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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 = x from x = 0 to x = 3 is rotated about the x-axis. Find the volume V of the solid (a cone).
y = x0 ≤ x ≤ 3
V =
2.IntegralsHard
Evaluate I = ∫₀1.5 x²·eˣ dx.
∫₀1.5 x²eˣ dx
I =
3.Sequences & seriesMedium
Apply the ratio test to Σk=1∞ (k!)²·2ᵏ/(2k)!. Find L = lim |ak+1/ak|.
ak = (k!)²·2ᵏ/(2k)!
L =
4.Sequences & seriesHard
Use the Maclaurin polynomial of degree 2 of f(x) = cos(3x) to approximate f(0.4). Give P₂(0.4).
f(x) = cos(3x)n = 2x = 0.4
P₂(0.4) =
5.Multivariable calculusMedium
For f(x, y) = −3x²y − 2x³ − 3y² + 4xy, find the maximum rate of change of f at the point (−1, 1), that is |∇f|.
f(x, y) = −3x²y − 2x³ − 3y² + 4xyP = (−1, 1)
|∇f| =
6.Multivariable calculusHard
Find the critical points of f(x, y) = −x³ + 12x − y² + 4y, classify them, and give the value fmax of f at its local maximum.
f(x, y) = −x³ + 12x − y² + 4y
fmax =
7.Differential equationsMedium
Solve y″ + 8y′ + 16y = 0 with y(0) = 5 and y′(0) = 4. Find y(0.5).
y″ + 8y′ + 16y = 0y(0) = 5y′(0) = 4
y(0.5) =
8.Differential equationsHard
A system obeys y″ + 2y′ + 10y = 10 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find y(1.5); angles in radians.
y″ + 2y′ + 10y = 10y(0) = 0y′(0) = 0
y(1.5) =
3Solutions
1.Integrals
Answer: V = 28.27
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∫₀3 1x² dx
Integrate
V = π·1·3³/3 = 28.27
Answer
V = 28.27
Check: a cone of radius 3 and height 3 has (1/3)πr²h = 28.27.
2.Integrals
Answer: I = 3.602
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
4.482·(1.25) − 1·2 = 3.602
Answer
I = 3.602
3.Sequences & series
Answer: L = 0.5
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.
Ratio of consecutive terms
|ak+1/ak| = 2·(k + 1)²/((2k + 2)(2k + 1)) → 2/4
Conclusion
L = 0.5 < 1
Answer
L = 0.5
L < 1, so the series converges.
4.Sequences & series
Answer: P₂(0.4) = 0.28
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
P₂(x) = 1 − 4.5x²
Substitute
P₂(0.4) = 1 − 0.72 = 0.28
Answer
P₂(0.4) = 0.28
The exact value is f(0.4) = 0.362358; the error 0.0823578 shrinks when you take more terms or a point closer to 0.
5.Multivariable calculus
Answer: |∇f| = 13.6
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.
Partial derivatives
fx = −6xy − 6x² + 4y → fx(−1, 1) = 4
fy = −3x² − 6y + 4x → fy(−1, 1) = −13
Length of the gradient
|∇f| = √(4² + (−13)²) = 13.6
Answer
|∇f| = 13.6
6.Multivariable calculus
Answer: fmax = 20
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 − 12) = 0
fy = −(2y − 4) = 0
(2, 2), (−2, 2)
Second-derivative test
(2, 2): fxx = −12, fyy = −2, fxy = 0, D = 24 → local maximum
(−2, 2): fxx = 12, fyy = −2, fxy = 0, D = −24 → saddle point
Value at the extremum
f(2, 2) = 20
Answer
fmax = 20
7.Differential equations
Answer: y(0.5) = 2.301
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 + 4)² = 0 → r = −4
Constants
A = y(0) = 5
y′(0) = B − 4A → B = 4 + 20 = 24
Evaluate
y(0.5) = (5 + 24·0.5)·e−2 = 2.301
Answer
y(0.5) = 2.301
8.Differential equations
Answer: y(1.5) = 1.12
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.