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IB Diploma Mathematics: Analysis & Approaches HL · Calculus

Mini-Lesson

Calculus

6 missions. Guess, learn, then check. Your place is saved.

Mission 1A car's distance graph curves. How fast is it at exactly 3 s?
Mission 2A tap's flow rate keeps changing. How much filled the bath?
Mission 3What shape of can uses the least metal for 330 ml?
Mission 4Is the derivative of $x^{2} \times x^{3}$ x² × x³ just $2 x \times 3 x^{2}$ 2x × 3x²?
Mission 5Under $y = \frac{1}{x}$ y = 1/x, is the area from 2 to 4 bigger than from 1 to 2?
Mission 6A particle goes back and forth. Is distance the same as displacement?

The specification points and sources are listed under the lesson: see Sources.

Mission 1

A car's distance graph curves. How fast is it at exactly 3 s?

Make a guess: A car's distance graph curves. How fast is it at exactly 3 s?

Calculus · differentiation

The gradient of a curve as a limit

A curve has no single gradient, but a chord between two points does.
Fix the point at $x = 3$ x = 3 on $y = x^{2}$ y = x² and move a second point closer. The chord gradients approach a limit:

Gap $h$ h	1	0.5	0.1	0.01
Chord gradient	7	6.5	6.1	6.01

That limit, 6, is the gradient of the tangent at $x = 3$ x = 3.
The gradient function is the derivative, written $\frac{d y}{d x}$ dy/dx or $f^{'} (x)$ f'(x). It is also a rate of change: $\frac{d s}{d t}$ ds/dt is how fast $s$ s changes with $t$ t.

Calculate

?Find the gradient of the chord joining the points on

y = x^{3}

where

x = 2

and

x = 2.1

. Give your answer to 2 decimal places.

Calculus · differentiation

Differentiating $a x^{n}$ axⁿ

For integer powers: if $f (x) = a x^{n}$ f(x) = axⁿ, then $f^{'} (x) = an x^{n - 1}$ f'(x) = anxⁿ⁻¹. Multiply by the power, then reduce the power by 1.
Differentiate a sum term by term. A constant term differentiates to 0, and $a x$ ax differentiates to $a$ a.
Rewrite fractions as negative powers first: $\frac{4}{x} = 4 x^{- 1}$ 4/x = 4x⁻¹ and $\frac{3}{x ^{2}} = 3 x^{- 2}$ 3/x² = 3x⁻².
The derivative at $x = a$ x = a, written $f^{'} (a)$ f'(a), is the gradient of the curve there.

Worked example Calculus · differentiation

Finding a gradient

Question

f (x) = 2 x^{3} - 5 x + \frac{4}{x}

. Find

f^{'} (x)

and the gradient of the curve

y = f (x)

x = 2

1
Rewrite with powers: $f (x) = 2 x^{3} - 5 x + 4 x^{- 1}$ f(x) = 2x³ − 5x + 4x⁻¹
2
Differentiate each term: $f^{'} (x) = 6 x^{2} - 5 - 4 x^{- 2}$ f'(x) = 6x² − 5 − 4x⁻²
3
Substitute $x = 2$ x = 2: $f^{'} (2) = 6 (4) - 5 - \frac{4}{4} = 24 - 5 - 1$ f'(2) = 6(4) − 5 − 4/4 = 24 − 5 − 1

Answer

f^{'} (x) = 6 x^{2} - 5 - \frac{4}{x ^{2}}

, and the gradient at

x = 2

is 18.

⚠️ Watch out

\frac{4}{x}

is not

4 x

. Rewrite it as

4 x^{- 1}

before you differentiate.

Quick check

?Find

\frac{d y}{d x}

when

y = \frac{3}{x ^{2}}

Calculus · differentiation

Increasing, decreasing, tangents and normals

If $f^{'} (x) > 0$ f'(x) > 0 the function is increasing; if $f^{'} (x) < 0$ f'(x) < 0 it is decreasing; if $f^{'} (x) = 0$ f'(x) = 0 the curve is momentarily flat.
The tangent at $x = a$ x = a has gradient $m = f^{'} (a)$ m = f'(a) and passes through $(a, f (a))$ (a, f(a)): $y - f (a) = m (x - a)$ y − f(a) = m(x − a).
The normal is perpendicular to the tangent, so its gradient is $- \frac{1}{m}$ −1/m.
On a GDC paper you can find $f^{'} (a)$ f'(a) numerically with your GDC, then write the line by hand.

Worked example Calculus · differentiation

Tangent and normal

Question

Find the equations of the tangent and the normal to

y = x^{3} - 4 x + 1

at the point where

x = 2

1
Find the point: $y = 8 - 8 + 1 = 1$ y = 8 − 8 + 1 = 1, so the point is $(2, 1)$ (2, 1)
2
Differentiate: $\frac{d y}{d x} = 3 x^{2} - 4$ dy/dx = 3x² − 4, so at $x = 2$ x = 2 the gradient is $12 - 4 = 8$ 12 − 4 = 8
3
Tangent: $y - 1 = 8 (x - 2)$ y − 1 = 8(x − 2), so $y = 8 x - 15$ y = 8x − 15
4
Normal: gradient $- \frac{1}{8}$ −1/8, so $y - 1 = - \frac{1}{8} (x - 2)$ y − 1 = −1/8(x − 2), which gives $8 y - 8 = - x + 2$ 8y − 8 = −x + 2

Answer

tangent

y = 8 x - 15

; normal

x + 8 y - 10 = 0

⚠️ Watch out

the normal's gradient is

- \frac{1}{m}

, not

- m

and not

\frac{1}{m}

Calculate

?The curve

y = x^{2} - 2 x + 3

has a normal at the point where

x = 3

. Find the

x

-coordinate of the point where this normal crosses the

x

-axis.

Sort it

f (x) = x^{3} - 12 x

. Decide what

f

is doing at each value of

x

Tap a statement, then tap the group it belongs in. Then press Check.

Mission check

?A car's distance is

s = 2 t^{2}

metres after

t

seconds. How fast is it going at exactly 3 s?

Mission 2

A tap's flow rate keeps changing. How much filled the bath?

Make a guess: A tap's flow rate keeps changing. How much filled the bath?

Calculus · integration

Anti-differentiation

Integration reverses differentiation. If $\frac{d y}{d x} = f (x)$ dy/dx = f(x), then $y = \int f (x) d x$ y = ∫ f(x) dx.
For integer $n \neq = - 1$ n ≠ −1:

\int a x^{n} d x = \frac{a x ^{n + 1}}{n + 1} + C

Raise the power by 1, then divide by the new power. Integrate a sum term by term.
Add the constant of integration $C$ C: every constant differentiates to 0, so the reverse step cannot know it.
A boundary condition (a known point on the curve) fixes the value of $C$ C.

Worked example Calculus · integration

Using a boundary condition

Question

\frac{d y}{d x} = 6 x^{2} - 4 x + 1

, and

y = 5

when

x = 2

. Find

y

in terms of

x

1
Integrate each term: $y = \frac{6 x ^{3}}{3} - \frac{4 x ^{2}}{2} + x + C = 2 x^{3} - 2 x^{2} + x + C$ y = 6x³/3 − 4x²/2 + x + C = 2x³ − 2x² + x + C
2
Substitute $x = 2$ x = 2, $y = 5$ y = 5: $5 = 16 - 8 + 2 + C$ 5 = 16 − 8 + 2 + C
3
Solve: $5 = 10 + C$ 5 = 10 + C, so $C = - 5$ C = −5

Answer

y = 2 x^{3} - 2 x^{2} + x - 5

⚠️ Watch out

the 1 integrates to

x

, not to 0. Only differentiating sends a constant to 0.

Quick check

?Find

\int (6 x^{2} + 3 x^{- 2}) d x

Calculate

f^{'} (x) = 4 x - 3

and

f (1) = 6

. Find

f (3)

Calculus · integration

Definite integrals and area

The area between $y = f (x)$ y = f(x) and the $x$ x-axis from $x = a$ x = a to $x = b$ x = b, where $f (x) > 0$ f(x) > 0, is the definite integral

Area = \int_{a b} f (x) d x

Write the integral first, with its limits, then use your GDC to evaluate it. Examiners award a mark for the correct expression.
If $f$ f is a rate of change, the area under its graph is the total change: the area under a velocity–time graph is distance, and under a flow-rate graph it is volume.

Worked example Calculus · integration

Area under a curve

Question

Find the area of the region enclosed by

y = 6 + x - x^{2}

and the

x

-axis.

1
Find where the curve meets the $x$ x-axis: $6 + x - x^{2} = 0$ 6 + x − x² = 0, so $(3 - x) (2 + x) = 0$ (3 − x)(2 + x) = 0, giving $x = - 2$ x = −2 and $x = 3$ x = 3
2
Check the curve is above the axis between them: at $x = 0$ x = 0, $y = 6 > 0$ y = 6 > 0
3
Write the expression: $Area = \int_{- 2 3} (6 + x - x^{2}) d x$ Area = ∫₋₂³ (6 + x − x²) dx
4
Evaluate with your GDC: $20.833 \dots$ 20.833…

Answer

area

= 20.8

square units (3 s.f.)

⚠️ Watch out

the limits are the

x

-intercepts, not the

y

-intercept.

Calculate

?Water flows into a tank at a rate of

r (t) = 12 + 0.6 t - 0.03 t^{2}

litres per minute, for

0 \leq t \leq 10

minutes. Use your GDC to find how much water flows in during these 10 minutes.

litres

Mission check

?You know a tap's flow rate (litres per minute) at every moment. How do you find how much water filled the bath?

Mission 3

What shape of can uses the least metal for 330 ml?

Make a guess: What shape of can uses the least metal for 330 ml?

Calculus · stationary points

Stationary points

A stationary point is where $f^{'} (x) = 0$ f'(x) = 0: the tangent is horizontal.
Local maximum: $f^{'}$ f' changes from positive to negative. Local minimum: $f^{'}$ f' changes from negative to positive.
To find them, solve $f^{'} (x) = 0$ f'(x) = 0, then test the sign of $f^{'}$ f' just either side.
On a GDC paper, your GDC can find a local maximum or minimum point directly.
A local maximum is not always the greatest value on a domain: check the endpoints too.

Worked example Calculus · stationary points

Finding and classifying stationary points

Question

Find the stationary points of

f (x) = x^{3} - 3 x^{2} - 9 x + 5

and determine their nature.

1
Differentiate: $f^{'} (x) = 3 x^{2} - 6 x - 9 = 3 (x - 3) (x + 1)$ f'(x) = 3x² − 6x − 9 = 3(x − 3)(x + 1)
2
Solve $f^{'} (x) = 0$ f'(x) = 0: $x = - 1$ x = −1 or $x = 3$ x = 3, giving $f (- 1) = 10$ f(−1) = 10 and $f (3) = - 22$ f(3) = −22
3
Test signs: $f^{'} (- 2) = 15 > 0$ f'(−2) = 15 > 0, $f^{'} (0) = - 9 < 0$ f'(0) = −9 < 0, $f^{'} (4) = 15 > 0$ f'(4) = 15 > 0
4
Read off: positive to negative at $x = - 1$ x = −1; negative to positive at $x = 3$ x = 3

Answer

local maximum

(- 1, 10)

; local minimum

(3, - 22)

⚠️ Watch out

always find the

y

-coordinate as well. "The point" means both coordinates.

Calculus · stationary points

The second derivative test and inflexion

The second derivative $f^{''} (x)$ f''(x) (or $\frac{d ^{2} y}{d x ^{2}}$ d²y/dx²) measures how the gradient changes.
At a stationary point: $f^{''} (x) < 0$ f''(x) < 0 gives a maximum (concave-down); $f^{''} (x) > 0$ f''(x) > 0 gives a minimum (concave-up).
If $f^{''} (x) = 0$ f''(x) = 0, the test fails: go back to the sign of $f^{'}$ f'. For $y = x^{4}$ y = x⁴, $f^{''} (0) = 0$ f''(0) = 0 but $(0, 0)$ (0, 0) is a minimum.
A point of inflexion is where the concavity changes: $f^{''} (x) = 0$ f''(x) = 0 and $f^{''}$ f'' changes sign. If $f^{'} (x) = 0$ f'(x) = 0 there too, it is a stationary point of inflexion.

Quick check

f (x) = x^{4} - 4 x^{3}

has

f^{'} (x) = 4 x^{2} (x - 3)

and

f^{''} (x) = 12 x^{2} - 24 x

. What is the nature of the stationary point at

x = 0

Calculate

f (x) = 2 x^{3} + 3 x^{2} - 12 x + 4

. Find the

y

-coordinate of the local maximum point.

Calculus · stationary points

Optimisation

1. Name the variables and write the quantity to optimise (area, cost, volume…).
2. Use the constraint (a fixed volume, a fixed length) to write it in one variable.
3. Differentiate and solve $\frac{d A}{d x} = 0$ dA/dx = 0 (or use your GDC on a GDC paper).
4. Show it is a maximum or minimum, then answer the question asked, with units, in context.

Worked example Calculus · stationary points

The largest open box

Question

An open box is made from a 30 cm square sheet by cutting a square of side

x

cm from each corner and folding up the sides. Find the maximum volume.

1
Write the volume: $V = x (30 - 2 x)^{2}$ V = x(30 − 2x)², for $0 < x < 15$ 0 < x < 15
2
Differentiate: $\frac{d V}{d x} = (30 - 2 x) (30 - 6 x)$ dV/dx = (30 − 2x)(30 − 6x)
3
Solve $\frac{d V}{d x} = 0$ dV/dx = 0: $x = 5$ x = 5 or $x = 15$ x = 15; $x = 15$ x = 15 gives $V = 0$ V = 0, so take $x = 5$ x = 5
4
Check: $V^{'} (4) = 132 > 0$ V'(4) = 132 > 0 and $V^{'} (6) = - 108 < 0$ V'(6) = −108 < 0, so $x = 5$ x = 5 gives a maximum
5
Evaluate: $V = 5 \times 2 0^{2}$ V = 5 × 20²

Answer

maximum volume

= 2000

cm³ (when

x = 5

cm)

⚠️ Watch out

reject values of

x

outside the domain or giving a zero box.

Calculate

?A closed cylindrical can of radius

r

cm and height

h

cm holds 330 cm³. Show that its surface area is

A = 2 π r^{2} + \frac{660}{r}

, then find the value of

r

that minimises

A

. Give your answer to 3 significant figures.

Calculate

f (x) = 2 x^{3} + 3 x^{2} - 12 x + 4

for

- 3 \leq x \leq 3

. Find the greatest value of

f (x)

in this domain.

Mission check

?A closed cylindrical can holds 330 ml. Which shape uses the least metal?

Mission 4

Is the derivative of $x^{2} \times x^{3}$ x² × x³ just $2 x \times 3 x^{2}$ 2x × 3x²?

Make a guess: Is the derivative of $x^{2} \times x^{3}$ x² × x³ just $2 x \times 3 x^{2}$ 2x × 3x²?

Calculus · differentiation rules

Standard derivatives and the chain rule

$f (x)$ f(x)	$x^{n}$ xⁿ, $n \in Q$ n ∈ ℚ	$sin x$ sin x	$cos x$ cos x	$e^{x}$ e^x	$ln x$ ln x
$f^{'} (x)$ f'(x)	$n x^{n - 1}$ nxⁿ⁻¹	$cos x$ cos x	$- sin x$ −sin x	$e^{x}$ e^x	$\frac{1}{x}$ 1/x

Also: the derivative of $tan x$ tan x is $\frac{1}{cos ^{2} x}$ 1/(cos² x).
Differentiate sums and multiples term by term. Angles are in radians.
Chain rule for a composite $y = f (u)$ y = f(u) with $u = g (x)$ u = g(x):

\frac{d y}{d x} = \frac{d y}{d u} \times \frac{d u}{d x}

In words: differentiate the outer function, keep the inside, then multiply by the derivative of the inside. So $sin (3 x - 1)$ sin (3x − 1) differentiates to $3 cos (3 x - 1)$ 3cos (3x − 1).

Worked example Calculus · differentiation rules

The chain rule with a power

Question

f (x) = (3 x^{2} - 5)^{4}

. Find

f^{'} (x)

and the gradient of the curve

y = f (x)

x = 1

1
Let $u = 3 x^{2} - 5$ u = 3x² − 5, so $y = u^{4}$ y = u⁴: $\frac{d y}{d u} = 4 u^{3}$ dy/du = 4u³ and $\frac{d u}{d x} = 6 x$ du/dx = 6x
2
Multiply: $f^{'} (x) = 4 (3 x^{2} - 5)^{3} \times 6 x = 24 x (3 x^{2} - 5)^{3}$ f'(x) = 4(3x² − 5)³ × 6x = 24x(3x² − 5)³
3
Substitute $x = 1$ x = 1: $24 \times 1 \times (3 - 5)^{3} = 24 \times (- 8)$ 24 × 1 × (3 − 5)³ = 24 × (−8)

Answer

f^{'} (x) = 24 x (3 x^{2} - 5)^{3}

, and the gradient at

x = 1

- 192

⚠️ Watch out

writing

4 (3 x^{2} - 5)^{3}

forgets the derivative of the inside,

6 x

, and gives

- 32

x = 1

Quick check

?Differentiate

y = e^{x^{2} + 2}

Calculus · differentiation rules

Product and quotient rules

For a product $y = uv$ y = uv, where $u$ u and $v$ v are functions of $x$ x:

\frac{d y}{d x} = u \frac{d v}{d x} + v \frac{d u}{d x}

For a quotient $y = \frac{u}{v}$ y = u/v:

\frac{d y}{d x} = \frac{v \frac{d u}{d x} - u \frac{d v}{d x}}{v ^{2}}

The derivative of a product is not the product of the derivatives.
In the quotient rule the order matters: $v$ v times $u^{'}$ u' comes first, then subtract.

Worked example Calculus · differentiation rules

The product rule

Question

Differentiate

y = x^{2} e^{3 x}

1
Set $u = x^{2}$ u = x² and $v = e^{3 x}$ v = e^3x: $\frac{d u}{d x} = 2 x$ du/dx = 2x and, by the chain rule, $\frac{d v}{d x} = 3 e^{3 x}$ dv/dx = 3e^3x
2
Apply the rule: $\frac{d y}{d x} = x^{2} \times 3 e^{3 x} + e^{3 x} \times 2 x$ dy/dx = x² × 3e^3x + e^3x × 2x
3
Factorise: $\frac{d y}{d x} = x e^{3 x} (3 x + 2)$ dy/dx = xe^3x(3x + 2)

Answer

\frac{d y}{d x} = x e^{3 x} (3 x + 2)

⚠️ Watch out

multiplying the two derivatives,

2 x \times 3 e^{3 x} = 6 x e^{3 x}

, is the classic slip.

Calculate

f (x) = \frac{2 x + 1}{x - 3}

x \neq = 3

. Find the exact value of

f^{'} (4)

Calculus · differentiation rules

The second derivative and concavity

Differentiate twice to get the second derivative, written $f^{''} (x)$ f''(x) or $\frac{d ^{2} y}{d x ^{2}}$ d²y/dx².
$f^{''} (x) > 0$ f''(x) > 0: the graph is concave-up. $f^{''} (x) < 0$ f''(x) < 0: it is concave-down.
At a stationary point, $f^{''} (x) > 0$ f''(x) > 0 means a minimum and $f^{''} (x) < 0$ f''(x) < 0 a maximum.
A point of inflexion is where the concavity changes: $f^{''} (x) = 0$ f''(x) = 0 and $f^{''}$ f'' changes sign. In context it is where a rate of change stops increasing and starts decreasing, or the reverse.
Where $f^{'}$ f' has a maximum or minimum, $f$ f has a point of inflexion.

Calculate

f (x) = x e^{- x}

. Show that

f^{''} (x) = (x - 2) e^{- x}

, and hence find the

y

-coordinate of the point of inflexion on the graph of

f

. Give your answer to 3 significant figures.

Mission check

?What is the derivative of

y = x^{2} \times x^{3}

Mission 5

Under $y = \frac{1}{x}$ y = 1/x, is the area from 2 to 4 bigger than from 1 to 2?

Make a guess: Under $y = \frac{1}{x}$ y = 1/x, is the area from 2 to 4 bigger than from 1 to 2?

Calculus · integration

Standard integrals

$f (x)$ f(x)	$x^{n}$ xⁿ, $n \neq = - 1$ n ≠ −1	$\frac{1}{x}$ 1/x	$e^{x}$ e^x	$sin x$ sin x	$cos x$ cos x
$\int f (x) d x$ ∫ f(x) dx	$\frac{x ^{n + 1}}{n + 1} + C$ (xⁿ⁺¹)/(n+1) + C	$ln ∣ x ∣ + C$ ln \| x\| + C	$e^{x} + C$ e^x + C	$- cos x + C$ −cos x + C	$sin x + C$ sin x + C

The modulus makes $ln ∣ x ∣$ ln | x| valid for $x < 0$ x < 0 too. SL questions usually have $x > 0$ x > 0, where it is just $ln x + C$ ln x + C, as in the AA guide.
Integration reverses differentiation, so always add the constant of integration $C$ C.
Linear inside: for $f (a x + b)$ f(ax + b), integrate as usual and divide by $a$ a. For example, $\int cos (2 x + 3) d x = \frac{1}{2} sin (2 x + 3) + C$ ∫ cos (2x + 3) dx = 1/2sin (2x + 3) + C.

Worked example Calculus · integration

Integrating a sum of standard functions

Question

Find

\int (4 x - \frac{3}{x} + e^{2 x}) d x

1
Write the root as a power: $4 x = 4 x^{\frac{1}{2}}$ 4√x = 4x^(1/2), which integrates to $\frac{4 x ^{3/2}}{3/2} = \frac{8}{3} x^{\frac{3}{2}}$ (4x^(3/2))/(3/2) = 8/3x^(3/2)
2
The reciprocal term: $- 3 \times ln ∣ x ∣$ −3 × ln |x|
3
Linear inside with $a = 2$ a = 2: $e^{2 x}$ e^2x integrates to $\frac{1}{2} e^{2 x}$ 1/2e^2x

Answer

\frac{8}{3} x^{\frac{3}{2}} - 3 ln ∣ x ∣ + \frac{1}{2} e^{2 x} + C

⚠️ Watch out

integrating multiplies by

\frac{1}{a}

. Writing

2 e^{2 x}

(as when differentiating) is wrong: check by differentiating your answer.

Quick check

?Find

\int sin (3 x) d x

Calculus · integration

Integration by inspection

Look for a function $g (x)$ g(x) and its derivative $g^{'} (x)$ g'(x) in the integrand:

\int k g^{'} (x) f (g (x)) d x

Guess the answer by reversing the chain rule, then adjust the constant. For example, $\int 2 x (x^{2} + 1)^{4} d x = \frac{1}{5} (x^{2} + 1)^{5} + C$ ∫ 2x(x² + 1)⁴ dx = 1/5(x² + 1)⁵ + C.
Or substitute $u = g (x)$ u = g(x), so $d u = g^{'} (x) d x$ du = g'(x) dx, and integrate in $u$ u.
A derivative over its function gives a log: $\int \frac{c o s x}{s i n x} d x = ln ∣ sin x ∣ + C$ ∫ (cos x)/(sin x) dx = ln |sin x| + C.

Calculate

?Find the exact value of

\int_{01} 6 x (x^{2} + 1)^{2} d x

Calculus · integration

Definite integrals and areas

A definite integral has no $+ C$ +C: $\int_{a b} f (x) d x = F (b) - F (a)$ ∫_a^b f(x) dx = F(b) − F(a), where $F^{'} = f$ F' = f.
Where the curve is below the $x$ x-axis, the integral is negative. Area is always positive.
So for an area, find where the curve crosses the axis, integrate each part separately and add the absolute values. Write the correct expression before you calculate.
Between two curves: area $= \int_{a b} (f (x) - g (x)) d x$ = ∫_a^b (f(x) − g(x)) dx, where $f$ f is the upper curve and $a$ a, $b$ b are where they meet.
Some definite integrals can only be found with technology.

Worked example Calculus · integration

An area partly below the axis

Question

Find the total area enclosed by

y = x^{2} - 2 x

, the

x

-axis and the line

x = 3

1
Roots: $x (x - 2) = 0$ x(x − 2) = 0, so $x = 0$ x = 0 or $x = 2$ x = 2. The curve is below the axis for $0 < x < 2$ 0 < x < 2
2
Antiderivative: $F (x) = \frac{x ^{3}}{3} - x^{2}$ F(x) = x³/3 − x²
3
Below the axis: $\int_{02} (x^{2} - 2 x) d x = F (2) - F (0) = \frac{8}{3} - 4 = - \frac{4}{3}$ ∫₀² (x² − 2x) dx = F(2) − F(0) = 8/3 − 4 = −4/3
4
Above the axis: $\int_{23} (x^{2} - 2 x) d x = F (3) - F (2) = 0 - (- \frac{4}{3}) = \frac{4}{3}$ ∫₂³ (x² − 2x) dx = F(3) − F(2) = 0 − (−4/3) = 4/3

Answer

area

= \frac{4}{3} + \frac{4}{3} = \frac{8}{3}

, about 2.67 square units

⚠️ Watch out

\int_{03} (x^{2} - 2 x) d x = 0

. The two parts cancel, so one integral across the root gives the wrong area.

Calculate

?Find the area of the region enclosed by the curves

y = 4 - x^{2}

and

y = x + 2

. Give your answer as a decimal.

Mission check

?Under

y = \frac{1}{x}

, how does the area from

x = 2

x = 4

compare with the area from

x = 1

x = 2

Mission 6

A particle goes back and forth. Is distance the same as displacement?

Make a guess: A particle goes back and forth. Is distance the same as displacement?

Calculus · kinematics

Displacement, velocity and acceleration

Displacement $s$ s is position measured from a fixed origin, with a sign for direction.
Velocity $v = \frac{d s}{d t}$ v = ds/dt and acceleration $a = \frac{d v}{d t} = \frac{d ^{2} s}{d t ^{2}}$ a = dv/dt = d²s/dt².
Speed is the magnitude of velocity, $∣ v ∣$ |v|.
At rest when $v = 0$ v = 0. The particle changes direction where $v$ v changes sign.
Speeding up when $v$ v and $a$ a have the same sign; slowing down when they have opposite signs.

Worked example Calculus · kinematics

When is the particle at rest?

Question

A particle moves in a straight line so that

s = t^{3} - 6 t^{2} + 9 t

metres at time

t

seconds,

t \geq 0

. Find when it is at rest and its acceleration at those times.

1
Differentiate: $v = 3 t^{2} - 12 t + 9$ v = 3t² − 12t + 9
2
Set $v = 0$ v = 0: $3 (t^{2} - 4 t + 3) = 3 (t - 1) (t - 3) = 0$ 3(t² − 4t + 3) = 3(t − 1)(t − 3) = 0, so $t = 1$ t = 1 or $t = 3$ t = 3
3
Differentiate again: $a = 6 t - 12$ a = 6t − 12
4
Substitute: $a (1) = - 6$ a(1) = −6 and $a (3) = 6$ a(3) = 6

Answer

at rest when

t = 1

s (

a = - 6

m s⁻²) and when

t = 3

s (

a = 6

m s⁻²)

⚠️ Watch out

at rest does not mean zero acceleration. Here

a \neq = 0

at both times, so the particle turns round.

Quick check

?At

t = 4

seconds a particle has velocity

- 2

m s⁻¹ and acceleration

3

m s⁻². How is it moving at that instant?

Calculus · kinematics

Displacement and total distance

Integrate velocity to get displacement:

displacement = \int_{t_{1} t_{2}} v (t) d t

Total distance travelled counts every part as positive:

distance = \int_{t_{1} t_{2}} ∣ v (t) ∣ d t

By hand: find where $v = 0$ v = 0, integrate each part and add the absolute values. With a GDC: integrate $∣ v (t) ∣$ |v(t)| directly.
Given $a$ a or $v$ v and a starting value, integrate and use the starting value to find the constant.

Worked example Calculus · kinematics

Distance versus displacement

Question

For the particle of screen 2,

v = 3 t^{2} - 12 t + 9

. Find its displacement and the total distance it travels for

0 \leq t \leq 4

1
Displacement: $\int_{04} v d t = [t^{3} - 6 t^{2} + 9 t]_{04} = 64 - 96 + 36 = 4$ ∫₀⁴ v dt = [t³ − 6t² + 9t]₀⁴ = 64 − 96 + 36 = 4
2
Split at the turns, $t = 1$ t = 1 and $t = 3$ t = 3: $\int_{01} v d t = 4$ ∫₀¹ v dt = 4, $\int_{13} v d t = - 4$ ∫₁³ v dt = −4, $\int_{34} v d t = 4$ ∫₃⁴ v dt = 4
3
Add the absolute values: $4 + 4 + 4 = 12$ 4 + 4 + 4 = 12

Answer

displacement 4 m, total distance 12 m

⚠️ Watch out

one integral from 0 to 4 gives the displacement only. The backward part cancels part of the forward motion.

Calculate

?A particle moves in a straight line with velocity

v = 6 cos t - t

m s⁻¹ for

0 \leq t \leq 4

seconds. Use your GDC to find the total distance travelled. Give your answer to 3 significant figures.

Calculate

?A particle moves in a straight line with acceleration

a = 6 t - 4

m s⁻² at time

t

seconds. When

t = 0

, its velocity is

- 2

m s⁻¹ and its displacement from the origin O is 5 m. Find its displacement from O when

t = 3

Mission check

?A particle has velocity

v = 2 t - 4

m s⁻¹ for

0 \leq t \leq 5

seconds. It ends 5 m from where it started. What total distance does it travel?

Recap

Big ideas

The derivative $\frac{d y}{d x}$ dy/dx or $f^{'} (x)$ f'(x) is the gradient of the curve (the limit of chord gradients) and the rate of change of $y$ y with $x$ x.
For integer $n$ n, $a x^{n}$ axⁿ differentiates to $an x^{n - 1}$ anxⁿ⁻¹; rewrite $\frac{a}{x ^{n}}$ a/xⁿ as $a x^{- n}$ ax⁻ⁿ first. $f^{'} (x) > 0$ f'(x) > 0 means increasing, $f^{'} (x) < 0$ f'(x) < 0 decreasing.
Tangent at $x = a$ x = a: gradient $m = f^{'} (a)$ m = f'(a) through $(a, f (a))$ (a, f(a)). Normal: gradient $- \frac{1}{m}$ −1/m through the same point.
$\int a x^{n} d x = \frac{a x ^{n + 1}}{n + 1} + C$ ∫ axⁿ dx = (axⁿ⁺¹)/(n+1) + C for $n \neq = - 1$ n ≠ −1; a boundary condition fixes $C$ C.
The area under $y = f (x) > 0$ y = f(x) > 0 from $a$ a to $b$ b is $\int_{a b} f (x) d x$ ∫_a^b f(x) dx: write the integral, then evaluate it with your GDC. The area under a rate graph is the total change.
Stationary points: solve $f^{'} (x) = 0$ f'(x) = 0. Classify by the sign of $f^{'}$ f' either side: $+$ + to $-$ − is a maximum, $-$ − to $+$ + a minimum.
Or use $f^{''}$ f'': negative gives a maximum, positive a minimum, zero means use the sign test. An inflexion point needs $f^{''}$ f'' to change sign.
Optimisation: use the constraint to get one variable, differentiate, set to zero, justify, answer in context. Check endpoints for the greatest or least value.
Standard derivatives: $x^{n} \to n x^{n - 1}$ xⁿ → nxⁿ⁻¹, $sin x \to cos x$ sin x → cos x, $cos x \to - sin x$ cos x → −sin x, $e^{x} \to e^{x}$ e^x → e^x, $ln x \to \frac{1}{x}$ ln x → 1/x.
Chain rule $\frac{d y}{d x} = \frac{d y}{d u} \frac{d u}{d x}$ dy/dx = dy/dudu/dx; product rule $u v^{'} + v u^{'}$ uv' + vu'; quotient rule $\frac{v u ^{'} - u v ^{'}}{v ^{2}}$ (vu' − uv')/v².
$f^{''} (x) > 0$ f''(x) > 0 concave-up, $f^{''} (x) < 0$ f''(x) < 0 concave-down; a point of inflexion is where the concavity changes.
$\int x^{n} d x = \frac{x ^{n + 1}}{n + 1} + C$ ∫ xⁿ dx = (xⁿ⁺¹)/(n+1) + C ( $n \neq = - 1$ n ≠ −1), $\int \frac{1}{x} d x = ln ∣ x ∣ + C$ ∫ 1/x dx = ln |x| + C; with $a x + b$ ax + b inside, divide by $a$ a.
Reverse the chain rule when you can see a function and its derivative.
Area: split at the roots and add absolute values.
Area between curves: integrate upper minus lower.
$v = \frac{d s}{d t}$ v = ds/dt, $a = \frac{d v}{d t}$ a = dv/dt; speed $= ∣ v ∣$ = |v|; the particle turns round where $v$ v changes sign.
Displacement $= \int v d t$ = ∫ v dt; distance $= \int ∣ v ∣ d t$ = ∫ |v| dt: split at the turning times, or use a GDC.

🏆

Mini-lesson complete!

⭐⭐⭐

You've worked through Calculus for IB Diploma Mathematics: Analysis & Approaches HL. 🎉

Mission 1A car's distance graph curves. How fast is it at exactly 3 s?
Mission 2A tap's flow rate keeps changing. How much filled the bath?
Mission 3What shape of can uses the least metal for 330 ml?
Mission 4Is the derivative of $x^{2} \times x^{3}$ x² × x³ just $2 x \times 3 x^{2}$ 2x × 3x²?
Mission 5Under $y = \frac{1}{x}$ y = 1/x, is the area from 2 to 4 bigger than from 1 to 2?
Mission 6A particle goes back and forth. Is distance the same as displacement?

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Specification points and sources for this lesson are listed at the bottom of the page.

Sources (6)

The specification points and sources behind this lesson.

1 IB Mathematics: analysis and approaches guide (first assessment 2021), SL 5.1–5.4
2 IB Mathematics: analysis and approaches guide (first assessment 2021), SL 5.5
3 IB Mathematics: analysis and approaches guide (first assessment 2021), SL 5.8 (with SL 5.2, 5.7)
4 IB Mathematics: analysis and approaches guide (first assessment 2021), SL 5.6, SL 5.7; AHL 5.14, 5.15
5 IB Mathematics: analysis and approaches guide (first assessment 2021), SL 5.10, SL 5.11; AHL 5.15, 5.17
6 IB Mathematics: analysis and approaches guide (first assessment 2021), SL 5.9

Calculus

A car's distance graph curves. How fast is it at exactly 3 s?

The gradient of a curve as a limit

Differentiating axnaxⁿ

Finding a gradient

Increasing, decreasing, tangents and normals

Tangent and normal

A tap's flow rate keeps changing. How much filled the bath?

Anti-differentiation

Using a boundary condition

Definite integrals and area

Area under a curve

What shape of can uses the least metal for 330 ml?

Stationary points

Finding and classifying stationary points

The second derivative test and inflexion

Optimisation

The largest open box

Is the derivative of x2×x3x² × x³ just 2x×3x22x × 3x²?

Standard derivatives and the chain rule

The chain rule with a power

Product and quotient rules

The product rule

The second derivative and concavity

Under y=x1​y = 1/x, is the area from 2 to 4 bigger than from 1 to 2?

Standard integrals

Integrating a sum of standard functions

Integration by inspection

Definite integrals and areas

An area partly below the axis

A particle goes back and forth. Is distance the same as displacement?

Displacement, velocity and acceleration

When is the particle at rest?

Displacement and total distance

Distance versus displacement

Big ideas

Mini-lesson complete!

📣 Smashed it? Share your score

Differentiating $a x^{n}$ axⁿ

Is the derivative of $x^{2} \times x^{3}$ x² × x³ just $2 x \times 3 x^{2}$ 2x × 3x²?

Under $y = \frac{1}{x}$ y = 1/x, is the area from 2 to 4 bigger than from 1 to 2?