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
xy3y = x

The region under y = x from x = 0 to x = 3 is rotated about the x-axis. Find the volume V of the solid (a cone).

y = x0 ≤ x ≤ 3
V =
2.IntegralsHard
xyy = x2​y = x

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

y = x2y = x
A =
3.Sequences & seriesMedium
AΣ (−1)ᵏ/kk = 1 … ∞BΣ 1/√kk = 1 … ∞CΣ k!/10ᵏk = 1 … ∞DΣ (7/4)ᵏk = 1 … ∞

Exactly one of the four series converges. Click it.

A: Σ (−1)ᵏ/kB: Σ 1/√kC: Σ k!/10ᵏD: Σ (7/4)ᵏ
  1. 1A: Σ (−1)ᵏ/k
  2. 2B: Σ 1/√k
  3. 3C: Σ k!/10ᵏ
  4. 4D: Σ (7/4)ᵏ
4.Sequences & seriesHard

f(x) = eˣ is approximated at x = 2 by its Maclaurin polynomial of degree n. With the Lagrange bound M·|x|ⁿ⁺¹/(n + 1)! and M = 8, find the smallest degree n that guarantees an error below 1 × 10⁻⁵.

f(x) = eˣx = 2M = 8ε = 1 × 10⁻⁵
n =
5.Multivariable calculusMedium

z = 5x³y − xy with x = −3t + 1 and y = −2t² + 1. Use the chain rule to find dz/dt at t = 0.

z = 5x³y − xyx = −3t + 1y = −2t² + 1t = 0
dz/dt =
6.Multivariable calculusHard
xyx = 1y = 3x13

Evaluate I = ∬T (x) dA over the triangle T bounded by y = 0, x = 1 and y = 3x.

f = x0 ≤ x ≤ 10 ≤ y ≤ 3x
I =
7.Differential equationsMedium
tyy(0) = 2y(1) = ?

Solve y″ + 8y′ + 16y = 0 with y(0) = 2 and y′(0) = -2. Find y(1).

y″ + 8y′ + 16y = 0y(0) = 2y′(0) = -2
y(1) =
8.Differential equationsHard
tyy(0) = 2y(1) = ?

Solve y″ + 4y′ + 13y = 0 with y(0) = 2 and y′(0) = 0 (released from rest). Find y(1); angles in radians.

y″ + 4y′ + 13y = 0y(0) = 2y′(0) = 0
y(1) =

3Solutions

1.Integrals

Answer: V = 28.27

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.

Disc method

V = π∫₀3 1x² dx

Integrate

V = π·1·3³/3 = 28.27

Answer

V = 28.27

Check: a cone of radius 3 and height 3 has (1/3)πr²h = 28.27.

2.Integrals

Answer: A = 0.1667

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

x2 = x → x = 0, x = 1

Integrate upper minus lower

A = [0.5x2 − 0.3333x3]01 = 0.1667

Answer

A = 0.1667
3.Sequences & series

Answer: A: Σ (−1)ᵏ/k

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 (−1)ᵏ/k → Alternating, with 1/k decreasing to 0: converges (conditionally).
B: Σ 1/√k → p-series with p = 1/2 ≤ 1: diverges.
C: Σ k!/10ᵏ → Ratio test: L = lim (k + 1)/10 = ∞, diverges.
D: Σ (7/4)ᵏ → Geometric with r = 7/4 > 1: the terms grow, so it diverges.

The answer

A: Σ (−1)ᵏ/k

Answer

A: Σ (−1)ᵏ/k
4.Sequences & series

Answer: n = 13

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)!bound2610.673245.33341202.13357200.7111650400.20327403200.0507983628800.01129936288000.00225710399168000.0004105114790016006.841 × 10⁻⁵1262270208001.052 × 10⁻⁵13871782912001.503 × 10⁻⁶

Bounds for increasing n

n = 12: 1.052 × 10⁻⁵ ≥ 1 × 10⁻⁵
n = 13: 1.503 × 10⁻⁶ < 1 × 10⁻⁵

Smallest degree

n = 13

Answer

n = 13
5.Multivariable calculus

Answer: dz/dt = −42

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) = 1, y(0) = 1
dx/dt = −3, dy/dt = −4t = 0

Partial derivatives

fx = 15x²y − y = 14
fy = 5x³ − x = 4

Chain rule

dz/dt = 14·(−3) + 4·0 = −42

Answer

dz/dt = −42
6.Multivariable calculus

Answer: I = 1

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).

Limits

0 ≤ x ≤ 1, 0 ≤ y ≤ 3x

Inner integral

∫03x (x) dy = x·3x = 3x²

Outer integral

I = 3·1³/3 = 1

Answer

I = 1
7.Differential equations

Answer: y(1) = 0.1465

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 + 4)² = 0 → r = −4

Constants

A = y(0) = 2
y′(0) = B − 4A → B = -2 + 8 = 6

Evaluate

y(1) = (2 + 6·1)·e−4 = 0.1465

Answer

y(1) = 0.1465
8.Differential equations

Answer: y(1) = −0.2425

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 = −2 ± 3j

Constants

A = 2, B = 2·2/3 = 1.333

Evaluate

y(1) = e−2(2·cos 3 + 1.333·sin 3) = −0.2425

Answer

y(1) = −0.2425

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