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
xy16y = x ln x

Evaluate I = ∫₁6 x·ln x dx.

∫₁6 x ln x dx
I =
2.IntegralsHard
xy−11

Find the area A of the region between the parabola y = 1 − x² and the x-axis.

y = 1 − x²
A =
3.Sequences & seriesMedium
k|aₖ₊₁/aₖ|123456789101112

Apply the ratio test to Σk=1∞ kᵏ/(2ᵏ·k!). Find L = lim |ak+1/ak|.

ak = kᵏ/(2ᵏ·k!)
L =
4.Sequences & seriesHard

f(x) = sin x is approximated at x = 0.5 by its Maclaurin polynomial of degree n. With the Lagrange bound M·|x|ⁿ⁺¹/(n + 1)! and M = 1, find the smallest degree n that guarantees an error below 0.0005.

f(x) = sin xx = 0.5M = 1ε = 0.0005
n =
5.Multivariable calculusMedium

z = x³e2y with x = 3t − 2 and y = −2t² − 1. Use the chain rule to find dz/dt at t = −1.

z = x³e2yx = 3t − 2y = −2t² − 1t = −1
dz/dt =
6.Multivariable calculusHard
xyx = 1.2y = x1.2

The integral I = ∫01.2 ∫y1.2 cos(x²) dx dy cannot be done in this order. Reverse the order of integration and evaluate it.

f = cos(x²)0 ≤ y ≤ x ≤ 1.2
I =
7.Differential equationsMedium
t (min)T (°C)9Ta​T = ?

A cup of coffee at 89 °C is left in a room at 20 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). With k = 0.06 min⁻¹, find its temperature T after 9 min.

T0 = 89 °CTa = 20 °Ck = 0.06 min⁻¹t = 9 min
T = °C
8.Differential equationsHard
ty1y(1) = ?

Solve y′ + 0.5y = 5e−2t with y(0) = 5. Find y(1).

y′ + 0.5y = 5e−2ty(0) = 5
y(1) =

3Solutions

1.Integrals

Answer: I = 23.5

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

(18·ln 6 − 9) − (0 − 0.25) = 23.5

Answer

I = 23.5
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.

Zeros

1 − x² = 0 → x = ±1

Integrate

A = [1x − x³/3]−11 = 0.6667 − (−0.6667) = 1.333

Answer

A = 1.333
3.Sequences & series

Answer: L = 1.359

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 + 1/k)ᵏ/2 → e/2

Conclusion

L = 1.359 > 1

Answer

L = 1.359

L > 1, so the series diverges.

4.Sequences & series

Answer: n = 4

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.

n(n + 1)!bound010.5120.125260.020833240.00260441200.0002604

Bounds for increasing n

n = 3: 0.002604 ≥ 0.0005
n = 4: 0.0002604 < 0.0005

Smallest degree

n = 4

Answer

n = 4
5.Multivariable calculus

Answer: dz/dt = −1.921

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.

Point and rates

x(−1) = −5, y(−1) = −3
dx/dt = 3, dy/dt = −4t = 4

Partial derivatives

fx = 3x²e2y = 0.1859
fy = 2x³e2y = −0.6197

Chain rule

dz/dt = 0.1859·3 + (−0.6197)·4 = −1.921

Answer

dz/dt = −1.921
6.Multivariable calculus

Answer: I = 0.4957

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).

Region

0 ≤ y ≤ x, 0 ≤ x ≤ 1.2

Reversed order

I = ∫01.2 ∫0x cos(x²) dy dx = ∫01.2 x·cos(x²) dx

Substitution u = x²

I = [sin(x²)/2]01.2 = 0.4957

Answer

I = 0.4957
7.Differential equations

Answer: T = 60.21 °C

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.

Solution of the cooling law

T(t) = 20 + 69·e−0.06t

Substitute t

T(9) = 20 + 69·0.5827 = 60.21 °C

Answer

T = 60.21 °C
8.Differential equations

Answer: y(1) = 4.603

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 = e0.5t
(e0.5ty)′ = 5e−1.5t

General solution

y = 5/(−1.5)·e−2t + C·e−0.5t = −3.333·e−2t + C·e−0.5t

Initial condition

5 = −3.333 + C → C = 8.333

Evaluate

y(1) = −3.333·0.1353 + 8.333·0.6065 = 4.603

Answer

y(1) = 4.603

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