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 1/x dx.

f(x) = 1/x[1, 2]
I =
2.IntegralsHard
xyy = x2​y = 3x

Find the area A of the region enclosed by y = x2 and y = 3x.

y = x2y = 3x
A =
3.Sequences & seriesMedium
xx₀ = 0??R

Find the radius of convergence R of the power series Σk=1∞ k!·xᵏ/kᵏ.

Σ k!·xᵏ/kᵏx₀ = 0
R =
4.Sequences & seriesHard

Recognise the series and find its sum: S = Σk=1∞ k·0.5ᵏ.

ak = k·0.5ᵏ
S =
5.Multivariable calculusMedium
xy−1−1P

Use the tangent plane of f(x, y) = 3y·exy at (−1, −1) to approximate f(−1.1, −1.1). Give the linear approximation L.

f(x, y) = 3y·exy(a, b) = (−1, −1)(x, y) = (−1.1, −1.1)
L =
6.Multivariable calculusHard
xy−202−2024AB

The marked points are the critical points of f(x, y) = −x³ + 3x − y² + 6y. Click the local maximum.

f(x, y) = −x³ + 3x − y² + 6y
  1. 1A (1, 3)
  2. 2B (−1, 3)
7.Differential equationsMedium
tyy(0) = 4y(1.5) = ?

Solve y″ + 4y′ + 3y = 0 with y(0) = 4 and y′(0) = -2. Find y(1.5).

y″ + 4y′ + 3y = 0y(0) = 4y′(0) = -2
y(1.5) =
8.Differential equationsHard
tyy(0) = 2y(1.5) = ?

Solve y″ + 3y′ + 6.25y = 0 with y(0) = 2 and y′(0) = 0 (released from rest). Find y(1.5); angles in radians.

y″ + 3y′ + 6.25y = 0y(0) = 2y′(0) = 0
y(1.5) =

3Solutions

1.Integrals

Answer: I = 0.6931

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) = ln x

Evaluate at the limits

I = (ln 2 − ln 1) = ln 2 = 0.6931

Answer

I = 0.6931
2.Integrals

Answer: A = 4.5

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 = 3x → x = 0, x = 3

Integrate upper minus lower

A = [1.5x2 − 0.3333x3]03 = 4.5

Answer

A = 4.5
3.Sequences & series

Answer: R = 2.718

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 + 1/k)ᵏ → e

Radius

R = 2.718

Answer

R = 2.718

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

4.Sequences & series

Answer: S = 2

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.

Standard series

Σ k·xᵏ = x/(1 − x)²

Substitute

x = 0.5
S = 0.5/(1 − 0.5)² = 2

Answer

S = 2
5.Multivariable calculus

Answer: L = −10.6

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(−1, −1) = −8.155
fx = 3y²exy → fx(−1, −1) = 8.155
fy = 3exy(1 + xy) → fy(−1, −1) = 16.31

Tangent plane

z = −8.155 + 8.155(x + 1) + 16.31(y + 1)
z = 8.155x + 16.31y + 16.31

Linear approximation

L = −8.155 + 8.155·(−0.1) + 16.31·(−0.1) = −10.6

Answer

L = −10.6

The exact value is f = −11.066499; the plane is close because the step is small.

6.Multivariable calculus

Answer: A (1, 3)

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 − 3) = 0
fy = −(2y − 6) = 0
(1, 3), (−1, 3)

Second-derivative test

(1, 3): fxx = −6, fyy = −2, fxy = 0, D = 12 → local maximum
(−1, 3): fxx = 6, fyy = −2, fxy = 0, D = −12 → saddle point

The point

A (1, 3): local maximum

Answer

A (1, 3)
7.Differential equations

Answer: y(1.5) = 1.105

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 = 4, −1A − 3B = -2
A = 5, B = −1

Evaluate

y(1.5) = 5·e−1.5 + (−1)·e−4.5 = 1.105

Answer

y(1.5) = 1.105
8.Differential equations

Answer: y(1.5) = −0.1864

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.

Complex roots

r = −1.5 ± 2j

Constants

A = 2, B = 1.5·2/2 = 1.5

Evaluate

y(1.5) = e−2.25(2·cos 3 + 1.5·sin 3) = −0.1864

Answer

y(1.5) = −0.1864

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