In the print window, choose “Save as PDF” to keep a copy on your device.
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 = 1 is rotated about the x-axis. Find the volume V of the solid (a cone).
y = x0 ≤ x ≤ 1
V =
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 converges. Click it.
Recognise the series and find its sum: S = Σk=0∞ (−1)ᵏ/k!.
ak = (−1)ᵏ/k!
S =
5.Multivariable calculusMedium
For f(x, y) = 2 ln(x² + 3y²), find the directional derivative Duf at (−1, 2) in the direction of v = (4, −2).
f(x, y) = 2 ln(x² + 3y²)P = (−1, 2)v = (4, −2)
Duf =
6.Multivariable calculusHard
The marked points are the critical points of f(x, y) = x³ − 12x + y². Click the local minimum.
f(x, y) = x³ − 12x + y²
1A (2, 0)
2B (−2, 0)
7.Differential equationsMedium
A can of soda at 3 °C is taken out of the fridge into a room at 26 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). With k = 0.1 min⁻¹, find its temperature T after 22 min.
T0 = 3 °CTa = 26 °Ck = 0.1 min⁻¹t = 22 min
T = °C
8.Differential equationsHard
Solve the initial value problem y′ + 3y/x = 6x for x > 0, with y(1) = 5. Find y(3).
y′ + 3y/x = 6xy(1) = 5
y(3) =
3Solutions
1.Integrals
Answer: V = 1.047
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 1x² dx
Integrate
V = π·1·1³/3 = 1.047
Answer
V = 1.047
Check: a cone of radius 1 and height 1 has (1/3)πr²h = 1.047.
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: B: Σ (−1)ᵏ/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 (k + 3)/k² → Behaves like 1/k for large k: diverges by comparison with the harmonic series.
B: Σ (−1)ᵏ/k → Alternating, with 1/k decreasing to 0: converges (conditionally).
C: Σ 2ᵏ/k² → Ratio test: L = 2 > 1, diverges.
D: Σ 1/(4k + 6) → Behaves like 1/(4k) for large k; compared with the harmonic series it diverges.
The answer
B: Σ (−1)ᵏ/k
Answer
B: Σ (−1)ᵏ/k
4.Sequences & series
Answer: S = 0.3679
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.
Standard series
Σ xᵏ/k! = eˣ
Substitute
x = −1
S = e−1 = 0.3679
Answer
S = 0.3679
5.Multivariable calculus
Answer: Duf = −1.101
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 = 4x/(x² + 3y²) → fx(−1, 2) = −0.3077
fy = 12y/(x² + 3y²) → fy(−1, 2) = 1.846
Unit vector
|v| = √(4² + (−2)²) = 4.472
u = (0.8944, −0.4472)
Dot product
Duf = (−0.3077)·0.8944 + 1.846·(−0.4472) = −1.101
Answer
Duf = −1.101
6.Multivariable calculus
Answer: A (2, 0)
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) = 0
(2, 0), (−2, 0)
Second-derivative test
(2, 0): fxx = 12, fyy = 2, fxy = 0, D = 24 → local minimum
(−2, 0): fxx = −12, fyy = 2, fxy = 0, D = −24 → saddle point
The point
A (2, 0): local minimum
Answer
A (2, 0)
7.Differential equations
Answer: T = 23.45 °C
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.
Solution of the cooling law
T(t) = 26 + (−23)·e−0.1t
Substitute t
T(22) = 26 + (−23)·0.1108 = 23.45 °C
Answer
T = 23.45 °C
8.Differential equations
Answer: y(3) = 10.94
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.