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
xyy = 0.5x2​y = x

Find the area A of the region enclosed by y = 0.5x2 and y = x.

y = 0.5x2y = x
A =
3.Sequences & seriesMedium
AΣ (k + 1)/k²k = 1 … ∞BΣ 1/(3ᵏ + k)k = 1 … ∞CΣ (1 + 1/k)ᵏk = 1 … ∞DΣ 1/√kk = 1 … ∞

Exactly one of the four series converges. Click it.

A: Σ (k + 1)/k²B: Σ 1/(3ᵏ + k)C: Σ (1 + 1/k)ᵏD: Σ 1/√k
  1. 1A: Σ (k + 1)/k²
  2. 2B: Σ 1/(3ᵏ + k)
  3. 3C: Σ (1 + 1/k)ᵏ
  4. 4D: Σ 1/√k
4.Sequences & seriesHard
xyfP₄1.5

Use the Maclaurin polynomial of degree 4 of f(x) = cos x to approximate f(1.5). Give P₄(1.5).

f(x) = cos xn = 4x = 1.5
P₄(1.5) =
5.Multivariable calculusMedium
xy31∇fP

For f(x, y) = 5xy − 4x² − 4y², find the maximum rate of change of f at the point (3, 1), that is |∇f|.

f(x, y) = 5xy − 4x² − 4y²P = (3, 1)
|∇f| =
6.Multivariable calculusHard
xy−20202(2, 2)(−2, 2)

Find the critical points of f(x, y) = −x³ + 12x − y² + 4y, classify them, and give the value fmax of f at its local maximum.

f(x, y) = −x³ + 12x − y² + 4y
fmax =
7.Differential equationsMedium
t (min)T (°C)12Ta​17 °C

A can of soda at 3 °C is taken out of the fridge into a room at 30 °C. After 12 min its temperature is 17 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). Find the constant k in min⁻¹.

T0 = 3 °CTa = 30 °CT(12) = 17 °C
k = min⁻¹
8.Differential equationsHard
t (min)m (kg)20 minm = ?

A tank holds 400 L of water in which 20 kg of salt is dissolved. Brine with 0.25 kg of salt per litre flows in at 5 L/min, and the well-mixed solution flows out at the same rate. How much salt m is in the tank after 20 min?

V = 400 Lm0 = 20 kgc = 0.25 kg/Lr = 5 L/min
m = kg

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: A = 0.6667

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.

Intersection points

0.5x2 = x → x = 0, x = 2

Integrate upper minus lower

A = [0.5x2 − 0.1667x3]02 = 0.6667

Answer

A = 0.6667
3.Sequences & series

Answer: B: Σ 1/(3ᵏ + 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 (k + 1)/k² → Behaves like 1/k for large k: diverges by comparison with the harmonic series.
B: Σ 1/(3ᵏ + k) → Smaller than (1/3)ᵏ, a convergent geometric series: converges by comparison.
C: Σ (1 + 1/k)ᵏ → The terms tend to e, not 0: diverges by the n-th term test.
D: Σ 1/√k → p-series with p = 1/2 ≤ 1: diverges.

The answer

B: Σ 1/(3ᵏ + k)

Answer

B: Σ 1/(3ᵏ + k)
4.Sequences & series

Answer: P₄(1.5) = 0.08594

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

P₄(x) = 1 − 0.5x² + 0.04167x⁴

Substitute

P₄(1.5) = 1 − 1.125 + 0.210938 = 0.08594

Answer

P₄(1.5) = 0.08594

The exact value is f(1.5) = 0.0707372; the error −0.0152003 shrinks when you take more terms or a point closer to 0.

5.Multivariable calculus

Answer: |∇f| = 20.25

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.

Partial derivatives

fx = 5y − 8x → fx(3, 1) = −19
fy = 5x − 8y → fy(3, 1) = 7

Length of the gradient

|∇f| = √((−19)² + 7²) = 20.25

Answer

|∇f| = 20.25
6.Multivariable calculus

Answer: fmax = 20

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 − 4) = 0
(2, 2), (−2, 2)

Second-derivative test

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

Value at the extremum

f(2, 2) = 20

Answer

fmax = 20
7.Differential equations

Answer: k = 0.06091 min⁻¹

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.

Ratio of temperature differences

e−12k = -13/-27

Solve for k

k = ln(-27/-13)/12 = 0.06091 min⁻¹

Answer

k = 0.06091 min⁻¹
8.Differential equations

Answer: m = 37.7 kg

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.

Set up the equation

dm/dt = 5·0.25 − (5/400)·m = 1.25 − 0.0125·m

Solve

The integrating factor is ert/V; the solution approaches the equilibrium c·V.

m(t) = 100 + (−80)·e−t/80

Substitute t

m(20) = 100 + (−80)·0.7788 = 37.7 kg

Answer

m = 37.7 kg

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