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
xy15

Evaluate I = ∫15 2/x dx.

f(x) = 2/x[1, 5]
I =
2.IntegralsHard
xy1y = x²eˣ

Evaluate I = ∫₀1 x²·eˣ dx.

∫₀1 x²eˣ dx
I =
3.Sequences & seriesMedium
xx₀ = −1??R

Find the radius of convergence R of the power series Σk=1∞ (2k)!·(x + 1)ᵏ/(k!)².

Σ (2k)!·(x + 1)ᵏ/(k!)²x₀ = −1
R =
4.Sequences & seriesHard
xyfP₂−1

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

f(x) = cos(−x)n = 2x = −1
P₂(−1) =
5.Multivariable calculusMedium
xy0246−20PQ

For f(x, y) = 2xy − 4x² − 4y², find the directional derivative Duf at P(2, −1) in the direction towards Q(6, 1).

f(x, y) = 2xy − 4x² − 4y²P = (2, −1)Q = (6, 1)
Duf =
6.Multivariable calculusHard
xy1.5

Use polar coordinates to evaluate I = ∬R (x² + y²) dA, where R is the disc x² + y² ≤ 2.25.

f = x² + y²0 ≤ r ≤ 1.50 ≤ θ ≤ 2π
I =
7.Differential equationsMedium
tyy(0) = 3y(0.5) = ?

Solve y″ + 6y′ + 8y = 0 with y(0) = 3 and y′(0) = -4. Find y(0.5).

y″ + 6y′ + 8y = 0y(0) = 3y′(0) = -4
y(0.5) =
8.Differential equationsHard
tyy(0) = 3y(1.5) = ?

Solve y″ + 3y′ + 18.25y = 0 with y(0) = 3 and y′(0) = 0 (released from rest). Find y(1.5); angles in radians.

y″ + 3y′ + 18.25y = 0y(0) = 3y′(0) = 0
y(1.5) =

3Solutions

1.Integrals

Answer: I = 3.219

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) = 2·ln x

Evaluate at the limits

I = 2·(ln 5 − ln 1) = 2·ln 5 = 3.219

Answer

I = 3.219
2.Integrals

Answer: I = 0.7183

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

2.718·(1) − 1·2 = 0.7183

Answer

I = 0.7183
3.Sequences & series

Answer: R = 0.25

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.

Coefficients

ck+1/ck = (2k + 2)(2k + 1)/(k + 1)² → 4, so R = 1/4

Radius

R = 0.25

Answer

R = 0.25

The ratio test says nothing at the two end points; they have to be checked separately.

4.Sequences & series

Answer: P₂(−1) = 0.5

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.

Maclaurin polynomial

P₂(x) = 1 − 0.5x²

Substitute

P₂(−1) = 1 − 0.5 = 0.5

Answer

P₂(−1) = 0.5

The exact value is f(−1) = 0.540302; the error 0.0403023 shrinks when you take more terms or a point closer to 0.

5.Multivariable calculus

Answer: Duf = −10.73

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 = 2y − 8x → fx(2, −1) = −18
fy = 2x − 8y → fy(2, −1) = 12

Unit vector

|v| = √(4² + 2²) = 4.472
u = (0.8944, 0.4472)

Dot product

Duf = (−18)·0.8944 + 12·0.4472 = −10.73

Answer

Duf = −10.73
6.Multivariable calculus

Answer: I = 7.952

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).

Polar form

I = ∫02π ∫01.5 r²·r dr dθ

Integrate

∫ r²·r dr = [r⁴/4]
I = 7.952

Answer

I = 7.952
7.Differential equations

Answer: y(0.5) = 1.336

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 + 2)(r + 4) = 0 → r = −2, −4

Constants from the initial values

A + B = 3, −2A − 4B = -4
A = 4, B = −1

Evaluate

y(0.5) = 4·e−1 + (−1)·e−2 = 1.336

Answer

y(0.5) = 1.336
8.Differential equations

Answer: y(1.5) = 0.2705

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 ± 4j

Constants

A = 3, B = 1.5·3/4 = 1.125

Evaluate

y(1.5) = e−2.25(3·cos 6 + 1.125·sin 6) = 0.2705

Answer

y(1.5) = 0.2705

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