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
xy2y = 0.5x

The region under y = 0.5x from x = 0 to x = 2 is rotated about the x-axis. Find the volume V of the solid (a cone).

y = 0.5x0 ≤ x ≤ 2
V =
2.IntegralsHard
xyy = x2​y = 4x

Find the area A of the region enclosed by y = x2 and y = 4x.

y = x2y = 4x
A =
3.Sequences & seriesMedium
xx₀ = 2??R

The power series Σk=1∞ k²·2ᵏ·(x − 2)ᵏ converges on an interval around x₀ = 2. Find the right end x of that interval.

Σ k²·2ᵏ·(x − 2)ᵏx₀ = 2
x =
4.Sequences & seriesHard

f(x) = eˣ is approximated by its Maclaurin polynomial of degree 2 at x = 0.3. Use the Lagrange error bound M·|x|ⁿ⁺¹/(n + 1)! with M = e0.3 = 1.35 to find the largest possible error |R₂|.

f(x) = eˣn = 2x = 0.3M = 1.35
|R₂| =
5.Multivariable calculusMedium
xy−23P

The tangent plane to z = 3x²y³ − 5y² − xy at the point where (x, y) = (−2, 3) can be written as z = px + qy + c. Find c.

f(x, y) = 3x²y³ − 5y² − xy(a, b) = (−2, 3)
c =
6.Multivariable calculusHard
xy−202−202(2, 1)(−2, 1)

Find the critical points of f(x, y) = −x³ + 12x − y² + 2y, classify them, and give the value fmax of f at its local maximum.

f(x, y) = −x³ + 12x − y² + 2y
fmax =
7.Differential equationsMedium
tyy(0) = 4y(1) = ?

Solve y″ + 7y′ + 12y = 0 with y(0) = 4 and y′(0) = -4. Find y(1).

y″ + 7y′ + 12y = 0y(0) = 4y′(0) = -4
y(1) =
8.Differential equationsHard
xy13(1, 2)y(3) = ?

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

y′ + 3y/x = 3y(1) = 2
y(3) =

3Solutions

1.Integrals

Answer: V = 2.094

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∫₀2 0.25x² dx

Integrate

V = π·0.25·2³/3 = 2.094

Answer

V = 2.094

Check: a cone of radius 1 and height 2 has (1/3)πr²h = 2.094.

2.Integrals

Answer: A = 10.67

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

x2 = 4x → x = 0, x = 4

Integrate upper minus lower

A = [2x2 − 0.3333x3]04 = 10.67

Answer

A = 10.67
3.Sequences & series

Answer: x = 2.5

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 = (1/2)·(k/(k + 1))² → 1/2

Radius

R = 0.5

Interval

1.5 < x < 2.5
x = 2.5

Answer

x = 2.5

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

4.Sequences & series

Answer: |R₂| = 0.006074

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 = e0.3 = 1.35

Lagrange bound

|R₂| ≤ 1.35·0.3³/3! = 0.006074

Answer

|R₂| = 0.006074

The true error is smaller: |f(0.3) − P₂(0.3)| = 0.004859.

5.Multivariable calculus

Answer: c = −1257

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(−2, 3) = 285
fx = 6xy³ − y → fx(−2, 3) = −327
fy = 9x²y² − 10y − x → fy(−2, 3) = 296

Tangent plane

z = 285 − 327(x + 2) + 296(y − 3)
z = −327x + 296y − 1257

Constant term

c = −1257

Answer

c = −1257
6.Multivariable calculus

Answer: fmax = 17

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).

Critical points

Solve fx = 0 and fy = 0 together.

fx = −(3x² − 12) = 0
fy = −(2y − 2) = 0
(2, 1), (−2, 1)

Second-derivative test

(2, 1): fxx = −12, fyy = −2, fxy = 0, D = 24 → local maximum
(−2, 1): fxx = 12, fyy = −2, fxy = 0, D = −24 → saddle point

Value at the extremum

f(2, 1) = 17

Answer

fmax = 17
7.Differential equations

Answer: y(1) = 0.4509

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 + 3)(r + 4) = 0 → r = −3, −4

Constants from the initial values

A + B = 4, −3A − 4B = -4
A = 12, B = −8

Evaluate

y(1) = 12·e−3 + (−8)·e−4 = 0.4509

Answer

y(1) = 0.4509
8.Differential equations

Answer: y(3) = 2.296

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∫3/x dx = x3

Integrate

(x3y)′ = 3x3 → x3y = 3x4/4 + C

Initial condition

2 = 3/4 + C → C = 1.25

Evaluate

y(3) = (3·34/4 + 1.25)/33 = 2.296

Answer

y(3) = 2.296

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