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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
Evaluate I = ∫02 e−x dx.
f(x) = e−x[0, 2]
I =
2.IntegralsHard
Find the area A of the region between the parabola y = 1 − x² and the x-axis.
y = 1 − x²
A =
3.Sequences & seriesMedium
Exactly one of the four series diverges. Click it.
Find the coefficient c₂ of (x − 1)² in the Taylor series of f(x) = ∛x about a = 1.
f(x) = ∛xa = 1k = 2
c₂ =
5.Multivariable calculusMedium
For f(x, y) = 3x³y³ − 5y³ − xy, find the direction of steepest ascent at (1, 1) as the angle θ from the positive x-axis, in degrees between −180° and 180°.
f(x, y) = 3x³y³ − 5y³ − xyP = (1, 1)
θ = °
6.Multivariable calculusHard
f(x, y) = x³ + y³ + 3xy has a critical point at (0, 0). Compute the discriminant D = fxxfyy − fxy² there.
f(x, y) = x³ + y³ + 3xyP = (0, 0)
D =
7.Differential equationsMedium
Solve y″ + 6y′ + 8y = 0 with y(0) = 1 and y′(0) = 2. Find y(1.5).
y″ + 6y′ + 8y = 0y(0) = 1y′(0) = 2
y(1.5) =
8.Differential equationsHard
A system obeys y″ + 2y′ + 26y = 130 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find y(1.5); angles in radians.
y″ + 2y′ + 26y = 130y(0) = 0y′(0) = 0
y(1.5) =
3Solutions
1.Integrals
Answer: I = 0.8647
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−2 − 1)/(−1) = (0.1353 − 1)/(−1) = 0.8647
Answer
I = 0.8647
2.Integrals
Answer: A = 1.333
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
1 − x² = 0 → x = ±1
Integrate
A = [1x − x³/3]−11 = 0.6667 − (−0.6667) = 1.333
Answer
A = 1.333
3.Sequences & series
Answer: D: Σ k/(k + 7)
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)ᵏ/k → Alternating, with 1/k decreasing to 0: converges (conditionally).
B: Σ 1/(3ᵏ + k) → Smaller than (1/3)ᵏ, a convergent geometric series: converges by comparison.
C: Σ (2/3)ᵏ → Geometric with r = 2/3, |r| < 1: converges.
D: Σ k/(k + 7) → The terms tend to 1, not 0: diverges by the n-th term test.
The answer
D: Σ k/(k + 7)
Answer
D: Σ k/(k + 7)
4.Sequences & series
Answer: c₂ = −0.1111
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 = −0.2222·11/3 − 2/2! = −0.1111
c₂ = −0.1111
Answer
c₂ = −0.1111
5.Multivariable calculus
Answer: θ = −41.19 °
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 = 9x²y³ − y → fx(1, 1) = 8
fy = 9x³y² − 15y² − x → fy(1, 1) = −7
Direction
θ = atan2(−7, 8) = −41.19°
Answer
θ = −41.19°
6.Multivariable calculus
Answer: D = −9
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 = 0, fyy = 0, fxy = 3
Discriminant
D = 0·0 − 3² = −9
→ saddle point
Answer
D = −9
7.Differential equations
Answer: y(1.5) = 0.1444
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.
Characteristic roots
(r + 2)(r + 4) = 0 → r = −2, −4
Constants from the initial values
A + B = 1, −2A − 4B = 2
A = 3, B = −2
Evaluate
y(1.5) = 3·e−3 + (−2)·e−6 = 0.1444
Answer
y(1.5) = 0.1444
8.Differential equations
Answer: y(1.5) = 4.404
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.