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
xy03

Evaluate I = ∫03 e−0.5x dx.

f(x) = e−0.5x[0, 3]
I =
2.IntegralsHard
xy−33

Find the area A of the region between the parabola y = 9 − x² and the x-axis.

y = 9 − x²
A =
3.Sequences & seriesMedium
xx₀ = −1??R

The power series Σk=1∞ k²·5ᵏ·(x + 1)ᵏ converges on an interval around x₀ = −1. Find the left end x of that interval.

Σ k²·5ᵏ·(x + 1)ᵏx₀ = −1
x =
4.Sequences & seriesHard

f(x) = ln(1 + x) is approximated by its Maclaurin polynomial of degree 3 at x = 0.5. Use the Lagrange error bound M·|x|ⁿ⁺¹/(n + 1)! with M = 3! = 6 to find the largest possible error |R₃|.

f(x) = ln(1 + x)n = 3x = 0.5M = 6
|R₃| =
5.Multivariable calculusMedium

z = −4x³y + 2x² − xy with x = −2t − 2 and y = −t² + 1. Use the chain rule to find dz/dt at t = 1.

z = −4x³y + 2x² − xyx = −2t − 2y = −t² + 1t = 1
dz/dt =
6.Multivariable calculusHard
xy05−2024(0, 0)(3, 3)

f(x, y) = x³ + y³ − 9xy has a critical point at (3, 3). Compute the discriminant D = fxxfyy − fxy² there.

f(x, y) = x³ + y³ − 9xyP = (3, 3)
D =
7.Differential equationsMedium
tyy(0) = 4y(0.5) = ?

Solve y″ + 5y′ + 6y = 0 with y(0) = 4 and y′(0) = 6. Find y(0.5).

y″ + 5y′ + 6y = 0y(0) = 4y′(0) = 6
y(0.5) =
8.Differential equationsHard
tyK = 10y(1) = ?

A system obeys y″ + 3y′ + 18.25y = 182.5 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find y(1); angles in radians.

y″ + 3y′ + 18.25y = 182.5y(0) = 0y′(0) = 0
y(1) =

3Solutions

1.Integrals

Answer: I = 1.554

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−0.5x/(−0.5)

Evaluate at the limits

I = (e−1.5 − 1)/(−0.5) = (0.2231 − 1)/(−0.5) = 1.554

Answer

I = 1.554
2.Integrals

Answer: A = 36

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

9 − x² = 0 → x = ±3

Integrate

A = [9x − x³/3]−33 = 18 − (−18) = 36

Answer

A = 36
3.Sequences & series

Answer: x = −1.2

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.

Coefficients

ck/ck+1 = (1/5)·(k/(k + 1))² → 1/5

Radius

R = 0.2

Interval

−1.2 < x < −0.8
x = −1.2

Answer

x = −1.2

The ratio test says nothing at the two end points; they have to be checked separately.

4.Sequences & series

Answer: |R₃| = 0.01563

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.

Bound on the derivative

M = 3! = 6

Lagrange bound

|R₃| ≤ 6·0.5⁴/4! = 0.01563

Answer

|R₃| = 0.01563

The true error is smaller: |f(0.5) − P₃(0.5)| = 0.0112.

5.Multivariable calculus

Answer: dz/dt = −488

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.

Point and rates

x(1) = −4, y(1) = 0
dx/dt = −2, dy/dt = −2t = −2

Partial derivatives

fx = −12x²y + 4x − y = −16
fy = −4x³ − x = 260

Chain rule

dz/dt = (−16)·(−2) + 260·(−2) = −488

Answer

dz/dt = −488
6.Multivariable calculus

Answer: D = 243

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 = 18, fyy = 18, fxy = −9

Discriminant

D = 18·18 − (−9)² = 243
→ local minimum

Answer

D = 243
7.Differential equations

Answer: y(0.5) = 3.498

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 + 3) = 0 → r = −2, −3

Constants from the initial values

A + B = 4, −2A − 3B = 6
A = 18, B = −14

Evaluate

y(0.5) = 18·e−1 + (−14)·e−1.5 = 3.498

Answer

y(0.5) = 3.498
8.Differential equations

Answer: y(1) = 12.09

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 = 10, r = −1.5 ± 4j

Constants from the initial values

y(0) = 10 + A = 0 → A = −10
y′(0) = −1.5A + 4B = 0 → B = −3.75

Evaluate

y(1) = 10·(1 − e−1.5(cos 4 + 0.375·sin 4)) = 12.09

Answer

y(1) = 12.09

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