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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
Rewrite the integrand as a sum of simple terms where possible.
Apply the power rule: ∫xⁿ dx = xⁿ⁺¹/(n + 1), and standard integrals for ex, 1/x, sin and cos.
Use substitution when you see a function and its derivative together.
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
Identify the type: arithmetic (constant difference), geometric (constant ratio), a standard series, a power series or a Taylor polynomial.
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.
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.
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
For a partial derivative, treat the other variable as a constant and use the one-variable rules.
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.
For extrema solve fx = 0 and fy = 0 together, then classify every point with D = fxx·fyy − fxy².
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
The region under y = √(2x) from x = 0 to x = 1 is rotated about the x-axis. Find the volume V of the solid (a paraboloid).
y = √(2x)0 ≤ x ≤ 1
V =
2.IntegralsHard
Find the area A of the region enclosed by y = 0.5x2 and y = 4x.
y = 0.5x2y = 4x
A =
3.Sequences & seriesMedium
Exactly one of the four series converges. Click it.
A: Σ (1/2)ᵏB: Σ 1/√kC: Σ (1 + 1/k)ᵏD: Σ 2ᵏ/k²
1A: Σ (1/2)ᵏ
2B: Σ 1/√k
3C: Σ (1 + 1/k)ᵏ
4D: Σ 2ᵏ/k²
4.Sequences & seriesHard
f(x) = eˣ is approximated at x = 0.5 by its Maclaurin polynomial of degree n. With the Lagrange bound M·|x|ⁿ⁺¹/(n + 1)! and M = 2, find the smallest degree n that guarantees an error below 1 × 10⁻⁵.
f(x) = eˣx = 0.5M = 2ε = 1 × 10⁻⁵
n =
5.Multivariable calculusMedium
For f(x, y) = 5x²y³ − 3x² − y² − xy, 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) = 5x²y³ − 3x² − y² − xyP = (3, 2)
θ = °
6.Multivariable calculusHard
f(x, y) = −x³ + 3x − y² − 4y has a critical point at (−1, −2). Compute the discriminant D = fxxfyy − fxy² there.
f(x, y) = −x³ + 3x − y² − 4yP = (−1, −2)
D =
7.Differential equationsMedium
Solve y″ + 6y′ + 9y = 0 with y(0) = 3 and y′(0) = -2. Find y(2).
y″ + 6y′ + 9y = 0y(0) = 3y′(0) = -2
y(2) =
8.Differential equationsHard
Solve the initial value problem y′ + y/x = 1 for x > 0, with y(1) = 5. Find y(2).
y′ + y/x = 1y(1) = 5
y(2) =
3Solutions
1.Integrals
Answer: V = 3.142
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∫₀1 2x dx
Integrate
V = π·2·1²/2 = 3.142
Answer
V = 3.142
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: A: Σ (1/2)ᵏ
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.
Test each series
A: Σ (1/2)ᵏ → Geometric with r = 1/2, |r| < 1: converges.
B: Σ 1/√k → p-series with p = 1/2 ≤ 1: diverges.
C: Σ (1 + 1/k)ᵏ → The terms tend to e, not 0: diverges by the n-th term test.
D: Σ 2ᵏ/k² → Ratio test: L = 2 > 1, diverges.
The answer
A: Σ (1/2)ᵏ
Answer
A: Σ (1/2)ᵏ
4.Sequences & series
Answer: n = 6
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.
Bounds for increasing n
n = 5: 4.34 × 10⁻⁵ ≥ 1 × 10⁻⁵
n = 6: 3.1 × 10⁻⁶ < 1 × 10⁻⁵
Smallest degree
n = 6
Answer
n = 6
5.Multivariable calculus
Answer: θ = 67.57 °
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 = 10xy³ − 6x − y → fx(3, 2) = 220
fy = 15x²y² − 2y − x → fy(3, 2) = 533
Direction
θ = atan2(533, 220) = 67.57°
Answer
θ = 67.57°
6.Multivariable calculus
Answer: D = −12
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).
Second derivatives
fxx = 6, fyy = −2, fxy = 0
Discriminant
D = 6·(−2) − 0² = −12
→ saddle point
Answer
D = −12
7.Differential equations
Answer: y(2) = 0.04214
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) = 3
y′(0) = B − 3A → B = -2 + 9 = 7
Evaluate
y(2) = (3 + 7·2)·e−6 = 0.04214
Answer
y(2) = 0.04214
8.Differential equations
Answer: y(2) = 3.25
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.