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
xy01

Evaluate I = ∫01 2x·(x² + 1)2 dx.

f(x) = 2x·(x² + 1)2[0, 1]
I =
2.IntegralsHard
xy2y = x²eˣ

Evaluate I = ∫₀2 x²·eˣ dx.

∫₀2 x²eˣ dx
I =
3.Sequences & seriesMedium
kSn​12345678910S = ?

Find the sum S = Σk=1∞ 2/(k(k + 3)).

ak = 2/(k(k + 3))
S =
4.Sequences & seriesHard
xyfP₂0.8

Use the Maclaurin polynomial of degree 2 of f(x) = e−x to approximate f(0.8). Give P₂(0.8).

f(x) = e−xn = 2x = 0.8
P₂(0.8) =
5.Multivariable calculusMedium
xy−23P

The tangent plane to z = 4x³y³ − 2x² + 2y³ − xy at the point where (x, y) = (−2, 3) can be written as z = px + qy + c. Find c.

f(x, y) = 4x³y³ − 2x² + 2y³ − xy(a, b) = (−2, 3)
c =
6.Multivariable calculusHard
xy−50−4−202AB

The marked points are the critical points of f(x, y) = x³ + y³ + 9xy. Click the local maximum.

f(x, y) = x³ + y³ + 9xy
  1. 1A (−3, −3)
  2. 2B (0, 0)
7.Differential equationsMedium
t (min)T (°C)8Ta​19 °C

A can of soda at 6 °C is taken out of the fridge into a room at 30 °C. After 8 min its temperature is 19 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). Find the constant k in min⁻¹.

T0 = 6 °CTa = 30 °CT(8) = 19 °C
k = min⁻¹
8.Differential equationsHard
xy12(1, −1)y(2) = ?

Solve the initial value problem y′ + 3y/x = 3x2 for x > 0, with y(1) = −1. Find y(2).

y′ + 3y/x = 3x2y(1) = −1
y(2) =

3Solutions

1.Integrals

Answer: I = 2.333

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.

Substitution

u = x² + 1, du = 2x dx
I = ∫12 u2 du

Integrate and evaluate

I = [u3/3]12 = (8 − 1)/3 = 2.333

Answer

I = 2.333
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: S = 1.222

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

2/(k(k + 3)) = (2/3)·(1/k − 1/(k + 3))

Cancel

S = (2/3)·(1 + 1/2 + 1/3) = 1.222

Answer

S = 1.222
4.Sequences & series

Answer: P₂(0.8) = 0.52

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

P₂(x) = 1 − x + 0.5x²

Substitute

P₂(0.8) = 1 − 0.8 + 0.32 = 0.52

Answer

P₂(0.8) = 0.52

The exact value is f(0.8) = 0.449329; the error −0.070671 shrinks when you take more terms or a point closer to 0.

5.Multivariable calculus

Answer: c = 4214

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, 3) = −812
fx = 12x²y³ − 4x − y → fx(−2, 3) = 1301
fy = 12x³y² + 6y² − x → fy(−2, 3) = −808

Tangent plane

z = −812 + 1301(x + 2) − 808(y − 3)
z = 1301x − 808y + 4214

Constant term

c = 4214

Answer

c = 4214
6.Multivariable calculus

Answer: A (−3, −3)

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 + 9y = 0
fy = 3y² + 9x = 0
(0, 0), (−3, −3)

Second-derivative test

(0, 0): fxx = 0, fyy = 0, fxy = 9, D = −81 → saddle point
(−3, −3): fxx = −18, fyy = −18, fxy = 9, D = 243 → local maximum

The point

A (−3, −3): local maximum

Answer

A (−3, −3)
7.Differential equations

Answer: k = 0.09752 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−8k = -11/-24

Solve for k

k = ln(-24/-11)/8 = 0.09752 min⁻¹

Answer

k = 0.09752 min⁻¹
8.Differential equations

Answer: y(2) = 3.813

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 = e∫3/x dx = x3

Integrate

(x3y)′ = 3x5 → x3y = 3x6/6 + C

Initial condition

−1 = 3/6 + C → C = −1.5

Evaluate

y(2) = (3·26/6 + (−1.5))/23 = 3.813

Answer

y(2) = 3.813

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