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
xy16y = x ln x

Evaluate I = ∫₁6 x·ln x dx.

∫₁6 x ln x dx
I =
2.IntegralsHard
xy1.667y = 5xy = 3x²

The region between the line y = 5x and the parabola y = 3x² is rotated about the x-axis. Find the volume V of the solid.

y = 5xy = 3x²
V =
3.Sequences & seriesMedium
xx₀ = 2??R

Find the radius of convergence R of the power series Σk=1∞ (x − 2)ᵏ/(k·4ᵏ).

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

Recognise the series and find its sum: S = Σk=0∞ 2ᵏ/k!.

ak = 2ᵏ/k!
S =
5.Multivariable calculusMedium

z = y·exy with x = −2t + 2 and y = t² − 2. Use the chain rule to find dz/dt at t = 2.

z = y·exyx = −2t + 2y = t² − 2t = 2
dz/dt =
6.Multivariable calculusHard
xyx = 4y = 2x48

Evaluate I = ∬T (4x + y) dA over the triangle T bounded by y = 0, x = 4 and y = 2x.

f = 4x + y0 ≤ x ≤ 40 ≤ y ≤ 2x
I =
7.Differential equationsMedium
t (min)T (°C)20Ta​16 °C

A can of soda at 6 °C is taken out of the fridge into a room at 24 °C. After 20 min its temperature is 16 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). Find the constant k in min⁻¹.

T0 = 6 °CTa = 24 °CT(20) = 16 °C
k = min⁻¹
8.Differential equationsHard
xy13(1, −3)y(3) = ?

Solve the initial value problem y′ + 3y/x = x for x > 0, with y(1) = −3. Find y(3).

y′ + 3y/x = xy(1) = −3
y(3) =

3Solutions

1.Integrals

Answer: I = 23.5

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 = ln x, du = dx/x; dv = x dx, v = x²/2

Antiderivative

∫x ln x dx = (x²/2)ln x − ∫(x/2) dx = (x²/2)ln x − x²/4

Limits

(18·ln 6 − 9) − (0 − 0.25) = 23.5

Answer

I = 23.5
2.Integrals

Answer: V = 48.48

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

5x = 3x² → x = 0, x = 1.667

Washer method

V = π∫₀1.667 ((5x)² − (3x²)²) dx

Integrate

V = π(25·1.667³/3 − 9·1.667⁵/5) = π·15.43 = 48.48

Answer

V = 48.48
3.Sequences & series

Answer: R = 4

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/ck+1 = 4·((k + 1)/k) → 4

Radius

R = 4

Answer

R = 4

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

4.Sequences & series

Answer: S = 7.389

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.

Standard series

Σ xᵏ/k! = eˣ

Substitute

x = 2
S = e2 = 7.389

Answer

S = 7.389
5.Multivariable calculus

Answer: dz/dt = −0.3663

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(2) = −2, y(2) = 2
dx/dt = −2, dy/dt = 2t = 4

Partial derivatives

fx = y²exy = 0.07326
fy = exy(1 + xy) = −0.05495

Chain rule

dz/dt = 0.07326·(−2) + (−0.05495)·4 = −0.3663

Answer

dz/dt = −0.3663
6.Multivariable calculus

Answer: I = 213.3

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).

Limits

0 ≤ x ≤ 4, 0 ≤ y ≤ 2x

Inner integral

∫02x (4x + y) dy = 4x·2x + (2x)²/2 = 10x²

Outer integral

I = 10·4³/3 = 213.3

Answer

I = 213.3
7.Differential equations

Answer: k = 0.04055 min⁻¹

First-order differential equations: dy/dt = k·y has the solution y = y₀·ekt (growth for k > 0, decay for k < 0); a half-life T gives k = ln 2/T. Newton's law of cooling, dT/dt = −k·(T − Tₐ), gives T(t) = Tₐ + (T₀ − Tₐ)·e−kt. A linear equation y′ + p(x)·y = q(x) is solved with the integrating factor μ = e∫p dx: (μy)′ = μq, so μy = ∫μq dx + C.

Ratio of temperature differences

e−20k = -8/-18

Solve for k

k = ln(-18/-8)/20 = 0.04055 min⁻¹

Answer

k = 0.04055 min⁻¹
8.Differential equations

Answer: y(3) = 1.681

First-order differential equations: dy/dt = k·y has the solution y = y₀·ekt (growth for k > 0, decay for k < 0); a half-life T gives k = ln 2/T. Newton's law of cooling, dT/dt = −k·(T − Tₐ), gives T(t) = Tₐ + (T₀ − Tₐ)·e−kt. A linear equation y′ + p(x)·y = q(x) is solved with the integrating factor μ = e∫p dx: (μy)′ = μq, so μy = ∫μq dx + C.

Integrating factor

μ = e∫3/x dx = x3

Integrate

(x3y)′ = x4 → x3y = 1x5/5 + C

Initial condition

−3 = 1/5 + C → C = −3.2

Evaluate

y(3) = (1·35/5 + (−3.2))/33 = 1.681

Answer

y(3) = 1.681

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