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
xyπy = x·cos x

Evaluate I = ∫₀π x·cos x dx (x in radians).

∫₀π x·cos x dx
I =
2.IntegralsHard
xyy = 0.5x2​y = 2x

Find the area A of the region enclosed by y = 0.5x2 and y = 2x.

y = 0.5x2y = 2x
A =
3.Sequences & seriesMedium
kSn​12345678910S = ?

Find the sum S = Σk=1∞ 6/(k(k + 4)).

ak = 6/(k(k + 4))
S =
4.Sequences & seriesHard

Find the coefficient c₃ of (x − 5)³ in the Taylor series of f(x) = 1/x about a = 5.

f(x) = 1/xa = 5k = 3
c₃ =
5.Multivariable calculusMedium

z = 4xy³ + 3x³ + 4y³ − 4xy with x = −2t + 2 and y = −2t² + 2. Use the chain rule to find dz/dt at t = 0.

z = 4xy³ + 3x³ + 4y³ − 4xyx = −2t + 2y = −2t² + 2t = 0
dz/dt =
6.Multivariable calculusHard
xy−202−202(0, 0)(1, 1)(−1, −1)

Find the critical points of f(x, y) = x⁴ + y⁴ − 4xy, classify them, and give the value fmin of f at its local minimum.

f(x, y) = x⁴ + y⁴ − 4xy
fmin =
7.Differential equationsMedium
t (min)T (°C)7Ta​T = ?

A can of soda at 4 °C is taken out of the fridge into a room at 26 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). With k = 0.07 min⁻¹, find its temperature T after 7 min.

T0 = 4 °CTa = 26 °Ck = 0.07 min⁻¹t = 7 min
T = °C
8.Differential equationsHard
t (min)m (kg)t = ?23 kg

A tank holds 400 L of water in which 5 kg of salt is dissolved. Brine with 0.1 kg of salt per litre flows in at 10 L/min, and the well-mixed solution flows out at the same rate. After how many minutes t does the tank contain 23 kg of salt?

V = 400 Lm0 = 5 kgc = 0.1 kg/Lr = 10 L/min
t = min

3Solutions

1.Integrals

Answer: I = −2

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 = x, du = dx; dv = cos x dx, v = sin x

Antiderivative

∫x cos x dx = x sin x − ∫sin x dx = x sin x + cos x

Limits

(π·sin π + cos π) − (0 + 1) = −2

Answer

I = −2
2.Integrals

Answer: A = 5.333

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 = 2x → x = 0, x = 4

Integrate upper minus lower

A = [x2 − 0.1667x3]04 = 5.333

Answer

A = 5.333
3.Sequences & series

Answer: S = 3.125

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.

Partial fractions

6/(k(k + 4)) = (6/4)·(1/k − 1/(k + 4))

Cancel

S = (6/4)·(1 + 1/2 + 1/3 + 1/4) = 3.125

Answer

S = 3.125
4.Sequences & series

Answer: c₃ = −0.0016

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⁽³⁾

f⁽³⁾(x) = −6/x⁴

Coefficient

c₃ = −6/(5⁴·6) = −0.0016
c₃ = −0.0016

Answer

c₃ = −0.0016
5.Multivariable calculus

Answer: dz/dt = −120

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.

Point and rates

x(0) = 2, y(0) = 2
dx/dt = −2, dy/dt = −4t = 0

Partial derivatives

fx = 4y³ + 9x² − 4y = 60
fy = 12xy² + 12y² − 4x = 136

Chain rule

dz/dt = 60·(−2) + 136·0 = −120

Answer

dz/dt = −120
6.Multivariable calculus

Answer: fmin = −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 = 4x³ − 4y = 0
fy = 4y³ − 4x = 0
(0, 0), (1, 1), (−1, −1)

Second-derivative test

(0, 0): fxx = 0, fyy = 0, fxy = −4, D = −16 → saddle point
(1, 1): fxx = 12, fyy = 12, fxy = −4, D = 128 → local minimum
(−1, −1): fxx = 12, fyy = 12, fxy = −4, D = 128 → local minimum

Value at the extremum

f(−1, −1) = −2

Answer

fmin = −2
7.Differential equations

Answer: T = 12.52 °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) = 26 + (−22)·e−0.07t

Substitute t

T(7) = 26 + (−22)·0.6126 = 12.52 °C

Answer

T = 12.52 °C
8.Differential equations

Answer: t = 28.89 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) = 40 + (−35)·e−t/40

Solve for t

e−t/40 = (40 − 23)/(40 − 5) = 0.4857
t = 40·ln(2.059) = 28.89 min

Answer

t = 28.89 min

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