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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 = ∫210 3/x dx.
f(x) = 3/x[2, 10]
I =
2.IntegralsHard
Evaluate I = ∫₀0.5 x²·eˣ dx.
∫₀0.5 x²eˣ dx
I =
3.Sequences & seriesMedium
Find the sum S = Σk=1∞ 6/(k(k + 3)).
ak = 6/(k(k + 3))
S =
4.Sequences & seriesHard
Use the Maclaurin polynomial of degree 3 of f(x) = 1/(1 − x) to approximate f(0.5). Give P₃(0.5).
f(x) = 1/(1 − x)n = 3x = 0.5
P₃(0.5) =
5.Multivariable calculusMedium
For f(x, y) = 2x³y − 2y³ − 3xy, find the directional derivative Duf at (3, −1) in the direction of v = (3, −4).
f(x, y) = 2x³y − 2y³ − 3xyP = (3, −1)v = (3, −4)
Duf =
6.Multivariable calculusHard
Find the critical points of f(x, y) = −x³ + 3x − y², classify them, and give the value fmax of f at its local maximum.
f(x, y) = −x³ + 3x − y²
fmax =
7.Differential equationsMedium
A forged steel part at 360 °C is left to cool in a workshop at 22 °C. After 49 min its temperature is 51 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). How long after the start, t, does it take to reach 31 °C?
T0 = 360 °CTa = 22 °CT(49) = 51 °CT = 31 °C
t = min
8.Differential equationsHard
Solve y″ + y′ + 16.25y = 0 with y(0) = 1 and y′(0) = 0 (released from rest). Find y(0.5); angles in radians.
y″ + y′ + 16.25y = 0y(0) = 1y′(0) = 0
y(0.5) =
3Solutions
1.Integrals
Answer: I = 4.828
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) = 3·ln x
Evaluate at the limits
I = 3·(ln 10 − ln 2) = 3·ln 5 = 4.828
Answer
I = 4.828
2.Integrals
Answer: I = 0.0609
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
1.649·(1.25) − 1·2 = 0.0609
Answer
I = 0.0609
3.Sequences & series
Answer: S = 3.667
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.
Partial fractions
6/(k(k + 3)) = (6/3)·(1/k − 1/(k + 3))
Cancel
S = (6/3)·(1 + 1/2 + 1/3) = 3.667
Answer
S = 3.667
4.Sequences & series
Answer: P₃(0.5) = 1.875
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
P₃(x) = 1 + x + x² + x³
Substitute
P₃(0.5) = 1 + 0.5 + 0.25 + 0.125 = 1.875
Answer
P₃(0.5) = 1.875
The exact value is f(0.5) = 2; the error 0.125 shrinks when you take more terms or a point closer to 0.
5.Multivariable calculus
Answer: Duf = −61.8
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 = 6x²y − 3y → fx(3, −1) = −51
fy = 2x³ − 6y² − 3x → fy(3, −1) = 39
Unit vector
|v| = √(3² + (−4)²) = 5
u = (0.6, −0.8)
Dot product
Duf = (−51)·0.6 + 39·(−0.8) = −61.8
Answer
Duf = −61.8
6.Multivariable calculus
Answer: fmax = 2
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 − 3) = 0
fy = −(2y) = 0
(1, 0), (−1, 0)
Second-derivative test
(1, 0): fxx = −6, fyy = −2, fxy = 0, D = 12 → local maximum
(−1, 0): fxx = 6, fyy = −2, fxy = 0, D = −12 → saddle point
Value at the extremum
f(1, 0) = 2
Answer
fmax = 2
7.Differential equations
Answer: t = 72.35 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(338/29)/49 = 0.05012 min⁻¹
Time to reach the target
t = ln(338/9)/0.05012 = 72.35 min
Answer
t = 72.35 min
8.Differential equations
Answer: y(0.5) = −0.2356
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.