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
xy4y = x

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

y = x0 ≤ x ≤ 4
V =
2.IntegralsHard
xy−22

Find the area A of the region between the parabola y = 4 − x² and the x-axis.

y = 4 − x²
A =
3.Sequences & seriesMedium
AΣ (k + 3)/k²k = 1 … ∞BΣ k!/10ᵏk = 1 … ∞CΣ 1/(3ᵏ + k)k = 1 … ∞DΣ (3/2)ᵏk = 1 … ∞

Exactly one of the four series converges. Click it.

A: Σ (k + 3)/k²B: Σ k!/10ᵏC: Σ 1/(3ᵏ + k)D: Σ (3/2)ᵏ
  1. 1A: Σ (k + 3)/k²
  2. 2B: Σ k!/10ᵏ
  3. 3C: Σ 1/(3ᵏ + k)
  4. 4D: Σ (3/2)ᵏ
4.Sequences & seriesHard

Recognise the series and find its sum: S = Σk=0∞ (−1)ᵏ/k!.

ak = (−1)ᵏ/k!
S =
5.Multivariable calculusMedium
xy−10123−2−10vP

For f(x, y) = xy + 4x² + y³, find the directional derivative Duf at (1, −1) in the direction of v = (0, 1).

f(x, y) = xy + 4x² + y³P = (1, −1)v = (0, 1)
Duf =
6.Multivariable calculusHard
xyy = 1.5y = x1.51.5

The integral I = ∫01.5 ∫x1.5 1/(1 + y²)² dy dx cannot be done in this order. Reverse the order of integration and evaluate it.

f = 1/(1 + y²)²0 ≤ x ≤ y ≤ 1.5
I =
7.Differential equationsMedium
t (min)T (°C)29Ta​T = ?

A can of soda at 3 °C is taken out of the fridge into a room at 28 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). With k = 0.09 min⁻¹, find its temperature T after 29 min.

T0 = 3 °CTa = 28 °Ck = 0.09 min⁻¹t = 29 min
T = °C
8.Differential equationsHard
ty1y(1) = ?

Solve y′ + y = 3e−0.5t with y(0) = 2. Find y(1).

y′ + y = 3e−0.5ty(0) = 2
y(1) =

3Solutions

1.Integrals

Answer: V = 67.02

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∫₀4 1x² dx

Integrate

V = π·1·4³/3 = 67.02

Answer

V = 67.02

Check: a cone of radius 4 and height 4 has (1/3)πr²h = 67.02.

2.Integrals

Answer: A = 10.67

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.

Zeros

4 − x² = 0 → x = ±2

Integrate

A = [4x − x³/3]−22 = 5.333 − (−5.333) = 10.67

Answer

A = 10.67
3.Sequences & series

Answer: C: Σ 1/(3ᵏ + 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 (k + 3)/k² → Behaves like 1/k for large k: diverges by comparison with the harmonic series.
B: Σ k!/10ᵏ → Ratio test: L = lim (k + 1)/10 = ∞, diverges.
C: Σ 1/(3ᵏ + k) → Smaller than (1/3)ᵏ, a convergent geometric series: converges by comparison.
D: Σ (3/2)ᵏ → Geometric with r = 3/2 > 1: the terms grow, so it diverges.

The answer

C: Σ 1/(3ᵏ + k)

Answer

C: Σ 1/(3ᵏ + k)
4.Sequences & series

Answer: S = 0.3679

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 = −1
S = e−1 = 0.3679

Answer

S = 0.3679
5.Multivariable calculus

Answer: Duf = 4

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 = y + 8x → fx(1, −1) = 7
fy = x + 3y² → fy(1, −1) = 4

Unit vector

|v| = √(0² + 1²) = 1
u = (0, 1)

Dot product

Duf = 7·0 + 4·1 = 4

Answer

Duf = 4
6.Multivariable calculus

Answer: I = 0.3462

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 ≤ x ≤ y, 0 ≤ y ≤ 1.5

Reversed order

I = ∫01.5 ∫0y 1/(1 + y²)² dx dy = ∫01.5 y·1/(1 + y²)² dy

Substitution u = y²

I = [−1/(2(1 + y²))]01.5 = 0.3462

Answer

I = 0.3462
7.Differential equations

Answer: T = 26.16 °C

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.

Solution of the cooling law

T(t) = 28 + (−25)·e−0.09t

Substitute t

T(29) = 28 + (−25)·0.07353 = 26.16 °C

Answer

T = 26.16 °C
8.Differential equations

Answer: y(1) = 2.168

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 = et
(ety)′ = 3e0.5t

General solution

y = 3/0.5·e−0.5t + C·e−t = 6·e−0.5t + C·e−t

Initial condition

2 = 6 + C → C = −4

Evaluate

y(1) = 6·0.6065 + (−4)·0.3679 = 2.168

Answer

y(1) = 2.168

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