Calculus 1: 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

Limits

Direct substitution
try x = a first; 0/0 means simplify
Factor and cancel
xⁿ − aⁿ = (x − a)(xⁿ⁻¹ + a·xⁿ⁻² + … + aⁿ⁻¹)
Conjugate
(√x − √a)(√x + √a) = x − a
Limits at infinity
divide by the highest power of x; equal degrees give the ratio of leading coefficients
Special limit
limx→∞ (1 + a/x)bx = eab
L'Hôpital's rule
lim f/g = lim f′/g′ (0/0 or ∞/∞)

Derivatives

Power rule
(xⁿ)′ = n·xⁿ⁻¹
Product rule
(uv)′ = u′v + uv′
Quotient rule
(u/v)′ = (u′v − uv′)/v²
Chain rule
f(g(x))′ = f′(g(x))·g′(x)
Common derivatives
(eˣ)′ = eˣ, (ln x)′ = 1/x, (sin x)′ = cos x, (cos x)′ = −sin x

Method

  1. Identify the structure: sum, product, quotient or composition.
  2. Apply the power rule to each power of x.
  3. Use the product, quotient or chain rule where needed.
  4. Simplify, and evaluate at the point if asked.

Optimization

Method
write Q(x) with one variable, solve Q′(x) = 0, check Q″ or the endpoints
Maximum
Q′(x) = 0, Q″(x) < 0
Minimum
Q′(x) = 0, Q″(x) > 0
Closed cylinder
V = πr²h, A = 2πr² + 2πrh
Profit
P = R − C, R = p·q, maximum where MR = MC

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).

2Practice problems

1.LimitsMedium
xyy = L

Find the limit L = limx→∞ (−x2 − 3x − 2)/(3x2 + 3x + 5).

f(x) = (−x2 − 3x − 2)/(3x2 + 3x + 5)x → ∞
L =
2.LimitsHard
xyx = 0

Find the limit L = limx→0 (1 − cos 3x)/(4x²), with x in radians.

f(x) = (1 − cos 3x)/(4x²)x → 0
L =
3.DerivativesMedium
xyx = 1

Let f(x) = (−2x + 1)3. Find f′(1).

f(x) = (−2x + 1)3x = 1
f′(1) =
4.DerivativesHard
xyx = 3

The tangent line to f(x) = 0.5x2 + 2x − 4 at x = 3 crosses the y-axis at y = b. Find b.

f(x) = 0.5x2 + 2x − 4x = 3
b =
5.OptimizationMedium
rh

A closed cylindrical can (with a top and a bottom) must hold 500 cm³. The metal used is proportional to the total surface area. Find the smallest possible surface area A.

V = 500 cm³
A = cm²
6.OptimizationHard
qP(q)maximum

A company sells q units per week. The price at which it can sell them is p = 140 − 2q euros per unit, and the cost is C(q) = 800 + 10q euros. Find the weekly output q that maximises profit.

p = 140 − 2qC = 800 + 10q
q = units
7.IntegralsMedium
xy4y = √(x)

The region under y = √(x) from x = 0 to x = 4 is rotated about the x-axis. Find the volume V of the solid (a paraboloid).

y = √(x)0 ≤ x ≤ 4
V =
8.IntegralsHard
xy4y = 2xy = 0.5x²

The region between the line y = 2x and the parabola y = 0.5x² is rotated about the x-axis. Find the volume V of the solid.

y = 2xy = 0.5x²
V =

3Solutions

1.Limits

Answer: L = −0.3333

Limits: first try direct substitution. If it gives 0/0, simplify first: factor and cancel the common factor, or multiply by the conjugate. As x → ∞, divide numerator and denominator by the highest power of x. L'Hôpital's rule: if f/g gives 0/0 or ∞/∞, then lim f/g = lim f′/g′, differentiating the numerator and the denominator separately (not with the quotient rule).

Divide by the highest power

(−1 + (−3)/x + (−2)/x2)/(3 + 3/x + 5/x2)

Let x → ∞

L = −1/3 = −0.3333

Answer

L = −0.3333

For equal degrees the limit is the ratio of the leading coefficients.

2.Limits

Answer: L = 1.125

Limits: first try direct substitution. If it gives 0/0, simplify first: factor and cancel the common factor, or multiply by the conjugate. As x → ∞, divide numerator and denominator by the highest power of x. L'Hôpital's rule: if f/g gives 0/0 or ∞/∞, then lim f/g = lim f′/g′, differentiating the numerator and the denominator separately (not with the quotient rule).

First application

3·sin 3x/(8x) → 0/0

Second application

9·cos 3x/8

Substitute x = 0

L = 9/8 = 1.125

Answer

L = 1.125
3.Derivatives

Answer: f′(1) = −6

Differentiation: the derivative f′(x) is the slope of the graph at x. Power rule: (xⁿ)′ = n·xⁿ⁻¹. Product rule: (uv)′ = u′v + uv′. Chain rule: (f(g(x)))′ = f′(g(x))·g′(x). Where f′(x) = 0 the graph has a horizontal tangent: a maximum, minimum or saddle point.

Chain rule

f′(x) = 3·(−2x + 1)2·(−2)

Substitute

f′(1) = 3·(−1)2·(−2) = −6

Answer

f′(1) = −6
4.Derivatives

Answer: b = −8.5

Differentiation: the derivative f′(x) is the slope of the graph at x. Power rule: (xⁿ)′ = n·xⁿ⁻¹. Product rule: (uv)′ = u′v + uv′. Chain rule: (f(g(x)))′ = f′(g(x))·g′(x). Where f′(x) = 0 the graph has a horizontal tangent: a maximum, minimum or saddle point.

Point and slope

f(3) = 6.5
f′(x) = x + 2 → f′(3) = 5

Tangent line at x = 0

b = f(x₀) − f′(x₀)·x₀ = 6.5 − 5·3 = −8.5

Answer

b = −8.5
5.Optimization

Answer: A = 348.7 cm²

Optimization: write the quantity to optimise as a function of one variable, using the constraint to eliminate the other. Set the derivative to zero and solve. Check that the critical point is a maximum or minimum (second derivative or sign change of the first derivative) and that it lies inside the allowed range; an endpoint such as a zero side length is not the answer.

Area as a function of r

h = 500/(πr²)
A(r) = 2πr² + 1000/r

Set the derivative to zero

A″(r) = 4π + 2000/r³ > 0, so this is a minimum.

4πr − 1000/r² = 0 → r = ∛(500/(2π)) = 4.301 cm

Height and area

h = 500/(π·4.301²) = 2r = 8.603 cm
A = 2π·4.301² + 1000/4.301 = 348.7 cm²

Answer

A = 348.7 cm²

The cheapest closed can is as tall as it is wide: h = 2r.

6.Optimization

Answer: q = 32.5 units

Optimization: write the quantity to optimise as a function of one variable, using the constraint to eliminate the other. Set the derivative to zero and solve. Check that the critical point is a maximum or minimum (second derivative or sign change of the first derivative) and that it lies inside the allowed range; an endpoint such as a zero side length is not the answer.

Profit function

P(q) = (140 − 2q)·q − (800 + 10q) = −2q2 + 130q − 800

Set the derivative to zero

P″(q) = −4 < 0, so this is a maximum.

q = 130/4 = 32.5

Price and profit

p = 140 − 2·32.5 = €75
P = 130·32.5 − 2·32.5² − 800 = €1313

Answer

q = 32.5
7.Integrals

Answer: V = 25.13

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∫₀4 x dx

Integrate

V = π·1·4²/2 = 25.13

Answer

V = 25.13
8.Integrals

Answer: V = 107.2

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

2x = 0.5x² → x = 0, x = 4

Washer method

V = π∫₀4 ((2x)² − (0.5x²)²) dx

Integrate

V = π(4·4³/3 − 0.25·4⁵/5) = π·34.13 = 107.2

Answer

V = 107.2

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