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
xy01

Evaluate I = ∫01 2x·(x² + 1)2 dx.

f(x) = 2x·(x² + 1)2[0, 1]
I =
2.IntegralsHard
xyy = 2x2​y = 4x

Find the area A of the region enclosed by y = 2x2 and y = 4x.

y = 2x2y = 4x
A =
3.Sequences & seriesMedium
AΣ 1/kk = 1 … ∞BΣ (−1)ᵏ/√kk = 1 … ∞CΣ (k + 5)/k²k = 1 … ∞DΣ k/(k + 2)k = 1 … ∞

Exactly one of the four series converges. Click it.

A: Σ 1/kB: Σ (−1)ᵏ/√kC: Σ (k + 5)/k²D: Σ k/(k + 2)
  1. 1A: Σ 1/k
  2. 2B: Σ (−1)ᵏ/√k
  3. 3C: Σ (k + 5)/k²
  4. 4D: Σ k/(k + 2)
4.Sequences & seriesHard

f(x) = eˣ is approximated at x = 1 by its Maclaurin polynomial of degree n. With the Lagrange bound M·|x|ⁿ⁺¹/(n + 1)! and M = 3, find the smallest degree n that guarantees an error below 0.0001.

f(x) = eˣx = 1M = 3ε = 0.0001
n =
5.Multivariable calculusMedium
xy12∇fP

For f(x, y) = 3xy³ + 5x³ − 2y³ + 2xy, find the direction of steepest ascent at (1, 2) as the angle θ from the positive x-axis, in degrees between −180° and 180°.

f(x, y) = 3xy³ + 5x³ − 2y³ + 2xyP = (1, 2)
θ = °
6.Multivariable calculusHard
xyx = 3y = 3x39

Evaluate I = ∬T (4x + 3y) dA over the triangle T bounded by y = 0, x = 3 and y = 3x.

f = 4x + 3y0 ≤ x ≤ 30 ≤ y ≤ 3x
I =
7.Differential equationsMedium
tyy(0) = 3y(2) = ?

Solve y″ + 2y′ + y = 0 with y(0) = 3 and y′(0) = -2. Find y(2).

y″ + 2y′ + y = 0y(0) = 3y′(0) = -2
y(2) =
8.Differential equationsHard
tyK = 2ymax​ = ?

A system obeys y″ + 3y′ + 27.25y = 54.5 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find the peak value ymax of its step response.

y″ + 3y′ + 27.25y = 54.5y(0) = 0y′(0) = 0
ymax =

3Solutions

1.Integrals

Answer: I = 2.333

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.

Substitution

u = x² + 1, du = 2x dx
I = ∫12 u2 du

Integrate and evaluate

I = [u3/3]12 = (8 − 1)/3 = 2.333

Answer

I = 2.333
2.Integrals

Answer: A = 2.667

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

2x2 = 4x → x = 0, x = 2

Integrate upper minus lower

A = [2x2 − 0.6667x3]02 = 2.667

Answer

A = 2.667
3.Sequences & series

Answer: B: 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 → A multiple of the harmonic series (p = 1): diverges.
B: Σ (−1)ᵏ/√k → Alternating, with 1/√k decreasing to 0: converges (conditionally).
C: Σ (k + 5)/k² → Behaves like 1/k for large k: diverges by comparison with the harmonic series.
D: Σ k/(k + 2) → The terms tend to 1, not 0: diverges by the n-th term test.

The answer

B: Σ (−1)ᵏ/√k

Answer

B: Σ (−1)ᵏ/√k
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)!bound013121.5260.53240.12541200.02557200.004167650400.00059527403207.44 × 10⁻⁵

Bounds for increasing n

n = 6: 0.0005952 ≥ 0.0001
n = 7: 7.44 × 10⁻⁵ < 0.0001

Smallest degree

n = 7

Answer

n = 7
5.Multivariable calculus

Answer: θ = 18.03 °

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 = 3y³ + 15x² + 2y → fx(1, 2) = 43
fy = 9xy² − 6y² + 2x → fy(1, 2) = 14

Direction

θ = atan2(14, 43) = 18.03°

Answer

θ = 18.03°
6.Multivariable calculus

Answer: I = 229.5

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 ≤ 3, 0 ≤ y ≤ 3x

Inner integral

∫03x (4x + 3y) dy = 4x·3x + 3·(3x)²/2 = 25.5x²

Outer integral

I = 25.5·3³/3 = 229.5

Answer

I = 229.5
7.Differential equations

Answer: y(2) = 0.6767

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 + 1)² = 0 → r = −1

Constants

A = y(0) = 3
y′(0) = B − 1A → B = -2 + 3 = 1

Evaluate

y(2) = (3 + 1·2)·e−2 = 0.6767

Answer

y(2) = 0.6767
8.Differential equations

Answer: ymax = 2.779

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 = 2, r = −1.5 ± 5j

Constants from the initial values

y(0) = 2 + A = 0 → A = −2
y′(0) = −1.5A + 5B = 0 → B = −0.6

Peak

t = π/5 = 0.6283
ymax = 2·(1 + e−1.5·π/5) = 2.779

Answer

ymax = 2.779

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