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
xy4y = √(2x)

The region under y = √(2x) from x = 0 to x = 4 is rotated about the x-axis. Find the volume V of the solid (a paraboloid).

y = √(2x)0 ≤ x ≤ 4
V =
2.IntegralsHard
xy1y = x²eˣ

Evaluate I = ∫₀1 x²·eˣ dx.

∫₀1 x²eˣ dx
I =
3.Sequences & seriesMedium
xx₀ = 0??R

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

Σ (1/2)ᵏ·xᵏ/(k + 1)x₀ = 0
R =
4.Sequences & seriesHard

Find the coefficient c₂ of (x − 8)² in the Taylor series of f(x) = ∛x about a = 8.

f(x) = ∛xa = 8k = 2
c₂ =
5.Multivariable calculusMedium
xy−2−2P

The tangent plane to z = 4xy² − 2y² + 2xy at the point where (x, y) = (−2, −2) can be written as z = px + qy + c. Find c.

f(x, y) = 4xy² − 2y² + 2xy(a, b) = (−2, −2)
c =
6.Multivariable calculusHard
xyy = 2y = x22

The integral I = ∫02 ∫x2 cos(y²) dy dx cannot be done in this order. Reverse the order of integration and evaluate it.

f = cos(y²)0 ≤ x ≤ y ≤ 2
I =
7.Differential equationsMedium
t (min)T (°C)41Ta​227 °C

A forged steel part at 480 °C is left to cool in a workshop at 27 °C. After 41 min its temperature is 227 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). Find the constant k in min⁻¹.

T0 = 480 °CTa = 27 °CT(41) = 227 °C
k = min⁻¹
8.Differential equationsHard
tyK = 1y(1.5) = ?

A system obeys y″ + 4y′ + 29y = 29 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find y(1.5); angles in radians.

y″ + 4y′ + 29y = 29y(0) = 0y′(0) = 0
y(1.5) =

3Solutions

1.Integrals

Answer: V = 50.27

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∫₀4 2x dx

Integrate

V = π·2·4²/2 = 50.27

Answer

V = 50.27
2.Integrals

Answer: I = 0.7183

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

2.718·(1) − 1·2 = 0.7183

Answer

I = 0.7183
3.Sequences & series

Answer: R = 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/(1/2))·(k + 2)/(k + 1) → 2

Radius

R = 2

Answer

R = 2

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

4.Sequences & series

Answer: c₂ = −0.003472

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 = −0.2222·81/3 − 2/2! = −0.003472
c₂ = −0.003472

Answer

c₂ = −0.003472
5.Multivariable calculus

Answer: c = 64

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, −2) = −32
fx = 4y² + 2y → fx(−2, −2) = 12
fy = 8xy − 4y + 2x → fy(−2, −2) = 36

Tangent plane

z = −32 + 12(x + 2) + 36(y + 2)
z = 12x + 36y + 64

Constant term

c = 64

Answer

c = 64
6.Multivariable calculus

Answer: I = −0.3784

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).

Region

0 ≤ x ≤ y, 0 ≤ y ≤ 2

Reversed order

I = ∫02 ∫0y cos(y²) dx dy = ∫02 y·cos(y²) dy

Substitution u = y²

I = [sin(y²)/2]02 = −0.3784

Answer

I = −0.3784
7.Differential equations

Answer: k = 0.01994 min⁻¹

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.

Ratio of temperature differences

e−41k = 200/453

Solve for k

k = ln(453/200)/41 = 0.01994 min⁻¹

Answer

k = 0.01994 min⁻¹
8.Differential equations

Answer: y(1.5) = 0.9641

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 = 1, r = −2 ± 5j

Constants from the initial values

y(0) = 1 + A = 0 → A = −1
y′(0) = −2A + 5B = 0 → B = −0.4

Evaluate

y(1.5) = 1·(1 − e−3(cos 7.5 + 0.4·sin 7.5)) = 0.9641

Answer

y(1.5) = 0.9641

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