Calculus Basics SL

IB Mathematics: Applications & Interpretation · Topic 5: Calculus

JMaths

www.jmaths.xyz

1

Limits & Introduction to the Derivative

The derivative measures the instantaneous rate of change of a function at a point. It gives the gradient of the tangent line to the curve.

Informal definition

\( f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \)

At SL, you investigate limits using a GDC table with decreasing values of \( h \). You do NOT need to evaluate limits algebraically.

TI-84 Plus CE — Investigating limits

Enter Y1 = (f(X+0.001) - f(X)) / 0.001 in [Y=]
Use [TABLE] to evaluate at specific \( x \)-values
Or: [MATH] → 8: nDeriv( → nDeriv(f(X), X, value) for numerical derivative

TI-Nspire CX II — Investigating limits

In Calculator: type nDeriv(f(x), x, value) for numerical derivative
In Graphs: [Menu] → Analyze Graph → dy/dx → click on curve at desired point

Casio fx-CG50 — Investigating limits

In Graph mode: draw the function, then
[G-SOLV] (SHIFT+F5) → \frac{d}{dx} → enter \( x \)-value → gives gradient at that point
Or in Run-Matrix: \frac{d}{dx}(f(x), value) using the [OPTN] calculus menu

2

Power Rule for Differentiation

Power rule (given in formula booklet)

If \( f(x) = ax^n \) then \( f'(x) = anx^{n-1} \)

Differentiate each term separately. Constants disappear: if \( f(x) = 5 \), then \( f'(x) = 0 \).

Worked Example

Find \( f'(x) \) for \( f(x) = 3x^4 - 2x^2 + 7x - 5 \)

\( f'(x) = 3(4)x^3 - 2(2)x^1 + 7(1)x^0 - 0 \)

\( f'(x) = 12x^3 - 4x + 7 \)

Answer: \( f'(x) = 12x^3 - 4x + 7 \)

✗

Common error: Wrong power rule application. "Bring the power down and reduce it by 1." For \( x^3 \), the derivative is \( 3x^2 \), NOT \( x^2 \) or \( 3x^3 \).

3

Equation of a Tangent Line

Tangent at point \( (a, f(a)) \)

\( y - f(a) = f'(a)(x - a) \)

Step 1: Find \( f'(a) \) = gradient. Step 2: Find the point \( (a, f(a)) \). Step 3: Substitute into the formula.

Worked Example

Find the equation of the tangent to \( f(x) = x^2 + 3x \) at \( x = 2 \).

\( f'(x) = 2x + 3 \Rightarrow f'(2) = 2(2) + 3 = 7 \) (gradient)

\( f(2) = 4 + 6 = 10 \) (point is (2, 10))

\( y - 10 = 7(x - 2) \Rightarrow y = 7x - 4 \)

Answer: \( y = 7x - 4 \)

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GDC check: Graph both the original curve and your tangent line. They should touch at exactly one point and the tangent should not cross the curve at that point.

Calculus Basics SL

IB Mathematics: Applications & Interpretation · Topic 5: Calculus

JMaths

www.jmaths.xyz

4

Stationary Points & Classification

Stationary points occur where \( f'(x) = 0 \). They can be a local maximum, local minimum, or point of inflexion.

Finding stationary points

Step 1: Find \( f'(x) \). Step 2: Solve \( f'(x) = 0 \). Step 3: Classify using the GDC (graph or sign of gradient either side).

Local maximum

\( f'(x) \) changes from + to −

Gradient goes from uphill to downhill.

Local minimum

\( f'(x) \) changes from − to +

Gradient goes from downhill to uphill.

TI-84 Plus CE — Stationary points

Graph Y1 = f(x) in [Y=], press [GRAPH]
[2nd][CALC] → 3: minimum → set left/right bounds → gives min point
[2nd][CALC] → 4: maximum → set left/right bounds → gives max point
Alternative: Graph Y2 = nDeriv(Y1, X, X) and find where Y2 = 0

TI-Nspire CX II — Stationary points

Graph the function, then:
[Menu] → Analyze Graph → Minimum (or Maximum)
Drag to set bounds → gives coordinates of the stationary point

Casio fx-CG50 — Stationary points

Graph the function: [DRAW] (F6)
[G-SOLV] (SHIFT+F5) → MIN or MAX
Gives coordinates of the stationary point directly

Worked Example

Find and classify the stationary points of \( f(x) = x^3 - 6x^2 + 9x + 1 \).

\( f'(x) = 3x^2 - 12x + 9 = 3(x^2 - 4x + 3) = 3(x - 1)(x - 3) \)

\( f'(x) = 0 \) when \( x = 1 \) or \( x = 3 \)

\( f(1) = 1 - 6 + 9 + 1 = 5 \Rightarrow (1, 5) \)

\( f(3) = 27 - 54 + 27 + 1 = 1 \Rightarrow (3, 1) \)

GDC graph shows: (1, 5) is a local max, (3, 1) is a local min.

Answer: Local maximum at (1, 5); local minimum at (3, 1)

5

Optimisation

Optimisation problems ask you to find the maximum or minimum value of a quantity (cost, area, profit, etc.) subject to a constraint.

Optimisation method

1. Write the quantity to optimise as a function of one variable.
2. Use the constraint to eliminate other variables.
3. Differentiate and set \( f'(x) = 0 \).
4. Solve and check it gives a max/min (GDC or sign test).
5. Answer the question in context with units.

Worked Example

A farmer has 80 m of fencing to make a rectangular pen against a wall. Find the dimensions that maximise the area.

Let width = \( x \). Three sides of fencing: \( 2x + l = 80 \), so \( l = 80 - 2x \)

\( A(x) = x(80 - 2x) = 80x - 2x^2 \)

\( A'(x) = 80 - 4x = 0 \Rightarrow x = 20 \)

\( l = 80 - 2(20) = 40 , A = 20 \times 40 = 800 \)

Answer: Width = 20 m, length = 40 m, maximum area = 800 m\( ^2 \)

✗

Common error: Forgetting to check the domain. In the example above, \( x \) must satisfy \( 0 < x < 40 \) (otherwise the pen has no area). Also check endpoints if the domain is closed.

Calculus Basics SL

IB Mathematics: Applications & Interpretation · Topic 5: Calculus

JMaths

www.jmaths.xyz

6

Integration (Reverse of Differentiation)

Integration is the reverse process of differentiation. It finds the original function from its derivative, or the area under a curve.

Power rule for integration

\( \int ax^n \, dx = \frac{ax^{n+1}}{n + 1} + C \)

\( n \neq -1 \) \( C \) = constant of integration

Definite integral (area)

\( \int_a^b f(x) \, dx \) = area between curve and \( x \)-axis from \( a \) to \( b \)

If curve is below \( x \)-axis, the integral is negative.

Worked Example — Indefinite

Find \( \int (6x^2 + 4x - 3) \, dx \)

\( = \frac{6x^3}{3} + \frac{4x^2}{2} - 3x + C \)

\( = 2x^3 + 2x^2 - 3x + C \)

Answer: \( 2x^3 + 2x^2 - 3x + C \)

Worked Example — Definite

Find the area under \( f(x) = x^2 + 1 \) from \( x = 0 \) to \( x = 3 \).

\( \int_0^3 (x^2 + 1) \, dx = \left[\frac{x^3}{3} + x\right]_0^3 \)

\( = (27/3 + 3) - (0 + 0) = 9 + 3 = 12 \)

Answer: Area = 12 units\( ^2 \)

TI-84 Plus CE — Definite integral

Method 1: [MATH] → 9: fnInt( → fnInt(f(X), X, a, b)
Method 2: Graph Y1 = f(x), then [2nd][CALC] → 7: ∫f(x)dx
Enter lower and upper bounds → gives shaded area value

TI-Nspire CX II — Definite integral

Calculator: type nInt(f(x), x, a, b)
Graphs: [Menu] → Analyze Graph → Integral → set bounds
Gives the numerical value of the definite integral

Casio fx-CG50 — Definite integral

Graph mode: draw the function, then
[G-SOLV] (SHIFT+F5) → ∫dx → enter lower and upper bounds
Or in Run-Matrix: ∫(f(x), a, b) using the [OPTN] calculus menu

✗

Common error: Forgetting \( + C \) for indefinite integrals. Every indefinite integral must include the constant of integration. You lose a mark every time you forget it.

7

Trapezoidal Rule

The trapezoidal rule approximates the area under a curve using trapezoids. Used when you have data points but no equation, or when asked to approximate.

Trapezoidal rule (given in formula booklet)

\( \int_a^b f(x) \, dx \approx \frac{h}{2} \left[ y_0 + 2(y_1 + y_2 + \ldots + y_{n-1}) + y_n \right] \)

where \( h = \frac{b - a}{n} \) (width of each strip) and \( n \) = number of strips

Worked Example

Use the trapezoidal rule with 4 strips to approximate \( \int_0^4 x^2 \, dx \).

\( h = (4 - 0) / 4 = 1 \) \( x \)-values: 0, 1, 2, 3, 4

\( y \)-values: 0, 1, 4, 9, 16

Area ≈ (1/2)[0 + 2(1 + 4 + 9) + 16]

= (1/2)[0 + 28 + 16] = (1/2)(44) = 22

Answer: Approximate area = 22 (exact value is \( 64/3 \approx 21.3 \))

8

Exam Traps & Key Reminders

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Wrong power rule for integration. "Add 1 to the power and divide by the new power." For \( x^3: integral = \( x^4/4 \), NOT \( x^4 \) or \( x^4/3 \).

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Area below the \( x \)-axis. If the curve is below the \( x \)-axis, the integral gives a negative value. For the actual area, take the absolute value or split into regions.

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Trapezoidal rule: \( n \) strips vs \( n + 1 \) values. 4 strips means 5 \( y \)-values (including both endpoints). Don't confuse the number of strips with the number of data points.

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At AI SL, use the GDC freely. Most integration questions expect you to use technology. Only show manual working if the question says "find the integral" or involves a simple power rule.

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Formula booklet: The power rule for differentiation and integration, the trapezoidal rule, and the tangent line formula are all given. Practise finding them quickly in the booklet under time pressure.

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3 sf rule: Give answers to 3 significant figures. For exact answers like 22 or 12, write them as given. Never round mid-calculation.