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
xy3y = 3x

The region under y = 3x from x = 0 to x = 3 is rotated about the x-axis. Find the volume V of the solid (a cone).

y = 3x0 ≤ x ≤ 3
V =
2.IntegralsHard
xy1y = x²eˣ

Evaluate I = ∫₀1 x²·eˣ dx.

∫₀1 x²eˣ dx
I =
3.Sequences & seriesMedium
AΣ 1/(2ᵏ + k)k = 1 … ∞BΣ (7/4)ᵏk = 1 … ∞CΣ (2/3)ᵏk = 1 … ∞DΣ 3ᵏ/k!k = 1 … ∞

Exactly one of the four series diverges. Click it.

A: Σ 1/(2ᵏ + k)B: Σ (7/4)ᵏC: Σ (2/3)ᵏD: Σ 3ᵏ/k!
  1. 1A: Σ 1/(2ᵏ + k)
  2. 2B: Σ (7/4)ᵏ
  3. 3C: Σ (2/3)ᵏ
  4. 4D: Σ 3ᵏ/k!
4.Sequences & seriesHard

Recognise the series and find its sum: S = Σk=1∞ k·0.2ᵏ.

ak = k·0.2ᵏ
S =
5.Multivariable calculusMedium
xy1−1∇fP

For f(x, y) = −2xe0.5y, find the maximum rate of change of f at the point (1, −1), that is |∇f|.

f(x, y) = −2xe0.5yP = (1, −1)
|∇f| =
6.Multivariable calculusHard
xy−202−202(0, 0)(1, 1)(−1, −1)

f(x, y) = x⁴ + y⁴ − 4xy has a critical point at (−1, −1). Compute the discriminant D = fxxfyy − fxy² there.

f(x, y) = x⁴ + y⁴ − 4xyP = (−1, −1)
D =
7.Differential equationsMedium
t (min)T (°C)24Ta​210 °C355 °C

A forged steel part at 500 °C is left to cool in a workshop at 30 °C. After 24 min its temperature is 210 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). How long after the start, t, does it take to reach 355 °C?

T0 = 500 °CTa = 30 °CT(24) = 210 °CT = 355 °C
t = min
8.Differential equationsHard
xy12(1, −2)y(2) = ?

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

y′ + y/x = xy(1) = −2
y(2) =

3Solutions

1.Integrals

Answer: V = 254.5

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∫₀3 9x² dx

Integrate

V = π·9·3³/3 = 254.5

Answer

V = 254.5

Check: a cone of radius 9 and height 3 has (1/3)πr²h = 254.5.

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: B: Σ (7/4)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 1/(2ᵏ + k) → Smaller than (1/2)ᵏ, a convergent geometric series: converges by comparison.
B: Σ (7/4)ᵏ → Geometric with r = 7/4 > 1: the terms grow, so it diverges.
C: Σ (2/3)ᵏ → Geometric with r = 2/3, |r| < 1: converges.
D: Σ 3ᵏ/k! → Ratio test: L = lim 3/(k + 1) = 0 < 1, converges.

The answer

B: Σ (7/4)ᵏ

Answer

B: Σ (7/4)ᵏ
4.Sequences & series

Answer: S = 0.3125

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 k·xᵏ = x/(1 − x)²

Substitute

x = 0.2
S = 0.2/(1 − 0.2)² = 0.3125

Answer

S = 0.3125
5.Multivariable calculus

Answer: |∇f| = 1.356

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(1, −1) = −1.213
fy = −xe0.5y → fy(1, −1) = −0.6065

Length of the gradient

|∇f| = √((−1.213)² + (−0.6065)²) = 1.356

Answer

|∇f| = 1.356
6.Multivariable calculus

Answer: D = 128

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).

Second derivatives

fxx = 12, fyy = 12, fxy = −4

Discriminant

D = 12·12 − (−4)² = 128
→ local minimum

Answer

D = 128
7.Differential equations

Answer: t = 9.225 min

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.

Cooling constant

k = ln(470/180)/24 = 0.03999 min⁻¹

Time to reach the target

t = ln(470/325)/0.03999 = 9.225 min

Answer

t = 9.225 min
8.Differential equations

Answer: y(2) = 0.1667

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)′ = x2 → x1y = 1x3/3 + C

Initial condition

−2 = 1/3 + C → C = −2.333

Evaluate

y(2) = (1·23/3 + (−2.333))/21 = 0.1667

Answer

y(2) = 0.1667

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