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
xy01

Evaluate I = ∫01 2x·(x² + 1)3 dx.

f(x) = 2x·(x² + 1)3[0, 1]
I =
2.IntegralsHard
xy1.5y = x²eˣ

Evaluate I = ∫₀1.5 x²·eˣ dx.

∫₀1.5 x²eˣ dx
I =
3.Sequences & seriesMedium
xx₀ = 0??R

Find the radius of convergence R of the power series Σk=1∞ 9ᵏx²ᵏ/k.

Σ 9ᵏx²ᵏ/kx₀ = 0
R =
4.Sequences & seriesHard

Find the coefficient c₂ of (x − 4)² in the Taylor series of f(x) = ln x about a = 4.

f(x) = ln xa = 4k = 2
c₂ =
5.Multivariable calculusMedium
xy−4−202024vP

For f(x, y) = 2y·e2xy, find the directional derivative Duf at (−1, 2) in the direction of v = (−4, 0).

f(x, y) = 2y·e2xyP = (−1, 2)v = (−4, 0)
Duf =
6.Multivariable calculusHard
xy−202−202(0, 0)(1, 1)(−1, −1)

Find the critical points of f(x, y) = x⁴ + y⁴ − 4xy, classify them, and give the value fmin of f at its local minimum.

f(x, y) = x⁴ + y⁴ − 4xy
fmin =
7.Differential equationsMedium
t (min)T (°C)27Ta​22 °C

A cup of coffee at 83 °C is left in a room at 19 °C. After 27 min its temperature is 22 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). Find the constant k in min⁻¹.

T0 = 83 °CTa = 19 °CT(27) = 22 °C
k = min⁻¹
8.Differential equationsHard
xy12(1, −2)y(2) = ?

Solve the initial value problem y′ + 3y/x = 2x for x > 0, with y(1) = −2. Find y(2).

y′ + 3y/x = 2xy(1) = −2
y(2) =

3Solutions

1.Integrals

Answer: I = 3.75

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.

Substitution

u = x² + 1, du = 2x dx
I = ∫12 u3 du

Integrate and evaluate

I = [u4/4]12 = (16 − 1)/4 = 3.75

Answer

I = 3.75
2.Integrals

Answer: I = 3.602

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

4.482·(1.25) − 1·2 = 3.602

Answer

I = 3.602
3.Sequences & series

Answer: R = 0.3333

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

The series is in x²: it converges for 9·x² < 1, so |x| < 1/3.

Radius

R = 0.3333

Answer

R = 0.3333

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

4.Sequences & series

Answer: c₂ = −0.03125

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.

Derivative f⁽²⁾

f⁽²⁾(x) = −1/x²

Coefficient

c₂ = f⁽²⁾(4)/2! = −1/(4²·2) = −0.03125
c₂ = −0.03125

Answer

c₂ = −0.03125
5.Multivariable calculus

Answer: Duf = −0.2931

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 = 4y²e2xy → fx(−1, 2) = 0.2931
fy = 2e2xy(1 + 2xy) → fy(−1, 2) = −0.1099

Unit vector

|v| = √((−4)² + 0²) = 4
u = (−1, 0)

Dot product

Duf = 0.2931·(−1) + (−0.1099)·0 = −0.2931

Answer

Duf = −0.2931
6.Multivariable calculus

Answer: fmin = −2

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 = 4x³ − 4y = 0
fy = 4y³ − 4x = 0
(0, 0), (1, 1), (−1, −1)

Second-derivative test

(0, 0): fxx = 0, fyy = 0, fxy = −4, D = −16 → saddle point
(1, 1): fxx = 12, fyy = 12, fxy = −4, D = 128 → local minimum
(−1, −1): fxx = 12, fyy = 12, fxy = −4, D = 128 → local minimum

Value at the extremum

f(−1, −1) = −2

Answer

fmin = −2
7.Differential equations

Answer: k = 0.1133 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−27k = 3/64

Solve for k

k = ln(64/3)/27 = 0.1133 min⁻¹

Answer

k = 0.1133 min⁻¹
8.Differential equations

Answer: y(2) = 1.3

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)′ = 2x4 → x3y = 2x5/5 + C

Initial condition

−2 = 2/5 + C → C = −2.4

Evaluate

y(2) = (2·25/5 + (−2.4))/23 = 1.3

Answer

y(2) = 1.3

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