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
xy1y = √(x)

The region under y = √(x) from x = 0 to x = 1 is rotated about the x-axis. Find the volume V of the solid (a paraboloid).

y = √(x)0 ≤ x ≤ 1
V =
2.IntegralsHard
xyy = 2x2​y = 4x

Find the area A of the region enclosed by y = 2x2 and y = 4x.

y = 2x2y = 4x
A =
3.Sequences & seriesMedium
kSn​12345678910

Find the sum S = Σk=139 6/(k(k + 1)).

ak = 6/(k(k + 1))n = 39
S =
4.Sequences & seriesHard
xyfP₃0.4

Use the Maclaurin polynomial of degree 3 of f(x) = e3x to approximate f(0.4). Give P₃(0.4).

f(x) = e3xn = 3x = 0.4
P₃(0.4) =
5.Multivariable calculusMedium
xy21P

The tangent plane to z = 3xe0.5y at the point where (x, y) = (2, 1) can be written as z = px + qy + c. Find c.

f(x, y) = 3xe0.5y(a, b) = (2, 1)
c =
6.Multivariable calculusHard
xyx = 4y = x44

Evaluate I = ∬T (x + 4y) dA over the triangle T bounded by y = 0, x = 4 and y = x.

f = x + 4y0 ≤ x ≤ 40 ≤ y ≤ x
I =
7.Differential equationsMedium
t (min)T (°C)26Ta​392 °C138 °C

A forged steel part at 500 °C is left to cool in a workshop at 29 °C. After 26 min its temperature is 392 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). How long after the start, t, does it take to reach 138 °C?

T0 = 500 °CTa = 29 °CT(26) = 392 °CT = 138 °C
t = min
8.Differential equationsHard
ty2y(2) = ?

Solve y′ + 3y = 6e−0.5t with y(0) = 2. Find y(2).

y′ + 3y = 6e−0.5ty(0) = 2
y(2) =

3Solutions

1.Integrals

Answer: V = 1.571

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.

Disc method

V = π∫₀1 x dx

Integrate

V = π·1·1²/2 = 1.571

Answer

V = 1.571
2.Integrals

Answer: A = 2.667

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.

Intersection points

2x2 = 4x → x = 0, x = 2

Integrate upper minus lower

A = [2x2 − 0.6667x3]02 = 2.667

Answer

A = 2.667
3.Sequences & series

Answer: S = 5.85

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

6/(k(k + 1)) = 6·(1/k − 1/(k + 1))

Cancel

S = 6·(1 − 1/40) = 5.85

Answer

S = 5.85
4.Sequences & series

Answer: P₃(0.4) = 3.208

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

P₃(x) = 1 + 3x + 4.5x² + 4.5x³

Substitute

P₃(0.4) = 1 + 1.2 + 0.72 + 0.288 = 3.208

Answer

P₃(0.4) = 3.208

The exact value is f(0.4) = 3.320117; the error 0.112117 shrinks when you take more terms or a point closer to 0.

5.Multivariable calculus

Answer: c = −4.946

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, 1) = 9.892
fx = 3·e0.5y → fx(2, 1) = 4.946
fy = 1.5xe0.5y → fy(2, 1) = 4.946

Tangent plane

z = 9.892 + 4.946(x − 2) + 4.946(y − 1)
z = 4.946x + 4.946y − 4.946

Constant term

c = −4.946

Answer

c = −4.946
6.Multivariable calculus

Answer: I = 64

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).

Limits

0 ≤ x ≤ 4, 0 ≤ y ≤ x

Inner integral

∫0x (x + 4y) dy = x·x + 4·(x)²/2 = 3x²

Outer integral

I = 3·4³/3 = 64

Answer

I = 64
7.Differential equations

Answer: t = 146.1 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(471/363)/26 = 0.01002 min⁻¹

Time to reach the target

t = ln(471/109)/0.01002 = 146.1 min

Answer

t = 146.1 min
8.Differential equations

Answer: y(2) = 0.8819

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)′ = 6e2.5t

General solution

y = 6/2.5·e−0.5t + C·e−3t = 2.4·e−0.5t + C·e−3t

Initial condition

2 = 2.4 + C → C = −0.4

Evaluate

y(2) = 2.4·0.3679 + (−0.4)·0.002479 = 0.8819

Answer

y(2) = 0.8819

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