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
xyπ/2y = x·sin x

Evaluate I = ∫₀π/2 x·sin x dx (x in radians).

∫₀π/2 x·sin x dx
I =
2.IntegralsHard
xyy = 2x2​y = x

Find the area A of the region enclosed by y = 2x2 and y = x.

y = 2x2y = x
A =
3.Sequences & seriesMedium
kSn​12345678910

Find the sum S = Σk=121 3/(k(k + 1)).

ak = 3/(k(k + 1))n = 21
S =
4.Sequences & seriesHard

f(x) = cos x is approximated at x = 2 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 0.001.

f(x) = cos xx = 2M = 1ε = 0.001
n =
5.Multivariable calculusMedium
xy33∇fP

For f(x, y) = 2xy² + 4x² − 4y³, find the maximum rate of change of f at the point (3, 3), that is |∇f|.

f(x, y) = 2xy² + 4x² − 4y³P = (3, 3)
|∇f| =
6.Multivariable calculusHard
xy−202−202(0, 0)(2, 2)(−2, −2)

f(x, y) = x⁴ + y⁴ − 16xy has a critical point at (0, 0). Compute the discriminant D = fxxfyy − fxy² there.

f(x, y) = x⁴ + y⁴ − 16xyP = (0, 0)
D =
7.Differential equationsMedium
t (min)T (°C)26Ta​121 °C58 °C

A forged steel part at 380 °C is left to cool in a workshop at 24 °C. After 26 min its temperature is 121 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). How long after the start, t, does it take to reach 58 °C?

T0 = 380 °CTa = 24 °CT(26) = 121 °CT = 58 °C
t = min
8.Differential equationsHard
t (min)m (kg)t = ?5 kg

A tank holds 100 L of pure water. Brine with 0.1 kg of salt per litre flows in at 20 L/min, and the well-mixed solution flows out at the same rate. After how many minutes t does the tank contain 5 kg of salt?

V = 100 Lm0 = 0 kgc = 0.1 kg/Lr = 20 L/min
t = min

3Solutions

1.Integrals

Answer: I = 1

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 = x, du = dx; dv = sin x dx, v = −cos x

Antiderivative

∫x sin x dx = −x cos x + ∫cos x dx = −x cos x + sin x

Limits

(−π/2·cos π/2 + sin π/2) − (0 + 0) = 1

Answer

I = 1
2.Integrals

Answer: A = 0.04167

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

2x2 = x → x = 0, x = 0.5

Integrate upper minus lower

A = [0.5x2 − 0.6667x3]00.5 = 0.04167

Answer

A = 0.04167
3.Sequences & series

Answer: S = 2.864

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.

Partial fractions

3/(k(k + 1)) = 3·(1/k − 1/(k + 1))

Cancel

S = 3·(1 − 1/22) = 2.864

Answer

S = 2.864
4.Sequences & series

Answer: n = 9

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)!bound012122261.3333240.666741200.266757200.08889650400.02547403200.00634983628800.001411936288000.0002822

Bounds for increasing n

n = 8: 0.001411 ≥ 0.001
n = 9: 0.0002822 < 0.001

Smallest degree

n = 9

Answer

n = 9
5.Multivariable calculus

Answer: |∇f| = 83.35

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 = 2y² + 8x → fx(3, 3) = 42
fy = 4xy − 12y² → fy(3, 3) = −72

Length of the gradient

|∇f| = √(42² + (−72)²) = 83.35

Answer

|∇f| = 83.35
6.Multivariable calculus

Answer: D = −256

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 = 0, fyy = 0, fxy = −16

Discriminant

D = 0·0 − (−16)² = −256
→ saddle point

Answer

D = −256
7.Differential equations

Answer: t = 46.96 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(356/97)/26 = 0.05001 min⁻¹

Time to reach the target

t = ln(356/34)/0.05001 = 46.96 min

Answer

t = 46.96 min
8.Differential equations

Answer: t = 3.466 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.

Result

m(t) = 10 + (−10)·e−t/5

Solve for t

e−t/5 = (10 − 5)/(10 − 0) = 0.5
t = 5·ln(2) = 3.466 min

Answer

t = 3.466 min

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