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
xy16y = x ln x

Evaluate I = ∫₁6 x·ln x dx.

∫₁6 x ln x dx
I =
2.IntegralsHard
xy0.5y = x²eˣ

Evaluate I = ∫₀0.5 x²·eˣ dx.

∫₀0.5 x²eˣ dx
I =
3.Sequences & seriesMedium
AΣ 3ᵏ/k!k = 1 … ∞BΣ (−1)ᵏ/√kk = 1 … ∞CΣ 1/√kk = 1 … ∞DΣ 1/(2ᵏ + k)k = 1 … ∞

Exactly one of the four series diverges. Click it.

A: Σ 3ᵏ/k!B: Σ (−1)ᵏ/√kC: Σ 1/√kD: Σ 1/(2ᵏ + k)
  1. 1A: Σ 3ᵏ/k!
  2. 2B: Σ (−1)ᵏ/√k
  3. 3C: Σ 1/√k
  4. 4D: Σ 1/(2ᵏ + k)
4.Sequences & seriesHard

f(x) = ln(1 + x) is approximated by its Maclaurin polynomial of degree 4 at x = 0.2. Use the Lagrange error bound M·|x|ⁿ⁺¹/(n + 1)! with M = 4! = 24 to find the largest possible error |R₄|.

f(x) = ln(1 + x)n = 4x = 0.2M = 24
|R₄| =
5.Multivariable calculusMedium
xy−4−20024vP

For f(x, y) = −2x³y² − 4x² + 5y² − 2xy, find the directional derivative Duf at (−2, 1) in the direction of v = (3, 4).

f(x, y) = −2x³y² − 4x² + 5y² − 2xyP = (−2, 1)v = (3, 4)
Duf =
6.Multivariable calculusHard
xyy = 1.5y = x1.51.5

The integral I = ∫01.5 ∫x1.5 cos(y²) dy dx cannot be done in this order. Reverse the order of integration and evaluate it.

f = cos(y²)0 ≤ x ≤ y ≤ 1.5
I =
7.Differential equationsMedium
t (min)T (°C)16Ta​T = ?

A forged steel part at 500 °C is left to cool in a workshop at 20 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). With k = 0.02 min⁻¹, find its temperature T after 16 min.

T0 = 500 °CTa = 20 °Ck = 0.02 min⁻¹t = 16 min
T = °C
8.Differential equationsHard
tyK = 1y(0.75) = ?

A system obeys y″ + 4y′ + 8y = 8 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find y(0.75); angles in radians.

y″ + 4y′ + 8y = 8y(0) = 0y′(0) = 0
y(0.75) =

3Solutions

1.Integrals

Answer: I = 23.5

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.

Choose u and dv

u = ln x, du = dx/x; dv = x dx, v = x²/2

Antiderivative

∫x ln x dx = (x²/2)ln x − ∫(x/2) dx = (x²/2)ln x − x²/4

Limits

(18·ln 6 − 9) − (0 − 0.25) = 23.5

Answer

I = 23.5
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: C: Σ 1/√k

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/k! → Ratio test: L = lim 3/(k + 1) = 0 < 1, converges.
B: Σ (−1)ᵏ/√k → Alternating, with 1/√k decreasing to 0: converges (conditionally).
C: Σ 1/√k → p-series with p = 1/2 ≤ 1: diverges.
D: Σ 1/(2ᵏ + k) → Smaller than (1/2)ᵏ, a convergent geometric series: converges by comparison.

The answer

C: Σ 1/√k

Answer

C: Σ 1/√k
4.Sequences & series

Answer: |R₄| = 6.4 × 10⁻⁵

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.

Bound on the derivative

M = 4! = 24

Lagrange bound

|R₄| ≤ 24·0.2⁵/5! = 6.4 × 10⁻⁵

Answer

|R₄| = 6.4 × 10⁻⁵

The true error is smaller: |f(0.2) − P₄(0.2)| = 5.489 × 10⁻⁵.

5.Multivariable calculus

Answer: Duf = 30.8

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 = −6x²y² − 8x − 2y → fx(−2, 1) = −10
fy = −4x³y + 10y − 2x → fy(−2, 1) = 46

Unit vector

|v| = √(3² + 4²) = 5
u = (0.6, 0.8)

Dot product

Duf = (−10)·0.6 + 46·0.8 = 30.8

Answer

Duf = 30.8
6.Multivariable calculus

Answer: I = 0.389

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 ≤ 1.5

Reversed order

I = ∫01.5 ∫0y cos(y²) dx dy = ∫01.5 y·cos(y²) dy

Substitution u = y²

I = [sin(y²)/2]01.5 = 0.389

Answer

I = 0.389
7.Differential equations

Answer: T = 368.6 °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) = 20 + 480·e−0.02t

Substitute t

T(16) = 20 + 480·0.7261 = 368.6 °C

Answer

T = 368.6 °C
8.Differential equations

Answer: y(0.75) = 0.7616

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 = 1, r = −2 ± 2j

Constants from the initial values

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

Evaluate

y(0.75) = 1·(1 − e−1.5(cos 1.5 + 1·sin 1.5)) = 0.7616

Answer

y(0.75) = 0.7616

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