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
xy02

Evaluate I = ∫02 ex dx.

f(x) = ex[0, 2]
I =
2.IntegralsHard
xy0.5y = x²eˣ

Evaluate I = ∫₀0.5 x²·eˣ dx.

∫₀0.5 x²eˣ dx
I =
3.Sequences & seriesMedium
AΣ 1/k⁴k = 1 … ∞BΣ (5/3)ᵏk = 1 … ∞CΣ (k + 4)/k²k = 1 … ∞DΣ k/(k + 8)k = 1 … ∞

Exactly one of the four series converges. Click it.

A: Σ 1/k⁴B: Σ (5/3)ᵏC: Σ (k + 4)/k²D: Σ k/(k + 8)
  1. 1A: Σ 1/k⁴
  2. 2B: Σ (5/3)ᵏ
  3. 3C: Σ (k + 4)/k²
  4. 4D: Σ k/(k + 8)
4.Sequences & seriesHard

f(x) = cos x is approximated at x = 2 by its Maclaurin polynomial of degree n. With the Lagrange bound M·|x|ⁿ⁺¹/(n + 1)! and M = 1, find the smallest degree n that guarantees an error below 1 × 10⁻⁶.

f(x) = cos xx = 2M = 1ε = 1 × 10⁻⁶
n =
5.Multivariable calculusMedium
xy−4−202012PQ

For f(x, y) = x²e3y, find the directional derivative Duf at P(0.5, 1) in the direction towards Q(−3.5, 1).

f(x, y) = x²e3yP = (0.5, 1)Q = (−3.5, 1)
Duf =
6.Multivariable calculusHard
xy−202−101(1, 0)(−1, 0)

Find the critical points of f(x, y) = −x³ + 3x − y², classify them, and give the value fmax of f at its local maximum.

f(x, y) = −x³ + 3x − y²
fmax =
7.Differential equationsMedium
t (min)T (°C)10Ta​T = ?

A can of soda at 6 °C is taken out of the fridge into a room at 25 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). With k = 0.05 min⁻¹, find its temperature T after 10 min.

T0 = 6 °CTa = 25 °Ck = 0.05 min⁻¹t = 10 min
T = °C
8.Differential equationsHard
tyK = 10y(0.5) = ?

A system obeys y″ + 3y′ + 11.25y = 112.5 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find y(0.5); angles in radians.

y″ + 3y′ + 11.25y = 112.5y(0) = 0y′(0) = 0
y(0.5) =

3Solutions

1.Integrals

Answer: I = 6.389

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) = ex/1

Evaluate at the limits

I = (e2 − 1)/1 = (7.389 − 1)/1 = 6.389

Answer

I = 6.389
2.Integrals

Answer: I = 0.0609

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

1.649·(1.25) − 1·2 = 0.0609

Answer

I = 0.0609
3.Sequences & series

Answer: A: Σ 1/k⁴

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/k⁴ → p-series with p = 4 > 1: converges.
B: Σ (5/3)ᵏ → Geometric with r = 5/3 > 1: the terms grow, so it diverges.
C: Σ (k + 4)/k² → Behaves like 1/k for large k: diverges by comparison with the harmonic series.
D: Σ k/(k + 8) → The terms tend to 1, not 0: diverges by the n-th term test.

The answer

A: Σ 1/k⁴

Answer

A: Σ 1/k⁴
4.Sequences & series

Answer: n = 13

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.

n(n + 1)!bound261.3333240.666741200.266757200.08889650400.02547403200.00634983628800.001411936288000.000282210399168005.131 × 10⁻⁵114790016008.551 × 10⁻⁶1262270208001.316 × 10⁻⁶13871782912001.879 × 10⁻⁷

Bounds for increasing n

n = 12: 1.316 × 10⁻⁶ ≥ 1 × 10⁻⁶
n = 13: 1.879 × 10⁻⁷ < 1 × 10⁻⁶

Smallest degree

n = 13

Answer

n = 13
5.Multivariable calculus

Answer: Duf = −20.09

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 = 2xe3y → fx(0.5, 1) = 20.09
fy = 3x²e3y → fy(0.5, 1) = 15.06

Unit vector

|v| = √((−4)² + 0²) = 4
u = (−1, 0)

Dot product

Duf = 20.09·(−1) + 15.06·0 = −20.09

Answer

Duf = −20.09
6.Multivariable calculus

Answer: fmax = 2

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 − 3) = 0
fy = −(2y) = 0
(1, 0), (−1, 0)

Second-derivative test

(1, 0): fxx = −6, fyy = −2, fxy = 0, D = 12 → local maximum
(−1, 0): fxx = 6, fyy = −2, fxy = 0, D = −12 → saddle point

Value at the extremum

f(1, 0) = 2

Answer

fmax = 2
7.Differential equations

Answer: T = 13.48 °C

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.

Solution of the cooling law

T(t) = 25 + (−19)·e−0.05t

Substitute t

T(10) = 25 + (−19)·0.6065 = 13.48 °C

Answer

T = 13.48 °C
8.Differential equations

Answer: y(0.5) = 7.31

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.

Steady value and roots

yss = 10, r = −1.5 ± 3j

Constants from the initial values

y(0) = 10 + A = 0 → A = −10
y′(0) = −1.5A + 3B = 0 → B = −5

Evaluate

y(0.5) = 10·(1 − e−0.75(cos 1.5 + 0.5·sin 1.5)) = 7.31

Answer

y(0.5) = 7.31

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