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
xy12

Evaluate I = ∫12 5/x dx.

f(x) = 5/x[1, 2]
I =
2.IntegralsHard
xy3y = x²eˣ

Evaluate I = ∫₀3 x²·eˣ dx.

∫₀3 x²eˣ dx
I =
3.Sequences & seriesMedium
xx₀ = −1??R

Find the radius of convergence R of the power series Σk=1∞ (2k)!·(x + 1)ᵏ/(k!)².

Σ (2k)!·(x + 1)ᵏ/(k!)²x₀ = −1
R =
4.Sequences & seriesHard
xyfP₂1

Use the Maclaurin polynomial of degree 2 of f(x) = cos x to approximate f(1). Give P₂(1).

f(x) = cos xn = 2x = 1
P₂(1) =
5.Multivariable calculusMedium
xy0.50.5P

Use the tangent plane of f(x, y) = 2x²ey at (0.5, 0.5) to approximate f(0.3, 0.6). Give the linear approximation L.

f(x, y) = 2x²ey(a, b) = (0.5, 0.5)(x, y) = (0.3, 0.6)
L =
6.Multivariable calculusHard
xyx = 1y = 3x13

Evaluate I = ∬T (x + 3y) dA over the triangle T bounded by y = 0, x = 1 and y = 3x.

f = x + 3y0 ≤ x ≤ 10 ≤ y ≤ 3x
I =
7.Differential equationsMedium
tyy(0) = 1y(2) = ?

Solve y″ + 4y′ + 3y = 0 with y(0) = 1 and y′(0) = 4. Find y(2).

y″ + 4y′ + 3y = 0y(0) = 1y′(0) = 4
y(2) =
8.Differential equationsHard
xy12(1, 3)y(2) = ?

Solve the initial value problem y′ + y/x = 2x2 for x > 0, with y(1) = 3. Find y(2).

y′ + y/x = 2x2y(1) = 3
y(2) =

3Solutions

1.Integrals

Answer: I = 3.466

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) = 5·ln x

Evaluate at the limits

I = 5·(ln 2 − ln 1) = 5·ln 2 = 3.466

Answer

I = 3.466
2.Integrals

Answer: I = 98.43

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

20.09·(5) − 1·2 = 98.43

Answer

I = 98.43
3.Sequences & series

Answer: R = 0.25

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+1/ck = (2k + 2)(2k + 1)/(k + 1)² → 4, so R = 1/4

Radius

R = 0.25

Answer

R = 0.25

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

4.Sequences & series

Answer: P₂(1) = 0.5

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.

Maclaurin polynomial

P₂(x) = 1 − 0.5x²

Substitute

P₂(1) = 1 − 0.5 = 0.5

Answer

P₂(1) = 0.5

The exact value is f(1) = 0.540302; the error 0.0403023 shrinks when you take more terms or a point closer to 0.

5.Multivariable calculus

Answer: L = 0.2473

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.

Value and slopes at the point

f(0.5, 0.5) = 0.8244
fx = 4xey → fx(0.5, 0.5) = 3.297
fy = 2x²ey → fy(0.5, 0.5) = 0.8244

Tangent plane

z = 0.8244 + 3.297(x − 0.5) + 0.8244(y − 0.5)
z = 3.297x + 0.8244y − 1.237

Linear approximation

L = 0.8244 + 3.297·(−0.2) + 0.8244·0.1 = 0.2473

Answer

L = 0.2473

The exact value is f = 0.327981; the plane is close because the step is small.

6.Multivariable calculus

Answer: I = 5.5

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).

Limits

0 ≤ x ≤ 1, 0 ≤ y ≤ 3x

Inner integral

∫03x (x + 3y) dy = x·3x + 3·(3x)²/2 = 16.5x²

Outer integral

I = 16.5·1³/3 = 5.5

Answer

I = 5.5
7.Differential equations

Answer: y(2) = 0.4675

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 + 1)(r + 3) = 0 → r = −1, −3

Constants from the initial values

A + B = 1, −1A − 3B = 4
A = 3.5, B = −2.5

Evaluate

y(2) = 3.5·e−2 + (−2.5)·e−6 = 0.4675

Answer

y(2) = 0.4675
8.Differential equations

Answer: y(2) = 5.25

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)′ = 2x3 → x1y = 2x4/4 + C

Initial condition

3 = 2/4 + C → C = 2.5

Evaluate

y(2) = (2·24/4 + 2.5)/21 = 5.25

Answer

y(2) = 5.25

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