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

Evaluate I = ∫12 5/x dx.

f(x) = 5/x[1, 2]
I =
2.IntegralsHard
xy1.5y = 3xy = 2x²

The region between the line y = 3x and the parabola y = 2x² is rotated about the x-axis. Find the volume V of the solid.

y = 3xy = 2x²
V =
3.Sequences & seriesMedium
xx₀ = −2??R

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

Σ (2k)!·(x + 2)ᵏ/(k!)²x₀ = −2
x =
4.Sequences & seriesHard

Recognise the series and find its sum: S = Σk=1∞ k·0.5ᵏ.

ak = k·0.5ᵏ
S =
5.Multivariable calculusMedium

z = −2xy + 4x³ + 3y² with x = −t and y = −2t² − 1. Use the chain rule to find dz/dt at t = 0.

z = −2xy + 4x³ + 3y²x = −ty = −2t² − 1t = 0
dz/dt =
6.Multivariable calculusHard
xy−202−202ABC

The marked points are the critical points of f(x, y) = x⁴ + y⁴ − 4xy. Click the saddle point.

f(x, y) = x⁴ + y⁴ − 4xy
  1. 1A (0, 0)
  2. 2B (−1, −1)
  3. 3C (1, 1)
7.Differential equationsMedium
tyy(0) = 2y(0.5) = ?

Solve y″ + 4y′ + 4y = 0 with y(0) = 2 and y′(0) = 0. Find y(0.5).

y″ + 4y′ + 4y = 0y(0) = 2y′(0) = 0
y(0.5) =
8.Differential equationsHard
tyK = 1ymax​ = ?

A system obeys y″ + 3y′ + 18.25y = 18.25 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find the peak value ymax of its step response.

y″ + 3y′ + 18.25y = 18.25y(0) = 0y′(0) = 0
ymax =

3Solutions

1.Integrals

Answer: I = 3.466

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 2 − ln 1) = 5·ln 2 = 3.466

Answer

I = 3.466
2.Integrals

Answer: V = 12.72

Volumes of revolution: rotating the region under y = f(x) from x = a to x = b about the x-axis makes a solid whose cross-section at x is a disc of radius f(x). Adding the discs gives V = π∫ₐᵇ f(x)² dx. When the region lies between an outer curve R(x) and an inner curve r(x), each slice is a washer: V = π∫ₐᵇ (R(x)² − r(x)²) dx. Square the radii before you subtract them.

Intersection points

3x = 2x² → x = 0, x = 1.5

Washer method

V = π∫₀1.5 ((3x)² − (2x²)²) dx

Integrate

V = π(9·1.5³/3 − 4·1.5⁵/5) = π·4.05 = 12.72

Answer

V = 12.72
3.Sequences & series

Answer: x = −2.25

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+1/ck = (2k + 2)(2k + 1)/(k + 1)² → 4, so R = 1/4

Radius

R = 0.25

Interval

−2.25 < x < −1.75
x = −2.25

Answer

x = −2.25

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

4.Sequences & series

Answer: S = 2

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.

Standard series

Σ k·xᵏ = x/(1 − x)²

Substitute

x = 0.5
S = 0.5/(1 − 0.5)² = 2

Answer

S = 2
5.Multivariable calculus

Answer: dz/dt = −2

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) = 0, y(0) = −1
dx/dt = −1, dy/dt = −4t = 0

Partial derivatives

fx = −2y + 12x² = 2
fy = −2x + 6y = −6

Chain rule

dz/dt = 2·(−1) + (−6)·0 = −2

Answer

dz/dt = −2
6.Multivariable calculus

Answer: A (0, 0)

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

The point

A (0, 0): saddle point

Answer

A (0, 0)
7.Differential equations

Answer: y(0.5) = 1.472

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 + 2)² = 0 → r = −2

Constants

A = y(0) = 2
y′(0) = B − 2A → B = 0 + 4 = 4

Evaluate

y(0.5) = (2 + 4·0.5)·e−1 = 1.472

Answer

y(0.5) = 1.472
8.Differential equations

Answer: ymax = 1.308

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.

Steady value and roots

yss = 1, r = −1.5 ± 4j

Constants from the initial values

y(0) = 1 + A = 0 → A = −1
y′(0) = −1.5A + 4B = 0 → B = −0.375

Peak

t = π/4 = 0.7854
ymax = 1·(1 + e−1.5·π/4) = 1.308

Answer

ymax = 1.308

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