > 0, decay for k < 0); a half-life T gives k = ln 2/T. Newton's law of cooling, dT/dt = −k·(T − Tₐ), gives T(t) = Tₐ + (T₀ − Tₐ)·e−kt. A linear equation y′ + p(x)·y = q(x) is solved with the integrating factor μ = e∫p dx: (μy)′ = μq, so μy = ∫μq dx + C.

Integrating factor

μ = e∫3/x dx = x3

Integrate

(x3y)′ = 2x3 → x3y = 2x4/4 + C

Initial condition

5 = 2/4 + C → C = 4.5

Evaluate

y(2) = (2·24/4 + 4.5)/23 = 1.563

Answer

y(2) = 1.563

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