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
xy1y = x·sin x

Evaluate I = ∫₀1 x·sin x dx (x in radians).

∫₀1 x·sin x dx
I =
2.IntegralsHard
xy−22

Find the area A of the region between the parabola y = 4 − x² and the x-axis.

y = 4 − x²
A =
3.Sequences & seriesMedium
xx₀ = 2??R

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

Σ (2k)!·(x − 2)ᵏ/(k!)²x₀ = 2
x =
4.Sequences & seriesHard
xyfP₄2

Use the Maclaurin polynomial of degree 4 of f(x) = eˣ to approximate f(2). Give P₄(2).

f(x) = eˣn = 4x = 2
P₄(2) =
5.Multivariable calculusMedium
xy−2024024vP

For f(x, y) = 4x³y² + 5x² − 5y² − xy, find the directional derivative Duf at (1, 2) in the direction of v = (−1, 3).

f(x, y) = 4x³y² + 5x² − 5y² − xyP = (1, 2)v = (−1, 3)
Duf =
6.Multivariable calculusHard
xyy = 2y = x22

The integral I = ∫02 ∫x2 1/(1 + y²)² dy dx cannot be done in this order. Reverse the order of integration and evaluate it.

f = 1/(1 + y²)²0 ≤ x ≤ y ≤ 2
I =
7.Differential equationsMedium
t (min)T (°C)16Ta​T = ?

A can of soda at 3 °C is taken out of the fridge into a room at 28 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). With k = 0.03 min⁻¹, find its temperature T after 16 min.

T0 = 3 °CTa = 28 °Ck = 0.03 min⁻¹t = 16 min
T = °C
8.Differential equationsHard
tyy(0) = 5y(0.75) = ?

Solve y″ + 2y′ + 26y = 0 with y(0) = 5 and y′(0) = 0 (released from rest). Find y(0.75); angles in radians.

y″ + 2y′ + 26y = 0y(0) = 5y′(0) = 0
y(0.75) =

3Solutions

1.Integrals

Answer: I = 0.3012

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.

Choose u and dv

u = x, du = dx; dv = sin x dx, v = −cos x

Antiderivative

∫x sin x dx = −x cos x + ∫cos x dx = −x cos x + sin x

Limits

(−1·cos 1 + sin 1) − (0 + 0) = 0.3012

Answer

I = 0.3012
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.

Zeros

4 − x² = 0 → x = ±2

Integrate

A = [4x − x³/3]−22 = 5.333 − (−5.333) = 10.67

Answer

A = 10.67
3.Sequences & series

Answer: x = 2.25

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

Interval

1.75 < x < 2.25
x = 2.25

Answer

x = 2.25

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

4.Sequences & series

Answer: P₄(2) = 7

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

P₄(x) = 1 + x + 0.5x² + 0.1667x³ + 0.04167x⁴

Substitute

P₄(2) = 1 + 2 + 2 + 1.333333 + 0.666667 = 7

Answer

P₄(2) = 7

The exact value is f(2) = 7.389056; the error 0.389056 shrinks when you take more terms or a point closer to 0.

5.Multivariable calculus

Answer: Duf = −22.45

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.

Gradient at the point

fx = 12x²y² + 10x − y → fx(1, 2) = 56
fy = 8x³y − 10y − x → fy(1, 2) = −5

Unit vector

|v| = √((−1)² + 3²) = 3.162
u = (−0.3162, 0.9487)

Dot product

Duf = 56·(−0.3162) + (−5)·0.9487 = −22.45

Answer

Duf = −22.45
6.Multivariable calculus

Answer: I = 0.4

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 1/(1 + y²)² dx dy = ∫02 y·1/(1 + y²)² dy

Substitution u = y²

I = [−1/(2(1 + y²))]02 = 0.4

Answer

I = 0.4
7.Differential equations

Answer: T = 12.53 °C

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.

Solution of the cooling law

T(t) = 28 + (−25)·e−0.03t

Substitute t

T(16) = 28 + (−25)·0.6188 = 12.53 °C

Answer

T = 12.53 °C
8.Differential equations

Answer: y(0.75) = −2.208

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 ± 5j

Constants

A = 5, B = 1·5/5 = 1

Evaluate

y(0.75) = e−0.75(5·cos 3.75 + 1·sin 3.75) = −2.208

Answer

y(0.75) = −2.208

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