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
xy1y = 2x

The region under y = 2x from x = 0 to x = 1 is rotated about the x-axis. Find the volume V of the solid (a cone).

y = 2x0 ≤ x ≤ 1
V =
2.IntegralsHard
xyy = x2​y = 4x

Find the area A of the region enclosed by y = x2 and y = 4x.

y = x2y = 4x
A =
3.Sequences & seriesMedium
AΣ (2/5)ᵏk = 1 … ∞BΣ 3/k³k = 1 … ∞CΣ k!/3ᵏk = 1 … ∞DΣ (−1)ᵏ/√kk = 1 … ∞

Exactly one of the four series diverges. Click it.

A: Σ (2/5)ᵏB: Σ 3/k³C: Σ k!/3ᵏD: Σ (−1)ᵏ/√k
  1. 1A: Σ (2/5)ᵏ
  2. 2B: Σ 3/k³
  3. 3C: Σ k!/3ᵏ
  4. 4D: Σ (−1)ᵏ/√k
4.Sequences & seriesHard

f(x) = sin x is approximated at x = 0.5 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 0.0001.

f(x) = sin xx = 0.5M = 1ε = 0.0001
n =
5.Multivariable calculusMedium
xy20∇fP

For f(x, y) = 2xe0.5y, find the maximum rate of change of f at the point (2, 0), that is |∇f|.

f(x, y) = 2xe0.5yP = (2, 0)
|∇f| =
6.Multivariable calculusHard
xy1.51

Use polar coordinates to evaluate I = ∬R e−(x² + y²) dA, where R is the ring 1 ≤ x² + y² ≤ 2.25.

f = e−(x² + y²)1 ≤ r ≤ 1.50 ≤ θ ≤ 2π
I =
7.Differential equationsMedium
t (min)T (°C)26Ta​T = ?

A cup of coffee at 86 °C is left in a room at 23 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). With k = 0.12 min⁻¹, find its temperature T after 26 min.

T0 = 86 °CTa = 23 °Ck = 0.12 min⁻¹t = 26 min
T = °C
8.Differential equationsHard
xy12(1, 2)y(2) = ?

Solve the initial value problem y′ + y/x = 5x2 for x > 0, with y(1) = 2. Find y(2).

y′ + y/x = 5x2y(1) = 2
y(2) =

3Solutions

1.Integrals

Answer: V = 4.189

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∫₀1 4x² dx

Integrate

V = π·4·1³/3 = 4.189

Answer

V = 4.189

Check: a cone of radius 2 and height 1 has (1/3)πr²h = 4.189.

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.

Intersection points

x2 = 4x → x = 0, x = 4

Integrate upper minus lower

A = [2x2 − 0.3333x3]04 = 10.67

Answer

A = 10.67
3.Sequences & series

Answer: C: Σ k!/3ᵏ

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 (2/5)ᵏ → Geometric with r = 2/5, |r| < 1: converges.
B: Σ 3/k³ → p-series with p = 3 > 1: converges.
C: Σ k!/3ᵏ → Ratio test: L = lim (k + 1)/3 = ∞, diverges.
D: Σ (−1)ᵏ/√k → Alternating, with 1/√k decreasing to 0: converges (conditionally).

The answer

C: Σ k!/3ᵏ

Answer

C: Σ k!/3ᵏ
4.Sequences & series

Answer: n = 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.

n(n + 1)!bound010.5120.125260.020833240.00260441200.000260457202.17 × 10⁻⁵

Bounds for increasing n

n = 4: 0.0002604 ≥ 0.0001
n = 5: 2.17 × 10⁻⁵ < 0.0001

Smallest degree

n = 5

Answer

n = 5
5.Multivariable calculus

Answer: |∇f| = 2.828

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.

Partial derivatives

fx = 2·e0.5y → fx(2, 0) = 2
fy = xe0.5y → fy(2, 0) = 2

Length of the gradient

|∇f| = √(2² + 2²) = 2.828

Answer

|∇f| = 2.828
6.Multivariable calculus

Answer: I = 0.8246

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 ∫11.5 e−r²·r dr dθ

Integrate

∫ e−r²·r dr = [−e−r²/2]
I = 0.8246

Answer

I = 0.8246
7.Differential equations

Answer: T = 25.78 °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) = 23 + 63·e−0.12t

Substitute t

T(26) = 23 + 63·0.04416 = 25.78 °C

Answer

T = 25.78 °C
8.Differential equations

Answer: y(2) = 10.38

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∫1/x dx = x1

Integrate

(x1y)′ = 5x3 → x1y = 5x4/4 + C

Initial condition

2 = 5/4 + C → C = 0.75

Evaluate

y(2) = (5·24/4 + 0.75)/21 = 10.38

Answer

y(2) = 10.38

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