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
xy13

Evaluate I = ∫13 3/x dx.

f(x) = 3/x[1, 3]
I =
2.IntegralsHard
xy2.5y = x²eˣ

Evaluate I = ∫₀2.5 x²·eˣ dx.

∫₀2.5 x²eˣ dx
I =
3.Sequences & seriesMedium
AΣ (2/3)ᵏk = 1 … ∞BΣ 1/(2k + 1)k = 1 … ∞CΣ 2ᵏ/k!k = 1 … ∞DΣ (−1)ᵏ/√kk = 1 … ∞

Exactly one of the four series diverges. Click it.

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

f(x) = eˣ is approximated at x = 0.5 by its Maclaurin polynomial of degree n. With the Lagrange bound M·|x|ⁿ⁺¹/(n + 1)! and M = 2, find the smallest degree n that guarantees an error below 0.0001.

f(x) = eˣx = 0.5M = 2ε = 0.0001
n =
5.Multivariable calculusMedium
xy23∇fP

For f(x, y) = −2xy² − 4x³ − 2y³ − 3xy, find the direction of steepest ascent at (2, 3) as the angle θ from the positive x-axis, in degrees between −180° and 180°.

f(x, y) = −2xy² − 4x³ − 2y³ − 3xyP = (2, 3)
θ = °
6.Multivariable calculusHard
xy−505−2024AB

The marked points are the critical points of f(x, y) = x³ − 27x + y² − 4y. Click the saddle point.

f(x, y) = x³ − 27x + y² − 4y
  1. 1A (3, 2)
  2. 2B (−3, 2)
7.Differential equationsMedium
t (min)T (°C)27Ta​T = ?

A cup of coffee at 81 °C is left in a room at 18 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). With k = 0.08 min⁻¹, find its temperature T after 27 min.

T0 = 81 °CTa = 18 °Ck = 0.08 min⁻¹t = 27 min
T = °C
8.Differential equationsHard
t (min)m (kg)t = ?82 kg

A tank holds 400 L of water in which 10 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. After how many minutes t does the tank contain 82 kg of salt?

V = 400 Lm0 = 10 kgc = 0.25 kg/Lr = 5 L/min
t = min

3Solutions

1.Integrals

Answer: I = 3.296

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) = 3·ln x

Evaluate at the limits

I = 3·(ln 3 − ln 1) = 3·ln 3 = 3.296

Answer

I = 3.296
2.Integrals

Answer: I = 37.59

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

12.18·(3.25) − 1·2 = 37.59

Answer

I = 37.59
3.Sequences & series

Answer: B: Σ 1/(2k + 1)

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 (2/3)ᵏ → Geometric with r = 2/3, |r| < 1: converges.
B: Σ 1/(2k + 1) → Behaves like 1/(2k) for large k; compared with the harmonic series it diverges.
C: Σ 2ᵏ/k! → Ratio test: L = lim 2/(k + 1) = 0 < 1, converges.
D: Σ (−1)ᵏ/√k → Alternating, with 1/√k decreasing to 0: converges (conditionally).

The answer

B: Σ 1/(2k + 1)

Answer

B: Σ 1/(2k + 1)
4.Sequences & series

Answer: n = 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.

n(n + 1)!bound011120.25260.041673240.00520841200.000520857204.34 × 10⁻⁵

Bounds for increasing n

n = 4: 0.0005208 ≥ 0.0001
n = 5: 4.34 × 10⁻⁵ < 0.0001

Smallest degree

n = 5

Answer

n = 5
5.Multivariable calculus

Answer: θ = −131.8 °

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.

Partial derivatives

fx = −2y² − 12x² − 3y → fx(2, 3) = −75
fy = −4xy − 6y² − 3x → fy(2, 3) = −84

Direction

θ = atan2(−84, −75) = −131.8°

Answer

θ = −131.8°
6.Multivariable calculus

Answer: B (−3, 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 − 27) = 0
fy = (2y − 4) = 0
(3, 2), (−3, 2)

Second-derivative test

(3, 2): fxx = 18, fyy = 2, fxy = 0, D = 36 → local minimum
(−3, 2): fxx = −18, fyy = 2, fxy = 0, D = −36 → saddle point

The point

B (−3, 2): saddle point

Answer

B (−3, 2)
7.Differential equations

Answer: T = 25.27 °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) = 18 + 63·e−0.08t

Substitute t

T(27) = 18 + 63·0.1153 = 25.27 °C

Answer

T = 25.27 °C
8.Differential equations

Answer: t = 128.8 min

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.

Result

m(t) = 100 + (−90)·e−t/80

Solve for t

e−t/80 = (100 − 82)/(100 − 10) = 0.2
t = 80·ln(5) = 128.8 min

Answer

t = 128.8 min

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