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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/3 x·cos x dx (x in radians).
∫₀π/3 x·cos x dx
I =
2.IntegralsHard
Evaluate I = ∫₀2 x²·eˣ dx.
∫₀2 x²eˣ dx
I =
3.Sequences & seriesMedium
Exactly one of the four series converges. Click it.
A: Σ k/(k + 3)B: Σ k³/2ᵏC: Σ 1/(2k + 2)D: Σ 2ᵏ/k²
1A: Σ k/(k + 3)
2B: Σ k³/2ᵏ
3C: Σ 1/(2k + 2)
4D: Σ 2ᵏ/k²
4.Sequences & seriesHard
f(x) = ln(1 + x) is approximated by its Maclaurin polynomial of degree 2 at x = 0.3. Use the Lagrange error bound M·|x|ⁿ⁺¹/(n + 1)! with M = 2! = 2 to find the largest possible error |R₂|.
f(x) = ln(1 + x)n = 2x = 0.3M = 2
|R₂| =
5.Multivariable calculusMedium
The tangent plane to z = −2x²ey at the point where (x, y) = (−1, 0) can be written as z = px + qy + c. Find c.
f(x, y) = −2x²ey(a, b) = (−1, 0)
c =
6.Multivariable calculusHard
Use polar coordinates to evaluate I = ∬R e−(x² + y²) dA, where R is the disc x² + y² ≤ 1.
f = e−(x² + y²)0 ≤ r ≤ 10 ≤ θ ≤ 2π
I =
7.Differential equationsMedium
A can of soda at 4 °C is taken out of the fridge into a room at 26 °C. After 11 min its temperature is 16 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). How long after the start, t, does it take to reach 12 °C?
T0 = 4 °CTa = 26 °CT(11) = 16 °CT = 12 °C
t = min
8.Differential equationsHard
Solve the initial value problem y′ + 3y/x = x for x > 0, with y(1) = 3. Find y(3).
y′ + 3y/x = xy(1) = 3
y(3) =
3Solutions
1.Integrals
Answer: I = 0.4069
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/3·sin π/3 + cos π/3) − (0 + 1) = 0.4069
Answer
I = 0.4069
2.Integrals
Answer: I = 12.78
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
7.389·(2) − 1·2 = 12.78
Answer
I = 12.78
3.Sequences & series
Answer: B: Σ k³/2ᵏ
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/(k + 3) → The terms tend to 1, not 0: diverges by the n-th term test.
B: Σ k³/2ᵏ → Ratio test: L = 1/2 < 1, converges.
C: Σ 1/(2k + 2) → Behaves like 1/(2k) for large k; compared with the harmonic series it diverges.
D: Σ 2ᵏ/k² → Ratio test: L = 2 > 1, diverges.
The answer
B: Σ k³/2ᵏ
Answer
B: Σ k³/2ᵏ
4.Sequences & series
Answer: |R₂| = 0.009
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.
Bound on the derivative
M = 2! = 2
Lagrange bound
|R₂| ≤ 2·0.3³/3! = 0.009
Answer
|R₂| = 0.009
The true error is smaller: |f(0.3) − P₂(0.3)| = 0.007364.
5.Multivariable calculus
Answer: c = 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(−1, 0) = −2
fx = −4xey → fx(−1, 0) = 4
fy = −2x²ey → fy(−1, 0) = −2
Tangent plane
z = −2 + 4(x + 1) − 2(y − 0)
z = 4x − 2y + 2
Constant term
c = 2
Answer
c = 2
6.Multivariable calculus
Answer: I = 1.986
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).
Polar form
I = ∫02π ∫01 e−r²·r dr dθ
Integrate
∫ e−r²·r dr = [−e−r²/2]
I = 1.986
Answer
I = 1.986
7.Differential equations
Answer: t = 6.306 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(-22/-10)/11 = 0.07168 min⁻¹
Time to reach the target
t = ln(-22/(−14))/0.07168 = 6.306 min
Answer
t = 6.306 min
8.Differential equations
Answer: y(3) = 1.904
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.