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·cos x

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

∫₀π/3 x·cos x dx
I =
2.IntegralsHard
xy2y = 2xy = x²

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

y = 2xy = x²
V =
3.Sequences & seriesMedium
k|aₖ₊₁/aₖ|123456789101112

Apply the ratio test to Σk=1∞ kᵏ/(3ᵏ·k!). Find L = lim |ak+1/ak|.

ak = kᵏ/(3ᵏ·k!)
L =
4.Sequences & seriesHard

f(x) = cos x is approximated at x = 1.5 by its Maclaurin polynomial of degree n. With the Lagrange bound M·|x|ⁿ⁺¹/(n + 1)! and M = 1, find the smallest degree n that guarantees an error below 0.001.

f(x) = cos xx = 1.5M = 1ε = 0.001
n =
5.Multivariable calculusMedium
xy−2024024vP

For f(x, y) = 5x³y³ − 2x² + y³ + xy, find the directional derivative Duf at (1, 2) in the direction of v = (4, −1).

f(x, y) = 5x³y³ − 2x² + y³ + xyP = (1, 2)v = (4, −1)
Duf =
6.Multivariable calculusHard
xyx = 1.2y = x1.2

The integral I = ∫01.2 ∫y1.2 sin(x²) dx dy cannot be done in this order. Reverse the order of integration and evaluate it.

f = sin(x²)0 ≤ y ≤ x ≤ 1.2
I =
7.Differential equationsMedium
tyy(0) = 1y(2) = ?

Solve y″ + 6y′ + 5y = 0 with y(0) = 1 and y′(0) = 4. Find y(2).

y″ + 6y′ + 5y = 0y(0) = 1y′(0) = 4
y(2) =
8.Differential equationsHard
tyK = 2y(1) = ?

A system obeys y″ + 3y′ + 18.25y = 36.5 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find y(1); angles in radians.

y″ + 3y′ + 18.25y = 36.5y(0) = 0y′(0) = 0
y(1) =

3Solutions

1.Integrals

Answer: I = 0.4069

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 = cos x dx, v = sin x

Antiderivative

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

Limits

(π/3·sin π/3 + cos π/3) − (0 + 1) = 0.4069

Answer

I = 0.4069
2.Integrals

Answer: V = 13.4

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 = 1x² → x = 0, x = 2

Washer method

V = π∫₀2 ((2x)² − (1x²)²) dx

Integrate

V = π(4·2³/3 − 1·2⁵/5) = π·4.267 = 13.4

Answer

V = 13.4
3.Sequences & series

Answer: L = 0.9061

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.

Ratio of consecutive terms

|ak+1/ak| = (1 + 1/k)ᵏ/3 → e/3

Conclusion

L = 0.9061 < 1

Answer

L = 0.9061

L < 1, so the series converges.

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)!bound011.5121.125260.56253240.210941200.0632857200.01582650400.003397403200.0006356

Bounds for increasing n

n = 6: 0.00339 ≥ 0.001
n = 7: 0.0006356 < 0.001

Smallest degree

n = 7

Answer

n = 7
5.Multivariable calculus

Answer: Duf = 96.77

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 = 15x²y³ − 4x + y → fx(1, 2) = 118
fy = 15x³y² + 3y² + x → fy(1, 2) = 73

Unit vector

|v| = √(4² + (−1)²) = 4.123
u = (0.9701, −0.2425)

Dot product

Duf = 118·0.9701 + 73·(−0.2425) = 96.77

Answer

Duf = 96.77
6.Multivariable calculus

Answer: I = 0.4348

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 ≤ 1.2

Reversed order

I = ∫01.2 ∫0x sin(x²) dy dx = ∫01.2 x·sin(x²) dx

Substitution u = x²

I = [−cos(x²)/2]01.2 = 0.4348

Answer

I = 0.4348
7.Differential equations

Answer: y(2) = 0.3044

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.

Characteristic roots

(r + 1)(r + 5) = 0 → r = −1, −5

Constants from the initial values

A + B = 1, −1A − 5B = 4
A = 2.25, B = −1.25

Evaluate

y(2) = 2.25·e−2 + (−1.25)·e−10 = 0.3044

Answer

y(2) = 0.3044
8.Differential equations

Answer: y(1) = 2.418

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 ± 4j

Constants from the initial values

y(0) = 2 + A = 0 → A = −2
y′(0) = −1.5A + 4B = 0 → B = −0.75

Evaluate

y(1) = 2·(1 − e−1.5(cos 4 + 0.375·sin 4)) = 2.418

Answer

y(1) = 2.418

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