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 e2x dx.

f(x) = e2x[0, 1]
I =
2.IntegralsHard
xy1.5y = x²eˣ

Evaluate I = ∫₀1.5 x²·eˣ dx.

∫₀1.5 x²eˣ dx
I =
3.Sequences & seriesMedium
AΣ 1/(k² + 4)k = 1 … ∞BΣ k²/2ᵏk = 1 … ∞CΣ 3ᵏ/k²k = 1 … ∞DΣ 3/k⁴k = 1 … ∞

Exactly one of the four series diverges. Click it.

A: Σ 1/(k² + 4)B: Σ k²/2ᵏC: Σ 3ᵏ/k²D: Σ 3/k⁴
  1. 1A: Σ 1/(k² + 4)
  2. 2B: Σ k²/2ᵏ
  3. 3C: Σ 3ᵏ/k²
  4. 4D: Σ 3/k⁴
4.Sequences & seriesHard
xyfP₅0.8

Use the Maclaurin polynomial of degree 5 of f(x) = sin(3x) to approximate f(0.8). Give P₅(0.8).

f(x) = sin(3x)n = 5x = 0.8
P₅(0.8) =
5.Multivariable calculusMedium
xy−4−202−2024vP

For f(x, y) = 5 sin(x + y), find the directional derivative Duf at (−1, 2) in the direction of v = (0, −4).

f(x, y) = 5 sin(x + y)P = (−1, 2)v = (0, −4)
Duf =
6.Multivariable calculusHard
xy−202−202(2, 1)(−2, 1)

Find the critical points of f(x, y) = −x³ + 12x − y² + 2y, classify them, and give the value fmax of f at its local maximum.

f(x, y) = −x³ + 12x − y² + 2y
fmax =
7.Differential equationsMedium
tyy(0) = 2y(0.5) = ?

Solve y″ + 2y′ + y = 0 with y(0) = 2 and y′(0) = 4. Find y(0.5).

y″ + 2y′ + y = 0y(0) = 2y′(0) = 4
y(0.5) =
8.Differential equationsHard
ty1.5y(1.5) = ?

Solve y′ + 0.5y = 2e−3t with y(0) = 4. Find y(1.5).

y′ + 0.5y = 2e−3ty(0) = 4
y(1.5) =

3Solutions

1.Integrals

Answer: I = 3.195

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) = e2x/2

Evaluate at the limits

I = (e2 − 1)/2 = (7.389 − 1)/2 = 3.195

Answer

I = 3.195
2.Integrals

Answer: I = 3.602

Integration by parts: ∫u·dv = u·v − ∫v·du, from the product rule. Choose u as the factor that gets simpler when you differentiate it (a power of x, or ln x), and dv as the factor you can integrate (eˣ, sin x, cos x, or x when u = ln x). For a definite integral, evaluate u·v at both limits: ∫ₐᵇ u dv = [u·v]ₐᵇ − ∫ₐᵇ v du. A factor x² needs two rounds.

First round

∫x²eˣ dx = x²eˣ − ∫2x eˣ dx

Second round

∫2x eˣ dx = 2x eˣ − 2eˣ
∫x²eˣ dx = eˣ(x² − 2x + 2)

Limits

4.482·(1.25) − 1·2 = 3.602

Answer

I = 3.602
3.Sequences & series

Answer: C: Σ 3ᵏ/k²

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 1/(k² + 4) → Smaller than 1/k², a convergent p-series: converges by comparison.
B: Σ k²/2ᵏ → Ratio test: L = 1/2 < 1, converges.
C: Σ 3ᵏ/k² → Ratio test: L = 3 > 1, diverges.
D: Σ 3/k⁴ → p-series with p = 4 > 1: converges.

The answer

C: Σ 3ᵏ/k²

Answer

C: Σ 3ᵏ/k²
4.Sequences & series

Answer: P₅(0.8) = 0.7596

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.

Maclaurin polynomial

P₅(x) = 3x − 4.5x³ + 2.025x⁵

Substitute

P₅(0.8) = 2.4 − 2.304 + 0.663552 = 0.7596

Answer

P₅(0.8) = 0.7596

The exact value is f(0.8) = 0.675463; the error −0.0840888 shrinks when you take more terms or a point closer to 0.

5.Multivariable calculus

Answer: Duf = −2.702

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 = 5 cos(x + y) → fx(−1, 2) = 2.702
fy = 5 cos(x + y) → fy(−1, 2) = 2.702

Unit vector

|v| = √(0² + (−4)²) = 4
u = (0, −1)

Dot product

Duf = 2.702·0 + 2.702·(−1) = −2.702

Answer

Duf = −2.702
6.Multivariable calculus

Answer: fmax = 17

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).

Critical points

Solve fx = 0 and fy = 0 together.

fx = −(3x² − 12) = 0
fy = −(2y − 2) = 0
(2, 1), (−2, 1)

Second-derivative test

(2, 1): fxx = −12, fyy = −2, fxy = 0, D = 24 → local maximum
(−2, 1): fxx = 12, fyy = −2, fxy = 0, D = −24 → saddle point

Value at the extremum

f(2, 1) = 17

Answer

fmax = 17
7.Differential equations

Answer: y(0.5) = 3.033

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 + 1)² = 0 → r = −1

Constants

A = y(0) = 2
y′(0) = B − 1A → B = 4 + 2 = 6

Evaluate

y(0.5) = (2 + 6·0.5)·e−0.5 = 3.033

Answer

y(0.5) = 3.033
8.Differential equations

Answer: y(1.5) = 2.258

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 = e0.5t
(e0.5ty)′ = 2e−2.5t

General solution

y = 2/(−2.5)·e−3t + C·e−0.5t = −0.8·e−3t + C·e−0.5t

Initial condition

4 = −0.8 + C → C = 4.8

Evaluate

y(1.5) = −0.8·0.01111 + 4.8·0.4724 = 2.258

Answer

y(1.5) = 2.258

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