Calculus 2: cram pack

Formula sheets, the key method steps for every topic and 8 practice problems, with full solutions at the back.

New problemsBack to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

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
xy01

Evaluate I = ∫01 e−x dx.

f(x) = e−x[0, 1]
I =
2.IntegralsHard
xyy = 0.5x2​y = 4x

Find the area A of the region enclosed by y = 0.5x2 and y = 4x.

y = 0.5x2y = 4x
A =
3.Sequences & seriesMedium
kSn​12345678910S = ?

Find the sum S = Σk=1∞ 12/(k(k + 4)).

ak = 12/(k(k + 4))
S =
4.Sequences & seriesHard

f(x) = cos x is approximated at x = 1 by its Maclaurin polynomial of degree n. With the Lagrange bound M·|x|ⁿ⁺¹/(n + 1)! and M = 1, find the smallest degree n that guarantees an error below 1 × 10⁻⁵.

f(x) = cos xx = 1M = 1ε = 1 × 10⁻⁵
n =
5.Multivariable calculusMedium
xy32∇fP

For f(x, y) = −3x³y − 4x³ − 5y³ − 4xy, find the direction of steepest ascent at (3, 2) as the angle θ from the positive x-axis, in degrees between −180° and 180°.

f(x, y) = −3x³y − 4x³ − 5y³ − 4xyP = (3, 2)
θ = °
6.Multivariable calculusHard
xy−202−202ABC

The marked points are the critical points of f(x, y) = x⁴ + y⁴ − 4xy. Click the saddle point.

f(x, y) = x⁴ + y⁴ − 4xy
  1. 1A (0, 0)
  2. 2B (−1, −1)
  3. 3C (1, 1)
7.Differential equationsMedium
tyy(0) = 1y(2) = ?

Solve y″ + 6y′ + 9y = 0 with y(0) = 1 and y′(0) = -1. Find y(2).

y″ + 6y′ + 9y = 0y(0) = 1y′(0) = -1
y(2) =
8.Differential equationsHard
tyy(0) = 1y(1) = ?

Solve y″ + 4y′ + 8y = 0 with y(0) = 1 and y′(0) = 0 (released from rest). Find y(1); angles in radians.

y″ + 4y′ + 8y = 0y(0) = 1y′(0) = 0
y(1) =

3Solutions

1.Integrals

Answer: I = 0.6321

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) = e−x/(−1)

Evaluate at the limits

I = (e−1 − 1)/(−1) = (0.3679 − 1)/(−1) = 0.6321

Answer

I = 0.6321
2.Integrals

Answer: A = 42.67

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

0.5x2 = 4x → x = 0, x = 8

Integrate upper minus lower

A = [2x2 − 0.1667x3]08 = 42.67

Answer

A = 42.67
3.Sequences & series

Answer: S = 6.25

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.

Partial fractions

12/(k(k + 4)) = (12/4)·(1/k − 1/(k + 4))

Cancel

S = (12/4)·(1 + 1/2 + 1/3 + 1/4) = 6.25

Answer

S = 6.25
4.Sequences & series

Answer: n = 8

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.

n(n + 1)!bound011120.5260.16673240.0416741200.00833357200.001389650400.00019847403202.48 × 10⁻⁵83628802.756 × 10⁻⁶

Bounds for increasing n

n = 7: 2.48 × 10⁻⁵ ≥ 1 × 10⁻⁵
n = 8: 2.756 × 10⁻⁶ < 1 × 10⁻⁵

Smallest degree

n = 8

Answer

n = 8
5.Multivariable calculus

Answer: θ = −151.2 °

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.

Partial derivatives

fx = −9x²y − 12x² − 4y → fx(3, 2) = −278
fy = −3x³ − 15y² − 4x → fy(3, 2) = −153

Direction

θ = atan2(−153, −278) = −151.2°

Answer

θ = −151.2°
6.Multivariable calculus

Answer: A (0, 0)

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 = 4x³ − 4y = 0
fy = 4y³ − 4x = 0
(0, 0), (1, 1), (−1, −1)

Second-derivative test

(0, 0): fxx = 0, fyy = 0, fxy = −4, D = −16 → saddle point
(1, 1): fxx = 12, fyy = 12, fxy = −4, D = 128 → local minimum
(−1, −1): fxx = 12, fyy = 12, fxy = −4, D = 128 → local minimum

The point

A (0, 0): saddle point

Answer

A (0, 0)
7.Differential equations

Answer: y(2) = 0.01239

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.

Double root

(r + 3)² = 0 → r = −3

Constants

A = y(0) = 1
y′(0) = B − 3A → B = -1 + 3 = 2

Evaluate

y(2) = (1 + 2·2)·e−6 = 0.01239

Answer

y(2) = 0.01239
8.Differential equations

Answer: y(1) = 0.06674

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 = −2 ± 2j

Constants

A = 1, B = 2·1/2 = 1

Evaluate

y(1) = e−2(1·cos 2 + 1·sin 2) = 0.06674

Answer

y(1) = 0.06674

ExamProblems · examproblems.com/courses/calculus-2