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
xy2y = √(4x)

The region under y = √(4x) from x = 0 to x = 2 is rotated about the x-axis. Find the volume V of the solid (a paraboloid).

y = √(4x)0 ≤ x ≤ 2
V =
2.IntegralsHard
xy1y = 2xy = 2x²

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

y = 2xy = 2x²
V =
3.Sequences & seriesMedium
AΣ (−1)ᵏ/kk = 1 … ∞BΣ 2ᵏ/k!k = 1 … ∞CΣ (7/4)ᵏk = 1 … ∞DΣ (4/5)ᵏk = 1 … ∞

Exactly one of the four series diverges. Click it.

A: Σ (−1)ᵏ/kB: Σ 2ᵏ/k!C: Σ (7/4)ᵏD: Σ (4/5)ᵏ
  1. 1A: Σ (−1)ᵏ/k
  2. 2B: Σ 2ᵏ/k!
  3. 3C: Σ (7/4)ᵏ
  4. 4D: Σ (4/5)ᵏ
4.Sequences & seriesHard
xyfP₄−1.5

Use the Maclaurin polynomial of degree 4 of f(x) = cos x to approximate f(−1.5). Give P₄(−1.5).

f(x) = cos xn = 4x = −1.5
P₄(−1.5) =
5.Multivariable calculusMedium
xy024−20PQ

For f(x, y) = −2xe2y, find the directional derivative Duf at P(0.5, −1) in the direction towards Q(3.5, 1).

f(x, y) = −2xe2yP = (0.5, −1)Q = (3.5, 1)
Duf =
6.Multivariable calculusHard
xyx = 0.8y = x0.8

The integral I = ∫00.8 ∫y0.8 ex² dx dy cannot be done in this order. Reverse the order of integration and evaluate it.

f = ex²0 ≤ y ≤ x ≤ 0.8
I =
7.Differential equationsMedium
tyy(0) = 3y(1.5) = ?

Solve y″ + 4y′ + 4y = 0 with y(0) = 3 and y′(0) = 1. Find y(1.5).

y″ + 4y′ + 4y = 0y(0) = 3y′(0) = 1
y(1.5) =
8.Differential equationsHard
tyy(0) = 5y(1.5) = ?

Solve y″ + 3y′ + 11.25y = 0 with y(0) = 5 and y′(0) = 0 (released from rest). Find y(1.5); angles in radians.

y″ + 3y′ + 11.25y = 0y(0) = 5y′(0) = 0
y(1.5) =

3Solutions

1.Integrals

Answer: V = 25.13

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∫₀2 4x dx

Integrate

V = π·4·2²/2 = 25.13

Answer

V = 25.13
2.Integrals

Answer: V = 1.676

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

2x = 2x² → x = 0, x = 1

Washer method

V = π∫₀1 ((2x)² − (2x²)²) dx

Integrate

V = π(4·1³/3 − 4·1⁵/5) = π·0.5333 = 1.676

Answer

V = 1.676
3.Sequences & series

Answer: C: Σ (7/4)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 2ᵏ/k! → Ratio test: L = lim 2/(k + 1) = 0 < 1, converges.
C: Σ (7/4)ᵏ → Geometric with r = 7/4 > 1: the terms grow, so it diverges.
D: Σ (4/5)ᵏ → Geometric with r = 4/5, |r| < 1: converges.

The answer

C: Σ (7/4)ᵏ

Answer

C: Σ (7/4)ᵏ
4.Sequences & series

Answer: P₄(−1.5) = 0.08594

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

P₄(x) = 1 − 0.5x² + 0.04167x⁴

Substitute

P₄(−1.5) = 1 − 1.125 + 0.210938 = 0.08594

Answer

P₄(−1.5) = 0.08594

The exact value is f(−1.5) = 0.0707372; the error −0.0152003 shrinks when you take more terms or a point closer to 0.

5.Multivariable calculus

Answer: Duf = −0.3754

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.

Gradient at the point

fx = −2·e2y → fx(0.5, −1) = −0.2707
fy = −4xe2y → fy(0.5, −1) = −0.2707

Unit vector

|v| = √(3² + 2²) = 3.606
u = (0.8321, 0.5547)

Dot product

Duf = (−0.2707)·0.8321 + (−0.2707)·0.5547 = −0.3754

Answer

Duf = −0.3754
6.Multivariable calculus

Answer: I = 0.4482

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).

Region

0 ≤ y ≤ x, 0 ≤ x ≤ 0.8

Reversed order

I = ∫00.8 ∫0x ex² dy dx = ∫00.8 x·ex² dx

Substitution u = x²

I = [ex²/2]00.8 = 0.4482

Answer

I = 0.4482
7.Differential equations

Answer: y(1.5) = 0.6721

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) = 3
y′(0) = B − 2A → B = 1 + 6 = 7

Evaluate

y(1.5) = (3 + 7·1.5)·e−3 = 0.6721

Answer

y(1.5) = 0.6721
8.Differential equations

Answer: y(1.5) = −0.3687

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 = 5, B = 1.5·5/3 = 2.5

Evaluate

y(1.5) = e−2.25(5·cos 4.5 + 2.5·sin 4.5) = −0.3687

Answer

y(1.5) = −0.3687

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