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
xy03

Evaluate I = ∫03 ex dx.

f(x) = ex[0, 3]
I =
2.IntegralsHard
xy4y = 4xy = x²

The region between the line y = 4x and the parabola y = x² is rotated about the x-axis. Find the volume V of the solid.

y = 4xy = x²
V =
3.Sequences & seriesMedium
k|aₖ|¹ᐟᵏ123456789101112

Apply the root test to Σk=1∞ ((3k + 3)/(5k + 4))ᵏ. Find L = lim |ak|1/k.

ak = ((3k + 3)/(5k + 4))ᵏ
L =
4.Sequences & seriesHard

f(x) = ln(1 + x) is approximated by its Maclaurin polynomial of degree 3 at x = 0.5. Use the Lagrange error bound M·|x|ⁿ⁺¹/(n + 1)! with M = 3! = 6 to find the largest possible error |R₃|.

f(x) = ln(1 + x)n = 3x = 0.5M = 6
|R₃| =
5.Multivariable calculusMedium

z = 2 ln(x² + 2y²) with x = −2t − 1 and y = −2t² − 2. Use the chain rule to find dz/dt at t = 0.

z = 2 ln(x² + 2y²)x = −2t − 1y = −2t² − 2t = 0
dz/dt =
6.Multivariable calculusHard
xyy = xy = x²11

Find the area I = ∬R 1 dA of the region R between the parabola y = x² and the line y = x.

y = x²y = xf = 1
I =
7.Differential equationsMedium
t (min)T (°C)12Ta​T = ?

A cup of coffee at 82 °C is left in a room at 19 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). With k = 0.06 min⁻¹, find its temperature T after 12 min.

T0 = 82 °CTa = 19 °Ck = 0.06 min⁻¹t = 12 min
T = °C
8.Differential equationsHard
ty2y(2) = ?

Solve y′ + 3y = 5e−t with y(0) = 4. Find y(2).

y′ + 3y = 5e−ty(0) = 4
y(2) =

3Solutions

1.Integrals

Answer: I = 19.09

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.

Antiderivative

F(x) = ex/1

Evaluate at the limits

I = (e3 − 1)/1 = (20.09 − 1)/1 = 19.09

Answer

I = 19.09
2.Integrals

Answer: V = 428.9

Volumes of revolution: rotating the region under y = f(x) from x = a to x = b about the x-axis makes a solid whose cross-section at x is a disc of radius f(x). Adding the discs gives V = π∫ₐᵇ f(x)² dx. When the region lies between an outer curve R(x) and an inner curve r(x), each slice is a washer: V = π∫ₐᵇ (R(x)² − r(x)²) dx. Square the radii before you subtract them.

Intersection points

4x = 1x² → x = 0, x = 4

Washer method

V = π∫₀4 ((4x)² − (1x²)²) dx

Integrate

V = π(16·4³/3 − 1·4⁵/5) = π·136.5 = 428.9

Answer

V = 428.9
3.Sequences & series

Answer: L = 0.6

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.

Root of the terms

|ak|1/k = (3k + 3)/(5k + 4) → 3/5

Conclusion

L = 0.6 < 1

Answer

L = 0.6

L < 1, so the series converges.

4.Sequences & series

Answer: |R₃| = 0.01563

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 = 3! = 6

Lagrange bound

|R₃| ≤ 6·0.5⁴/4! = 0.01563

Answer

|R₃| = 0.01563

The true error is smaller: |f(0.5) − P₃(0.5)| = 0.0112.

5.Multivariable calculus

Answer: dz/dt = 0.8889

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(0) = −1, y(0) = −2
dx/dt = −2, dy/dt = −4t = 0

Partial derivatives

fx = 4x/(x² + 2y²) = −0.4444
fy = 8y/(x² + 2y²) = −1.778

Chain rule

dz/dt = (−0.4444)·(−2) + (−1.778)·0 = 0.8889

Answer

dz/dt = 0.8889
6.Multivariable calculus

Answer: I = 0.1667

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 = x → x = 0, x = 1

Inner integral

∫x²x 1 dy = x − x²

Outer integral

I = ∫01 (x − x²) dx = 0.1667

Answer

I = 0.1667
7.Differential equations

Answer: T = 49.67 °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) = 19 + 63·e−0.06t

Substitute t

T(12) = 19 + 63·0.4868 = 49.67 °C

Answer

T = 49.67 °C
8.Differential equations

Answer: y(2) = 0.3421

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 = e3t
(e3ty)′ = 5e2t

General solution

y = 5/2·e−t + C·e−3t = 2.5·e−t + C·e−3t

Initial condition

4 = 2.5 + C → C = 1.5

Evaluate

y(2) = 2.5·0.1353 + 1.5·0.002479 = 0.3421

Answer

y(2) = 0.3421

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