Calculus 2: cram pack

Formula sheets, the key method steps for every topic and 8 practice problems, with full solutions at the back.

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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

  1. Rewrite the integrand as a sum of simple terms where possible.
  2. Apply the power rule: ∫xⁿ dx = xⁿ⁺¹/(n + 1), and standard integrals for ex, 1/x, sin and cos.
  3. Use substitution when you see a function and its derivative together.
  4. 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

  1. Identify the type: arithmetic (constant difference), geometric (constant ratio), a standard series, a power series or a Taylor polynomial.
  2. 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.
  3. 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.
  4. 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

  1. For a partial derivative, treat the other variable as a constant and use the one-variable rules.
  2. 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.
  3. For extrema solve fx = 0 and fy = 0 together, then classify every point with D = fxx·fyy − fxy².
  4. 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
xy13y = x ln x

Evaluate I = ∫₁3 x·ln x dx.

∫₁3 x ln x dx
I =
2.IntegralsHard
xy−33

Find the area A of the region between the parabola y = 9 − x² and the x-axis.

y = 9 − x²
A =
3.Sequences & seriesMedium
AΣ k/3ᵏk = 1 … ∞BΣ (3/4)ᵏk = 1 … ∞CΣ (5/4)ᵏk = 1 … ∞DΣ (−1)ᵏ/kk = 1 … ∞

Exactly one of the four series diverges. Click it.

A: Σ k/3ᵏB: Σ (3/4)ᵏC: Σ (5/4)ᵏD: Σ (−1)ᵏ/k
  1. 1A: Σ k/3ᵏ
  2. 2B: Σ (3/4)ᵏ
  3. 3C: Σ (5/4)ᵏ
  4. 4D: Σ (−1)ᵏ/k
4.Sequences & seriesHard

f(x) = sin x is approximated by its Maclaurin polynomial of degree 4 at x = −0.5. Use the Lagrange error bound M·|x|ⁿ⁺¹/(n + 1)! with M = 1 to find the largest possible error |R₄|.

f(x) = sin xn = 4x = −0.5M = 1
|R₄| =
5.Multivariable calculusMedium
xy−22∇fP

For f(x, y) = −x³y² + 3x² + 4y² + 2xy, find the maximum rate of change of f at the point (−2, 2), that is |∇f|.

f(x, y) = −x³y² + 3x² + 4y² + 2xyP = (−2, 2)
|∇f| =
6.Multivariable calculusHard
xyy = 4xy = x²416

Evaluate I = ∬R xy dA, where R is the region between the parabola y = x² and the line y = 4x.

y = x²y = 4xf = xy
I =
7.Differential equationsMedium
t (min)T (°C)5Ta​10 °C

A can of soda at 4 °C is taken out of the fridge into a room at 22 °C. After 5 min its temperature is 10 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). Find the constant k in min⁻¹.

T0 = 4 °CTa = 22 °CT(5) = 10 °C
k = min⁻¹
8.Differential equationsHard
ty1.5y(1.5) = ?

Solve y′ + 2y = 6e−0.5t with y(0) = 5. Find y(1.5).

y′ + 2y = 6e−0.5ty(0) = 5
y(1.5) =

3Solutions

1.Integrals

Answer: I = 2.944

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 = ln x, du = dx/x; dv = x dx, v = x²/2

Antiderivative

∫x ln x dx = (x²/2)ln x − ∫(x/2) dx = (x²/2)ln x − x²/4

Limits

(4.5·ln 3 − 2.25) − (0 − 0.25) = 2.944

Answer

I = 2.944
2.Integrals

Answer: A = 36

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.

Zeros

9 − x² = 0 → x = ±3

Integrate

A = [9x − x³/3]−33 = 18 − (−18) = 36

Answer

A = 36
3.Sequences & series

Answer: C: Σ (5/4)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 k/3ᵏ → Ratio test: L = 1/3 < 1, converges.
B: Σ (3/4)ᵏ → Geometric with r = 3/4, |r| < 1: converges.
C: Σ (5/4)ᵏ → Geometric with r = 5/4 > 1: the terms grow, so it diverges.
D: Σ (−1)ᵏ/k → Alternating, with 1/k decreasing to 0: converges (conditionally).

The answer

C: Σ (5/4)ᵏ

Answer

C: Σ (5/4)ᵏ
4.Sequences & series

Answer: |R₄| = 0.0002604

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 = 1

Lagrange bound

|R₄| ≤ 1·0.5⁵/5! = 0.0002604

Answer

|R₄| = 0.0002604

The true error is smaller: |f(−0.5) − P₄(−0.5)| = 0.0002589.

5.Multivariable calculus

Answer: |∇f| = 71.22

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.

Partial derivatives

fx = −3x²y² + 6x + 2y → fx(−2, 2) = −56
fy = −2x³y + 8y + 2x → fy(−2, 2) = 44

Length of the gradient

|∇f| = √((−56)² + 44²) = 71.22

Answer

|∇f| = 71.22
6.Multivariable calculus

Answer: I = 170.7

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).

Intersections

x² = 4x → x = 0, x = 4

Inner integral

∫x²4x xy dy = x·(16x² − x⁴)/2

Outer integral

I = ∫04 (16x³ − x⁵)/2 dx = 170.7

Answer

I = 170.7
7.Differential equations

Answer: k = 0.08109 min⁻¹

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.

Ratio of temperature differences

e−5k = -12/-18

Solve for k

k = ln(-18/-12)/5 = 0.08109 min⁻¹

Answer

k = 0.08109 min⁻¹
8.Differential equations

Answer: y(1.5) = 1.939

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.

Integrating factor

μ = e2t
(e2ty)′ = 6e1.5t

General solution

y = 6/1.5·e−0.5t + C·e−2t = 4·e−0.5t + C·e−2t

Initial condition

5 = 4 + C → C = 1

Evaluate

y(1.5) = 4·0.4724 + 1·0.04979 = 1.939

Answer

y(1.5) = 1.939

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