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→∞ (√(4x2 − 10x) − 2x).

f(x) = √(4x2 − 10x) − 2xx → ∞
L =
2.LimitsHard
xyx = 0

Find the limit L = limx→0 ln(1 + 3x)/sin(x), with x in radians.

f(x) = ln(1 + 3x)/sin(x)x → 0
L =
3.DerivativesMedium
xyx = 1

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

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

Let f(x) = x3 − 10.5x2 + 36x + 4. Find the x-coordinate x of the local maximum.

f(x) = x3 − 10.5x2 + 36x + 4
x =
5.OptimizationMedium
rh

A closed cylindrical can (with a top and a bottom) must hold 400 cm³. The metal used is proportional to the total surface area. Find the height h of the can that uses the least material.

V = 400 cm³
h = cm
6.OptimizationHard
p (€)R(p)15now

A theatre sells 340 tickets per show at €15 each. Market research shows that every €1 increase in the ticket price means 10 fewer tickets sold (and every €1 decrease, 10 more). Find the ticket price p that maximises revenue.

340 × €1510 / €1
p = €
7.IntegralsMedium
xy1y = 0.5x

The region under y = 0.5x from x = 0 to x = 1 is rotated about the x-axis. Find the volume V of the solid (a cone).

y = 0.5x0 ≤ x ≤ 1
V =
8.IntegralsHard
xy2y = 6xy = 3x²

The region between the line y = 6x and the parabola y = 3x² is rotated about the x-axis. Find the volume V of the solid.

y = 6xy = 3x²
V =

3Solutions

1.Limits

Answer: L = −2.5

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) = −10x/(√(4x2 − 10x) + 2x)

Divide by x

f(x) = −10/(√(4 + (−10)/x) + 2)

Let x → ∞

L = −10/(2 + 2) = −2.5

Answer

L = −2.5
2.Limits

Answer: L = 3

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).

L'Hôpital's rule

L = limx→0 [3/(1 + 3x)]/[cos x]

Substitute x = 0

L = 3/1 = 3

Answer

L = 3
3.Derivatives

Answer: f′(1) = 1.382

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(x) + x·cos(x)

Substitute

f′(1) = 0.8415 + 1·0.5403 = 1.382

Answer

f′(1) = 1.382
4.Derivatives

Answer: x = 3

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 − 21x + 36 = 3·(x2 − 7x + 12)

Solve f′(x) = 0

x = 3 or x = 4

Which one is which

f″(x) = 6x − 21: negative at a maximum, positive at a minimum.

f″(3) = −3, f″(4) = 3
x = 3

Answer

x = 3
5.Optimization

Answer: h = 7.986 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 r

h = 400/(πr²)
A(r) = 2πr² + 800/r

Set the derivative to zero

A″(r) = 4π + 1600/r³ > 0, so this is a minimum.

4πr − 800/r² = 0 → r = ∛(400/(2π)) = 3.993 cm

Height and area

h = 400/(π·3.993²) = 2r = 7.986 cm
A = 2π·3.993² + 800/3.993 = 300.5 cm²

Answer

h = 7.986 cm

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

6.Optimization

Answer: p = 24.5 €

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) = 490 − 10p
R(p) = 490p − 10p²

Set the derivative to zero

R″(p) = −20 < 0, so this is a maximum.

p = 490/20 = €24.5

Maximum revenue

n = 490 − 10·24.5 = 245
R = 24.5·245 = €6003

Answer

p = €24.5
7.Integrals

Answer: V = 0.2618

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 0.25x² dx

Integrate

V = π·0.25·1³/3 = 0.2618

Answer

V = 0.2618

Check: a cone of radius 0.5 and height 1 has (1/3)πr²h = 0.2618.

8.Integrals

Answer: V = 120.6

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

6x = 3x² → x = 0, x = 2

Washer method

V = π∫₀2 ((6x)² − (3x²)²) dx

Integrate

V = π(36·2³/3 − 9·2⁵/5) = π·38.4 = 120.6

Answer

V = 120.6

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