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
xyπ/3y = x·sin x

Evaluate I = ∫₀π/3 x·sin x dx (x in radians).

∫₀π/3 x·sin x dx
I =
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
kSn​12345678910S = ?

Find the sum S = Σk=1∞ 5/(k(k + 4)).

ak = 5/(k(k + 4))
S =
4.Sequences & seriesHard
xyfP₂−0.2

Use the Maclaurin polynomial of degree 2 of f(x) = 1/(1 − 2x) to approximate f(−0.2). Give P₂(−0.2).

f(x) = 1/(1 − 2x)n = 2x = −0.2
P₂(−0.2) =
5.Multivariable calculusMedium
xy−2−1∇fP

For f(x, y) = x³y³ − 4x² − 4y², find the direction of steepest ascent at (−2, −1) as the angle θ from the positive x-axis, in degrees between −180° and 180°.

f(x, y) = x³y³ − 4x² − 4y²P = (−2, −1)
θ = °
6.Multivariable calculusHard
xy−202−4−202AB

The marked points are the critical points of f(x, y) = −x³ + 3x − y² − 6y. Click the local maximum.

f(x, y) = −x³ + 3x − y² − 6y
  1. 1A (1, −3)
  2. 2B (−1, −3)
7.Differential equationsMedium
tyy(0) = 2y(0.5) = ?

Solve y″ + 6y′ + 9y = 0 with y(0) = 2 and y′(0) = -1. Find y(0.5).

y″ + 6y′ + 9y = 0y(0) = 2y′(0) = -1
y(0.5) =
8.Differential equationsHard
tyK = 2y(0.25) = ?

A system obeys y″ + 2y′ + 26y = 52 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find y(0.25); angles in radians.

y″ + 2y′ + 26y = 52y(0) = 0y′(0) = 0
y(0.25) =

3Solutions

1.Integrals

Answer: I = 0.3424

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 = sin x dx, v = −cos x

Antiderivative

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

Limits

(−π/3·cos π/3 + sin π/3) − (0 + 0) = 0.3424

Answer

I = 0.3424
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: S = 2.604

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

5/(k(k + 4)) = (5/4)·(1/k − 1/(k + 4))

Cancel

S = (5/4)·(1 + 1/2 + 1/3 + 1/4) = 2.604

Answer

S = 2.604
4.Sequences & series

Answer: P₂(−0.2) = 0.76

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.

Maclaurin polynomial

P₂(x) = 1 + 2x + 4x²

Substitute

P₂(−0.2) = 1 − 0.4 + 0.16 = 0.76

Answer

P₂(−0.2) = 0.76

The exact value is f(−0.2) = 0.714286; the error −0.0457143 shrinks when you take more terms or a point closer to 0.

5.Multivariable calculus

Answer: θ = −75.96 °

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 = 3x²y³ − 8x → fx(−2, −1) = 4
fy = 3x³y² − 8y → fy(−2, −1) = −16

Direction

θ = atan2(−16, 4) = −75.96°

Answer

θ = −75.96°
6.Multivariable calculus

Answer: A (1, −3)

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 − 3) = 0
fy = −(2y + 6) = 0
(1, −3), (−1, −3)

Second-derivative test

(1, −3): fxx = −6, fyy = −2, fxy = 0, D = 12 → local maximum
(−1, −3): fxx = 6, fyy = −2, fxy = 0, D = −12 → saddle point

The point

A (1, −3): local maximum

Answer

A (1, −3)
7.Differential equations

Answer: y(0.5) = 1.004

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) = 2
y′(0) = B − 3A → B = -1 + 6 = 5

Evaluate

y(0.5) = (2 + 5·0.5)·e−1.5 = 1.004

Answer

y(0.5) = 1.004
8.Differential equations

Answer: y(0.25) = 1.213

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 ± 5j

Constants from the initial values

y(0) = 2 + A = 0 → A = −2
y′(0) = −1A + 5B = 0 → B = −0.4

Evaluate

y(0.25) = 2·(1 − e−0.25(cos 1.25 + 0.2·sin 1.25)) = 1.213

Answer

y(0.25) = 1.213

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