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
xy02

Evaluate I = ∫02 e−x dx.

f(x) = e−x[0, 2]
I =
2.IntegralsHard
xy−11

Find the area A of the region between the parabola y = 1 − x² and the x-axis.

y = 1 − x²
A =
3.Sequences & seriesMedium
AΣ (−1)ᵏ/kk = 1 … ∞BΣ 1/(3ᵏ + k)k = 1 … ∞CΣ (2/3)ᵏk = 1 … ∞DΣ k/(k + 7)k = 1 … ∞

Exactly one of the four series diverges. Click it.

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

Find the coefficient c₂ of (x − 1)² in the Taylor series of f(x) = ∛x about a = 1.

f(x) = ∛xa = 1k = 2
c₂ =
5.Multivariable calculusMedium
xy11∇fP

For f(x, y) = 3x³y³ − 5y³ − xy, find the direction of steepest ascent at (1, 1) as the angle θ from the positive x-axis, in degrees between −180° and 180°.

f(x, y) = 3x³y³ − 5y³ − xyP = (1, 1)
θ = °
6.Multivariable calculusHard
xy−20−2−101(0, 0)(−1, −1)

f(x, y) = x³ + y³ + 3xy has a critical point at (0, 0). Compute the discriminant D = fxxfyy − fxy² there.

f(x, y) = x³ + y³ + 3xyP = (0, 0)
D =
7.Differential equationsMedium
tyy(0) = 1y(1.5) = ?

Solve y″ + 6y′ + 8y = 0 with y(0) = 1 and y′(0) = 2. Find y(1.5).

y″ + 6y′ + 8y = 0y(0) = 1y′(0) = 2
y(1.5) =
8.Differential equationsHard
tyK = 5y(1.5) = ?

A system obeys y″ + 2y′ + 26y = 130 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find y(1.5); angles in radians.

y″ + 2y′ + 26y = 130y(0) = 0y′(0) = 0
y(1.5) =

3Solutions

1.Integrals

Answer: I = 0.8647

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.

Antiderivative

F(x) = e−x/(−1)

Evaluate at the limits

I = (e−2 − 1)/(−1) = (0.1353 − 1)/(−1) = 0.8647

Answer

I = 0.8647
2.Integrals

Answer: A = 1.333

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

1 − x² = 0 → x = ±1

Integrate

A = [1x − x³/3]−11 = 0.6667 − (−0.6667) = 1.333

Answer

A = 1.333
3.Sequences & series

Answer: D: Σ k/(k + 7)

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)ᵏ/k → Alternating, with 1/k decreasing to 0: converges (conditionally).
B: Σ 1/(3ᵏ + k) → Smaller than (1/3)ᵏ, a convergent geometric series: converges by comparison.
C: Σ (2/3)ᵏ → Geometric with r = 2/3, |r| < 1: converges.
D: Σ k/(k + 7) → The terms tend to 1, not 0: diverges by the n-th term test.

The answer

D: Σ k/(k + 7)

Answer

D: Σ k/(k + 7)
4.Sequences & series

Answer: c₂ = −0.1111

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) = −0.2222·x1/3 − 2

Coefficient

c₂ = −0.2222·11/3 − 2/2! = −0.1111
c₂ = −0.1111

Answer

c₂ = −0.1111
5.Multivariable calculus

Answer: θ = −41.19 °

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 = 9x²y³ − y → fx(1, 1) = 8
fy = 9x³y² − 15y² − x → fy(1, 1) = −7

Direction

θ = atan2(−7, 8) = −41.19°

Answer

θ = −41.19°
6.Multivariable calculus

Answer: D = −9

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 = 3

Discriminant

D = 0·0 − 3² = −9
→ saddle point

Answer

D = −9
7.Differential equations

Answer: y(1.5) = 0.1444

Second-order linear equations: for y″ + b·y′ + c·y = 0 try y = ert: this gives the characteristic equation r² + b·r + c = 0. Two real roots r₁ ≠ r₂ give y = A·er₁t + B·er₂t (overdamped); a double root r gives y = (A + Bt)·ert (critically damped); complex roots −α ± jω give y = e−αt(A cos ωt + B sin ωt) (underdamped, oscillating). The initial values y(0) and y′(0) fix A and B. Written as y″ + 2ζωny′ + ωn²y = 0, ωn is the natural frequency and ζ the damping ratio: ζ < 1 oscillates. With a constant input, y″ + by′ + cy = c·K, the solution is the steady value K plus the free response.

Characteristic roots

(r + 2)(r + 4) = 0 → r = −2, −4

Constants from the initial values

A + B = 1, −2A − 4B = 2
A = 3, B = −2

Evaluate

y(1.5) = 3·e−3 + (−2)·e−6 = 0.1444

Answer

y(1.5) = 0.1444
8.Differential equations

Answer: y(1.5) = 4.404

Second-order linear equations: for y″ + b·y′ + c·y = 0 try y = ert: this gives the characteristic equation r² + b·r + c = 0. Two real roots r₁ ≠ r₂ give y = A·er₁t + B·er₂t (overdamped); a double root r gives y = (A + Bt)·ert (critically damped); complex roots −α ± jω give y = e−αt(A cos ωt + B sin ωt) (underdamped, oscillating). The initial values y(0) and y′(0) fix A and B. Written as y″ + 2ζωny′ + ωn²y = 0, ωn is the natural frequency and ζ the damping ratio: ζ < 1 oscillates. With a constant input, y″ + by′ + cy = c·K, the solution is the steady value K plus the free response.

Steady value and roots

yss = 5, r = −1 ± 5j

Constants from the initial values

y(0) = 5 + A = 0 → A = −5
y′(0) = −1A + 5B = 0 → B = −1

Evaluate

y(1.5) = 5·(1 − e−1.5(cos 7.5 + 0.2·sin 7.5)) = 4.404

Answer

y(1.5) = 4.404

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