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
xy28

Evaluate I = ∫28 5/x dx.

f(x) = 5/x[2, 8]
I =
2.IntegralsHard
xy−11

Find the area A of the region between the parabola y = 1 − x² and the x-axis.

y = 1 − x²
A =
3.Sequences & seriesMedium
xx₀ = 2??R

The power series Σk=1∞ 5ᵏ·(x − 2)ᵏ/(k + 1) converges on an interval around x₀ = 2. Find the right end x of that interval.

Σ 5ᵏ·(x − 2)ᵏ/(k + 1)x₀ = 2
x =
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 1 × 10⁻⁶.

f(x) = eˣx = 0.5M = 2ε = 1 × 10⁻⁶
n =
5.Multivariable calculusMedium
xy−202012vP

For f(x, y) = 4x²y³ + 2x³ + 3y³ + 4xy, find the directional derivative Duf at (1, 1) in the direction of v = (−4, 1).

f(x, y) = 4x²y³ + 2x³ + 3y³ + 4xyP = (1, 1)v = (−4, 1)
Duf =
6.Multivariable calculusHard
xy−505−4−202(3, −2)(−3, −2)

Find the critical points of f(x, y) = x³ − 27x + y² + 4y, classify them, and give the value fmin of f at its local minimum.

f(x, y) = x³ − 27x + y² + 4y
fmin =
7.Differential equationsMedium
tyy(0) = 5y(0.5) = ?

Solve y″ + 6y′ + 9y = 0 with y(0) = 5 and y′(0) = 4. Find y(0.5).

y″ + 6y′ + 9y = 0y(0) = 5y′(0) = 4
y(0.5) =
8.Differential equationsHard
tyy(0) = 1y(0.75) = ?

Solve y″ + 2y′ + 5y = 0 with y(0) = 1 and y′(0) = 0 (released from rest). Find y(0.75); angles in radians.

y″ + 2y′ + 5y = 0y(0) = 1y′(0) = 0
y(0.75) =

3Solutions

1.Integrals

Answer: I = 6.931

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) = 5·ln x

Evaluate at the limits

I = 5·(ln 8 − ln 2) = 5·ln 4 = 6.931

Answer

I = 6.931
2.Integrals

Answer: A = 1.333

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.

Zeros

1 − x² = 0 → x = ±1

Integrate

A = [1x − x³/3]−11 = 0.6667 − (−0.6667) = 1.333

Answer

A = 1.333
3.Sequences & series

Answer: x = 2.2

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.

Coefficients

ck/ck+1 = (1/5)·(k + 2)/(k + 1) → 0.2

Radius

R = 0.2

Interval

1.8 < x < 2.2
x = 2.2

Answer

x = 2.2

The ratio test says nothing at the two end points; they have to be checked separately.

4.Sequences & series

Answer: n = 7

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 × 10⁻⁵650403.1 × 10⁻⁶7403201.938 × 10⁻⁷

Bounds for increasing n

n = 6: 3.1 × 10⁻⁶ ≥ 1 × 10⁻⁶
n = 7: 1.938 × 10⁻⁷ < 1 × 10⁻⁶

Smallest degree

n = 7

Answer

n = 7
5.Multivariable calculus

Answer: Duf = −11.4

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 = 8xy³ + 6x² + 4y → fx(1, 1) = 18
fy = 12x²y² + 9y² + 4x → fy(1, 1) = 25

Unit vector

|v| = √((−4)² + 1²) = 4.123
u = (−0.9701, 0.2425)

Dot product

Duf = 18·(−0.9701) + 25·0.2425 = −11.4

Answer

Duf = −11.4
6.Multivariable calculus

Answer: fmin = −58

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

Value at the extremum

f(3, −2) = −58

Answer

fmin = −58
7.Differential equations

Answer: y(0.5) = 3.235

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.

Double root

(r + 3)² = 0 → r = −3

Constants

A = y(0) = 5
y′(0) = B − 3A → B = 4 + 15 = 19

Evaluate

y(0.5) = (5 + 19·0.5)·e−1.5 = 3.235

Answer

y(0.5) = 3.235
8.Differential equations

Answer: y(0.75) = 0.269

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.

Complex roots

r = −1 ± 2j

Constants

A = 1, B = 1·1/2 = 0.5

Evaluate

y(0.75) = e−0.75(1·cos 1.5 + 0.5·sin 1.5) = 0.269

Answer

y(0.75) = 0.269

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