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 + 8x) − x).

f(x) = √(x2 + 8x) − xx → ∞
L =
2.LimitsHard
xyx = 0

Find the limit L = limx→0 (1 − cos 5x)/(3x²), with x in radians.

f(x) = (1 − cos 5x)/(3x²)x → 0
L =
3.DerivativesMedium
xyx = 1

Let f(x) = x·sin(3x), with x in radians. Find f′(1).

f(x) = x·sin(3x)x = 1 rad
f′(1) =
4.DerivativesHard
xy

Let f(x) = x3 + 7.5x2 + 18x + 3. Find the x-coordinate x of the local minimum.

f(x) = x3 + 7.5x2 + 18x + 3
x =
5.OptimizationMedium
xhxno lid

An open box (no lid) with a square base must hold 32000 cm³. Use the least possible material, that is the smallest area of the base plus the four sides. Find the height h.

V = 32000 cm³
h = cm
6.OptimizationHard
p (€)R(p)15now

A theatre sells 675 tickets per show at €15 each. Market research shows that every €1 increase in the ticket price means 25 fewer tickets sold (and every €1 decrease, 25 more). Find the maximum revenue R per show.

675 × €1525 / €1
R = €
7.IntegralsMedium
xy2y = 3x

The region under y = 3x from x = 0 to x = 2 is rotated about the x-axis. Find the volume V of the solid (a cone).

y = 3x0 ≤ x ≤ 2
V =
8.IntegralsHard
xyy = x2​y = 2x

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

y = x2y = 2x
A =

3Solutions

1.Limits

Answer: L = 4

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).

Multiply by the conjugate

f(x) = 8x/(√(x2 + 8x) + x)

Divide by x

f(x) = 8/(√(1 + 8/x) + 1)

Let x → ∞

L = 8/(1 + 1) = 4

Answer

L = 4
2.Limits

Answer: L = 4.167

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

5·sin 5x/(6x) → 0/0

Second application

25·cos 5x/6

Substitute x = 0

L = 25/6 = 4.167

Answer

L = 4.167
3.Derivatives

Answer: f′(1) = −2.829

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.

Product and chain rule

f′(x) = sin(3x) + 3x·cos(3x)

Substitute

f′(1) = 0.1411 + 3·(−0.99) = −2.829

Answer

f′(1) = −2.829
4.Derivatives

Answer: x = −2

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.

Derivative

f′(x) = 3x2 + 15x + 18 = 3·(x2 + 5x + 6)

Solve f′(x) = 0

x = −3 or x = −2

Which one is which

f″(x) = 6x + 15: negative at a maximum, positive at a minimum.

f″(−3) = −3, f″(−2) = 3
x = −2

Answer

x = −2
5.Optimization

Answer: h = 20 cm

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 x

S(x) = x² + 128000/x

Set the derivative to zero

S″(x) = 2 + 256000/x³ > 0, so this is a minimum.

x³ = 64000 → x = 40 cm

Height and area

h = 32000/40² = 20 cm
S = 40² + 128000/40 = 4800 cm²

Answer

h = 20 cm

The best open box is half as tall as it is wide.

6.Optimization

Answer: R = 11025 €

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.

Revenue as a function of price

n(p) = 1050 − 25p
R(p) = 1050p − 25p²

Set the derivative to zero

R″(p) = −50 < 0, so this is a maximum.

p = 1050/50 = €21

Maximum revenue

n = 1050 − 25·21 = 525
R = 21·525 = €11025

Answer

R = €11025
7.Integrals

Answer: V = 75.4

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∫₀2 9x² dx

Integrate

V = π·9·2³/3 = 75.4

Answer

V = 75.4

Check: a cone of radius 6 and height 2 has (1/3)πr²h = 75.4.

8.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.

Intersection points

x2 = 2x → x = 0, x = 2

Integrate upper minus lower

A = [x2 − 0.3333x3]02 = 1.333

Answer

A = 1.333

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