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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 = ∫₀π x·cos x dx (x in radians).
∫₀π x·cos x dx
I =
2.IntegralsHard
Find the area A of the region enclosed by y = x2 and y = 2x.
y = x2y = 2x
A =
3.Sequences & seriesMedium
Apply the ratio test to Σk=1∞ (k!)²·1ᵏ/(2k)!. Find L = lim |ak+1/ak|.
ak = (k!)²·1ᵏ/(2k)!
L =
4.Sequences & seriesHard
Use the Maclaurin polynomial of degree 3 of f(x) = sin x to approximate f(1.5). Give P₃(1.5).
f(x) = sin xn = 3x = 1.5
P₃(1.5) =
5.Multivariable calculusMedium
Use the tangent plane of f(x, y) = 2xy³ − 5x² + 5y³ at (2, 1) to approximate f(2.2, 1.2). Give the linear approximation L.
Evaluate I = ∬T (x + 3y) dA over the triangle T bounded by y = 0, x = 3 and y = 3x.
f = x + 3y0 ≤ x ≤ 30 ≤ y ≤ 3x
I =
7.Differential equationsMedium
A can of soda at 4 °C is taken out of the fridge into a room at 22 °C. After 17 min its temperature is 17 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). How long after the start, t, does it take to reach 11 °C?
T0 = 4 °CTa = 22 °CT(17) = 17 °CT = 11 °C
t = min
8.Differential equationsHard
A tank holds 100 L of water in which 20 kg of salt is dissolved. Brine with 0.25 kg of salt per litre flows in at 10 L/min, and the well-mixed solution flows out at the same rate. How much salt m is in the tank after 20 min?
V = 100 Lm0 = 20 kgc = 0.25 kg/Lr = 10 L/min
m = kg
3Solutions
1.Integrals
Answer: I = −2
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 = cos x dx, v = sin x
Antiderivative
∫x cos x dx = x sin x − ∫sin x dx = x sin x + cos x
Limits
(π·sin π + cos π) − (0 + 1) = −2
Answer
I = −2
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.
Intersection points
x2 = 2x → x = 0, x = 2
Integrate upper minus lower
A = [x2 − 0.3333x3]02 = 1.333
Answer
A = 1.333
3.Sequences & series
Answer: L = 0.25
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| = 1·(k + 1)²/((2k + 2)(2k + 1)) → 1/4
Conclusion
L = 0.25 < 1
Answer
L = 0.25
L < 1, so the series converges.
4.Sequences & series
Answer: P₃(1.5) = 0.9375
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) = x − 0.1667x³
Substitute
P₃(1.5) = 1.5 − 0.5625 = 0.9375
Answer
P₃(1.5) = 0.9375
The exact value is f(1.5) = 0.997495; the error 0.059995 shrinks when you take more terms or a point closer to 0.
5.Multivariable calculus
Answer: L = −9.2
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.
Value and slopes at the point
f(2, 1) = −11
fx = 2y³ − 10x → fx(2, 1) = −18
fy = 6xy² + 15y² → fy(2, 1) = 27
Tangent plane
z = −11 − 18(x − 2) + 27(y − 1)
z = −18x + 27y − 2
Linear approximation
L = −11 + (−18)·0.2 + 27·0.2 = −9.2
Answer
L = −9.2
The exact value is f = −7.9568; the plane is close because the step is small.
6.Multivariable calculus
Answer: I = 148.5
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).
Limits
0 ≤ x ≤ 3, 0 ≤ y ≤ 3x
Inner integral
∫03x (x + 3y) dy = x·3x + 3·(3x)²/2 = 16.5x²
Outer integral
I = 16.5·3³/3 = 148.5
Answer
I = 148.5
7.Differential equations
Answer: t = 6.536 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(-18/-5)/17 = 0.07535 min⁻¹
Time to reach the target
t = ln(-18/(−11))/0.07535 = 6.536 min
Answer
t = 6.536 min
8.Differential equations
Answer: m = 24.32 kg
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.
Set up the equation
dm/dt = 10·0.25 − (10/100)·m = 2.5 − 0.1·m
Solve
The integrating factor is ert/V; the solution approaches the equilibrium c·V.