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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 2x·(x² + 1)3 dx.
f(x) = 2x·(x² + 1)3[0, 1]
I =
2.IntegralsHard
The region between the line y = 4x and the parabola y = x² is rotated about the x-axis. Find the volume V of the solid.
y = 4xy = x²
V =
3.Sequences & seriesMedium
Exactly one of the four series diverges. Click it.
A: Σ (−1)ᵏ/kB: Σ 1/k³C: Σ 1/(2ᵏ + k)D: Σ (7/4)ᵏ
1A: Σ (−1)ᵏ/k
2B: Σ 1/k³
3C: Σ 1/(2ᵏ + k)
4D: Σ (7/4)ᵏ
4.Sequences & seriesHard
f(x) = cos x is approximated at x = 2 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 0.001.
f(x) = cos xx = 2M = 1ε = 0.001
n =
5.Multivariable calculusMedium
z = ln(x² + 2y²) with x = −2t − 2 and y = −t². Use the chain rule to find dz/dt at t = −1.
z = ln(x² + 2y²)x = −2t − 2y = −t²t = −1
dz/dt =
6.Multivariable calculusHard
The marked points are the critical points of f(x, y) = −x³ + 12x − y² + 6y. Click the local maximum.
f(x, y) = −x³ + 12x − y² + 6y
1A (2, 3)
2B (−2, 3)
7.Differential equationsMedium
A forged steel part at 420 °C is left to cool in a workshop at 27 °C. After 25 min its temperature is 213 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). How long after the start, t, does it take to reach 296 °C?
T0 = 420 °CTa = 27 °CT(25) = 213 °CT = 296 °C
t = min
8.Differential equationsHard
Solve the initial value problem y′ + 2y/x = 4x2 for x > 0, with y(1) = −3. Find y(3).
y′ + 2y/x = 4x2y(1) = −3
y(3) =
3Solutions
1.Integrals
Answer: I = 3.75
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.
Substitution
u = x² + 1, du = 2x dx
I = ∫12 u3 du
Integrate and evaluate
I = [u4/4]12 = (16 − 1)/4 = 3.75
Answer
I = 3.75
2.Integrals
Answer: V = 428.9
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.
Intersection points
4x = 1x² → x = 0, x = 4
Washer method
V = π∫₀4 ((4x)² − (1x²)²) dx
Integrate
V = π(16·4³/3 − 1·4⁵/5) = π·136.5 = 428.9
Answer
V = 428.9
3.Sequences & series
Answer: D: Σ (7/4)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)ᵏ/k → Alternating, with 1/k decreasing to 0: converges (conditionally).
B: Σ 1/k³ → p-series with p = 3 > 1: converges.
C: Σ 1/(2ᵏ + k) → Smaller than (1/2)ᵏ, a convergent geometric series: converges by comparison.
D: Σ (7/4)ᵏ → Geometric with r = 7/4 > 1: the terms grow, so it diverges.
The answer
D: Σ (7/4)ᵏ
Answer
D: Σ (7/4)ᵏ
4.Sequences & series
Answer: n = 9
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 = 8: 0.001411 ≥ 0.001
n = 9: 0.0002822 < 0.001
Smallest degree
n = 9
Answer
n = 9
5.Multivariable calculus
Answer: dz/dt = −4
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.
Point and rates
x(−1) = 0, y(−1) = −1
dx/dt = −2, dy/dt = −2t = 2
Partial derivatives
fx = 2x/(x² + 2y²) = 0
fy = 4y/(x² + 2y²) = −2
Chain rule
dz/dt = 0·(−2) + (−2)·2 = −4
Answer
dz/dt = −4
6.Multivariable calculus
Answer: A (2, 3)
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 − 6) = 0
(2, 3), (−2, 3)
Second-derivative test
(2, 3): fxx = −12, fyy = −2, fxy = 0, D = 24 → local maximum
(−2, 3): fxx = 12, fyy = −2, fxy = 0, D = −24 → saddle point
The point
A (2, 3): local maximum
Answer
A (2, 3)
7.Differential equations
Answer: t = 12.67 min
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.
Cooling constant
k = ln(393/186)/25 = 0.02992 min⁻¹
Time to reach the target
t = ln(393/269)/0.02992 = 12.67 min
Answer
t = 12.67 min
8.Differential equations
Answer: y(3) = 21.18
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.