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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 = ∫₀1 x·sin x dx (x in radians).
∫₀1 x·sin x dx
I =
2.IntegralsHard
Find the area A of the region enclosed by y = x2 and y = x.
y = x2y = x
A =
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
f(x) = sin 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) = sin xx = 1M = 1ε = 1 × 10⁻⁵
n =
5.Multivariable calculusMedium
For f(x, y) = −4xy + x³ − y², find the directional derivative Duf at P(2, −1) in the direction towards Q(2, −5).
f(x, y) = −4xy + x³ − y²P = (2, −1)Q = (2, −5)
Duf =
6.Multivariable calculusHard
Find the critical points of f(x, y) = x³ − 12x + y² − 4y, classify them, and give the value fmin of f at its local minimum.
f(x, y) = x³ − 12x + y² − 4y
fmin =
7.Differential equationsMedium
A forged steel part at 440 °C is left to cool in a workshop at 21 °C. After 54 min its temperature is 69 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). Find the constant k in min⁻¹.
T0 = 440 °CTa = 21 °CT(54) = 69 °C
k = min⁻¹
8.Differential equationsHard
Solve y′ + y = 4e−3t with y(0) = 3. Find y(0.5).
y′ + y = 4e−3ty(0) = 3
y(0.5) =
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 = 0.1667
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
x2 = x → x = 0, x = 1
Integrate upper minus lower
A = [0.5x2 − 0.3333x3]01 = 0.1667
Answer
A = 0.1667
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: 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.
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: Duf = 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.
Gradient at the point
fx = −4y + 3x² → fx(2, −1) = 16
fy = −4x − 2y → fy(2, −1) = −6
Unit vector
|v| = √(0² + (−4)²) = 4
u = (0, −1)
Dot product
Duf = 16·0 + (−6)·(−1) = 6
Answer
Duf = 6
6.Multivariable calculus
Answer: fmin = −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 minimum
(−2, 2): fxx = −12, fyy = 2, fxy = 0, D = −24 → saddle point
Value at the extremum
f(2, 2) = −20
Answer
fmin = −20
7.Differential equations
Answer: k = 0.04012 min⁻¹
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.
Ratio of temperature differences
e−54k = 48/419
Solve for k
k = ln(419/48)/54 = 0.04012 min⁻¹
Answer
k = 0.04012 min⁻¹
8.Differential equations
Answer: y(0.5) = 2.586
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.