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
xy12.5y = x ln x

Evaluate I = ∫₁2.5 x·ln x dx.

∫₁2.5 x ln x dx
I =
2.IntegralsHard
xyy = 2x2​y = 3x

Find the area A of the region enclosed by y = 2x2 and y = 3x.

y = 2x2y = 3x
A =
3.Sequences & seriesMedium
AΣ ln k / kk = 1 … ∞BΣ 1/(k² + 2)k = 1 … ∞CΣ 2/kk = 1 … ∞DΣ (3/2)ᵏk = 1 … ∞

Exactly one of the four series converges. Click it.

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

Find the coefficient c₄ of (x − 2)⁴ in the Taylor series of f(x) = ln x about a = 2.

f(x) = ln xa = 2k = 4
c₄ =
5.Multivariable calculusMedium
xy02−2−10PQ

For f(x, y) = 5x³y³ + 5x² − xy, find the directional derivative Duf at P(1, −1) in the direction towards Q(−1, −1).

f(x, y) = 5x³y³ + 5x² − xyP = (1, −1)Q = (−1, −1)
Duf =
6.Multivariable calculusHard
xyx = 1.5y = x1.5

The integral I = ∫01.5 ∫y1.5 ex² dx dy cannot be done in this order. Reverse the order of integration and evaluate it.

f = ex²0 ≤ y ≤ x ≤ 1.5
I =
7.Differential equationsMedium
t (min)T (°C)21Ta​T = ?

A can of soda at 5 °C is taken out of the fridge into a room at 23 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). With k = 0.06 min⁻¹, find its temperature T after 21 min.

T0 = 5 °CTa = 23 °Ck = 0.06 min⁻¹t = 21 min
T = °C
8.Differential equationsHard
t (min)m (kg)70 minm = ?

A tank holds 500 L of water in which 20 kg of salt is dissolved. Brine with 0.05 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 70 min?

V = 500 Lm0 = 20 kgc = 0.05 kg/Lr = 5 L/min
m = kg

3Solutions

1.Integrals

Answer: I = 1.551

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

(3.125·ln 2.5 − 1.5625) − (0 − 0.25) = 1.551

Answer

I = 1.551
2.Integrals

Answer: A = 1.125

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

2x2 = 3x → x = 0, x = 1.5

Integrate upper minus lower

A = [1.5x2 − 0.6667x3]01.5 = 1.125

Answer

A = 1.125
3.Sequences & series

Answer: B: Σ 1/(k² + 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 ln k / k → Larger than 1/k for k ≥ 3: diverges by comparison with the harmonic series.
B: Σ 1/(k² + 2) → Smaller than 1/k², a convergent p-series: converges by comparison.
C: Σ 2/k → A multiple of the harmonic series (p = 1): diverges.
D: Σ (3/2)ᵏ → Geometric with r = 3/2 > 1: the terms grow, so it diverges.

The answer

B: Σ 1/(k² + 2)

Answer

B: Σ 1/(k² + 2)
4.Sequences & series

Answer: c₄ = −0.01563

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⁽⁴⁾

f⁽⁴⁾(x) = −6/x⁴

Coefficient

c₄ = f⁽⁴⁾(2)/4! = −6/(2⁴·24) = −0.01563
c₄ = −0.01563

Answer

c₄ = −0.01563
5.Multivariable calculus

Answer: Duf = 4

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 = 15x²y³ + 10x − y → fx(1, −1) = −4
fy = 15x³y² − x → fy(1, −1) = 14

Unit vector

|v| = √((−2)² + 0²) = 2
u = (−1, 0)

Dot product

Duf = (−4)·(−1) + 14·0 = 4

Answer

Duf = 4
6.Multivariable calculus

Answer: I = 4.244

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 ≤ y ≤ x, 0 ≤ x ≤ 1.5

Reversed order

I = ∫01.5 ∫0x ex² dy dx = ∫01.5 x·ex² dx

Substitution u = x²

I = [ex²/2]01.5 = 4.244

Answer

I = 4.244
7.Differential equations

Answer: T = 17.89 °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) = 23 + (−18)·e−0.06t

Substitute t

T(21) = 23 + (−18)·0.2837 = 17.89 °C

Answer

T = 17.89 °C
8.Differential equations

Answer: m = 22.52 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 = 5·0.05 − (5/500)·m = 0.25 − 0.01·m

Solve

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

m(t) = 25 + (−5)·e−t/100

Substitute t

m(70) = 25 + (−5)·0.4966 = 22.52 kg

Answer

m = 22.52 kg

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