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
xy03

Evaluate I = ∫03 e−x dx.

f(x) = e−x[0, 3]
I =
2.IntegralsHard
xy2.5y = x²eˣ

Evaluate I = ∫₀2.5 x²·eˣ dx.

∫₀2.5 x²eˣ dx
I =
3.Sequences & seriesMedium
k|aₖ|¹ᐟᵏ123456789101112

Apply the root test to Σk=1∞ ((5k + 4)/(6k + 5))ᵏ. Find L = lim |ak|1/k.

ak = ((5k + 4)/(6k + 5))ᵏ
L =
4.Sequences & seriesHard

f(x) = eˣ is approximated by its Maclaurin polynomial of degree 3 at x = 0.2. Use the Lagrange error bound M·|x|ⁿ⁺¹/(n + 1)! with M = e0.2 = 1.221 to find the largest possible error |R₃|.

f(x) = eˣn = 3x = 0.2M = 1.221
|R₃| =
5.Multivariable calculusMedium

z = −2x²e2y with x = 3t + 1 and y = 2t² − 1. Use the chain rule to find dz/dt at t = 0.

z = −2x²e2yx = 3t + 1y = 2t² − 1t = 0
dz/dt =
6.Multivariable calculusHard
xy−505−202(3, 0)(−3, 0)

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

f(x, y) = x³ − 27x + y²
fmin =
7.Differential equationsMedium
tyy(0) = 1y(1.5) = ?

Solve y″ + 2y′ + y = 0 with y(0) = 1 and y′(0) = 0. Find y(1.5).

y″ + 2y′ + y = 0y(0) = 1y′(0) = 0
y(1.5) =
8.Differential equationsHard
tyy(0) = 2y(0.5) = ?

Solve y″ + 3y′ + 11.25y = 0 with y(0) = 2 and y′(0) = 0 (released from rest). Find y(0.5); angles in radians.

y″ + 3y′ + 11.25y = 0y(0) = 2y′(0) = 0
y(0.5) =

3Solutions

1.Integrals

Answer: I = 0.9502

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) = e−x/(−1)

Evaluate at the limits

I = (e−3 − 1)/(−1) = (0.04979 − 1)/(−1) = 0.9502

Answer

I = 0.9502
2.Integrals

Answer: I = 37.59

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.

First round

∫x²eˣ dx = x²eˣ − ∫2x eˣ dx

Second round

∫2x eˣ dx = 2x eˣ − 2eˣ
∫x²eˣ dx = eˣ(x² − 2x + 2)

Limits

12.18·(3.25) − 1·2 = 37.59

Answer

I = 37.59
3.Sequences & series

Answer: L = 0.8333

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.

Root of the terms

|ak|1/k = (5k + 4)/(6k + 5) → 5/6

Conclusion

L = 0.8333 < 1

Answer

L = 0.8333

L < 1, so the series converges.

4.Sequences & series

Answer: |R₃| = 8.143 × 10⁻⁵

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.

Bound on the derivative

M = e0.2 = 1.221

Lagrange bound

|R₃| ≤ 1.221·0.2⁴/4! = 8.143 × 10⁻⁵

Answer

|R₃| = 8.143 × 10⁻⁵

The true error is smaller: |f(0.2) − P₃(0.2)| = 6.942 × 10⁻⁵.

5.Multivariable calculus

Answer: dz/dt = −1.624

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 = −4xe2y = −0.5413
fy = −4x²e2y = −0.5413

Chain rule

dz/dt = (−0.5413)·3 + (−0.5413)·0 = −1.624

Answer

dz/dt = −1.624
6.Multivariable calculus

Answer: fmin = −54

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) = 0
(3, 0), (−3, 0)

Second-derivative test

(3, 0): fxx = 18, fyy = 2, fxy = 0, D = 36 → local minimum
(−3, 0): fxx = −18, fyy = 2, fxy = 0, D = −36 → saddle point

Value at the extremum

f(3, 0) = −54

Answer

fmin = −54
7.Differential equations

Answer: y(1.5) = 0.5578

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) = 1
y′(0) = B − 1A → B = 0 + 1 = 1

Evaluate

y(1.5) = (1 + 1·1.5)·e−1.5 = 0.5578

Answer

y(1.5) = 0.5578
8.Differential equations

Answer: y(0.5) = 0.538

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 = −1.5 ± 3j

Constants

A = 2, B = 1.5·2/3 = 1

Evaluate

y(0.5) = e−0.75(2·cos 1.5 + 1·sin 1.5) = 0.538

Answer

y(0.5) = 0.538

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