IB Mathematics AA – Topic 5: Calculus

Comprehensive Guide to Integral Calculus

Introduction to Integral Calculus

Integral calculus is the mathematics of accumulation and total change. While differentiation breaks functions down into instantaneous rates of change, integration builds them back up—calculating total distance from velocity, total area under curves, volumes of complex solids, and accumulated quantities over time. Integration is the inverse process of differentiation, providing powerful tools for solving real-world problems.

Key concepts: The indefinite integral (antiderivative) reverses differentiation and includes a constant of integration. The definite integral calculates the exact accumulated value between two limits, represented geometrically as the signed area under a curve. Integration techniques—including substitution and integration by parts—enable us to find antiderivatives of complex functions. The Fundamental Theorem of Calculus elegantly connects differentiation and integration.

Why integration matters: Integration enables calculation of quantities that can't be measured directly. Engineers use it to find centers of mass and moments of inertia, physicists to calculate work done by varying forces, economists to compute consumer surplus, and biologists to model total population change. Finding areas between curves and volumes of revolution provides practical tools for design and analysis across all technical fields.

In this guide: We'll master all integration rules and techniques with complete proficiency, calculate areas under curves and between multiple curves using definite integrals, find volumes of solids of revolution around axes, apply integration to real-world problems including kinematics and rates of change, and understand the deep connection between differentiation and integration through the Fundamental Theorem—all essential for IB exam success.

1. Integration Rules and Techniques

Basic Integration Rules

Essential Integration Formulas

Power Rule

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

(for \(n \neq -1\))

Standard Functions

\(\int e^x \, dx = e^x + C\)

\(\int \frac{1}{x} \, dx = \ln|x| + C\) (for \(x \neq 0\))

\(\int \sin x \, dx = -\cos x + C\)

\(\int \cos x \, dx = \sin x + C\)

\(\int \sec^2 x \, dx = \tan x + C\)

Constant Multiple and Sum Rules

\(\int kf(x) \, dx = k\int f(x) \, dx\)

\(\int [f(x) \pm g(x)] \, dx = \int f(x) \, dx \pm \int g(x) \, dx\)

Definite Integrals

Fundamental Theorem of Calculus

\(\int_a^b f(x) \, dx = F(b) - F(a)\)

where \(F'(x) = f(x)\) (F is antiderivative of f)

No constant C needed for definite integrals!

Key Properties:

\(\int_a^b f(x) \, dx = -\int_b^a f(x) \, dx\)

\(\int_a^b f(x) \, dx = \int_a^c f(x) \, dx + \int_c^b f(x) \, dx\)

Integration by Substitution

Chain Rule in Reverse:

\(\int f(g(x))g'(x) \, dx = F(g(x)) + C\)

Strategy:

Let \(u = g(x)\) (the inside function)
Find \(du = g'(x) \, dx\)
Substitute to get integral in terms of \(u\)
Integrate
Substitute back to \(x\)

Example: \(\int (3x+1)^5 \, dx\) — let \(u = 3x+1\), \(du = 3dx\)

⚠ Integration Pitfalls:

Power rule exception: For \(n = -1\), use \(\ln|x|\), not power rule
Don't forget +C: Always include constant for indefinite integrals
Limits matter: For definite integrals, evaluate at upper minus lower limit
Chain rule needs extra factor: Must have derivative of inside function present

2. Areas Under and Between Curves

Area Under a Curve

Area Formula

Area between curve \(y = f(x)\) and x-axis from \(x = a\) to \(x = b\):

\(A = \int_a^b f(x) \, dx\)

Important:

If \(f(x) > 0\): area is positive
If \(f(x) < 0\): integral gives negative value
For actual area: use \(A = \int_a^b |f(x)| \, dx\)
Or split at x-intercepts and add absolute values

Area Between Two Curves

Area Between Curves

Area between \(y = f(x)\) (upper) and \(y = g(x)\) (lower):

\(A = \int_a^b [f(x) - g(x)] \, dx\)

Subtract lower from upper

Steps:

Sketch both curves to identify which is upper/lower
Find intersection points (these become limits)
Set up integral: upper minus lower
Evaluate definite integral

💡 Area Tips:

Always sketch the region first
Check which function is "on top" in the interval
If curves cross, split into separate integrals
Use GDC to verify your answer

Example 1: Area Between Curves (IB-Style)

Problem: Find the area enclosed between the curves \(y = x^2\) and \(y = 2x + 3\)

Solution:

Step 1: Find intersection points

Set equations equal: \(x^2 = 2x + 3\)

\(x^2 - 2x - 3 = 0\)

\((x-3)(x+1) = 0\)

\(x = -1\) or \(x = 3\)

Intersection points: \((-1, 1)\) and \((3, 9)\)

Step 2: Determine which is upper curve

Test a point between \(x = -1\) and \(x = 3\), say \(x = 0\):

Line: \(y = 2(0) + 3 = 3\)

Parabola: \(y = 0^2 = 0\)

Line \(y = 2x + 3\) is above parabola \(y = x^2\)

Step 3: Set up integral

\(A = \int_{-1}^3 [(2x+3) - x^2] \, dx\)

\(= \int_{-1}^3 (2x + 3 - x^2) \, dx\)

Step 4: Integrate

\(= \left[x^2 + 3x - \frac{x^3}{3}\right]_{-1}^3\)

Step 5: Evaluate

At \(x = 3\): \(9 + 9 - 9 = 9\)

At \(x = -1\): \(1 - 3 - (-\frac{1}{3}) = 1 - 3 + \frac{1}{3} = -\frac{5}{3}\)

\(A = 9 - (-\frac{5}{3}) = 9 + \frac{5}{3} = \frac{27 + 5}{3} = \frac{32}{3}\)

Area = \(\frac{32}{3}\) square units (or 10.67 square units)

3. Volumes of Solids of Revolution

Revolution About the x-axis

Volume of Revolution (x-axis)

When region under \(y = f(x)\) from \(x = a\) to \(x = b\) is rotated about x-axis:

\(V = \pi\int_a^b [f(x)]^2 \, dx\)

or \(V = \pi\int_a^b y^2 \, dx\)

Think: sum of circular disks with radius \(y\) and thickness \(dx\)

Revolution About the y-axis

Volume of Revolution (y-axis):

\(V = \pi\int_c^d x^2 \, dy\)

Must express \(x\) as a function of \(y\) first

Example 2: Volume of Revolution

Problem: The region bounded by \(y = \sqrt{x}\), \(x = 0\), and \(x = 4\) is rotated about the x-axis. Find the volume.

Solution:

Step 1: Identify the setup

Function: \(y = \sqrt{x}\)

Limits: \(a = 0\), \(b = 4\)

Rotating about x-axis

Step 2: Apply volume formula

\(V = \pi\int_0^4 (\sqrt{x})^2 \, dx\)

\(= \pi\int_0^4 x \, dx\)

Step 3: Integrate

\(= \pi\left[\frac{x^2}{2}\right]_0^4\)

Step 4: Evaluate

\(= \pi\left[\frac{16}{2} - 0\right]\)

\(= \pi \times 8 = 8\pi\)

Volume = \(8\pi\) cubic units (or 25.1 cubic units)

4. Applications of Integration

Kinematics

Motion Relationships:

Velocity from acceleration: \(v(t) = \int a(t) \, dt\)
Position from velocity: \(s(t) = \int v(t) \, dt\)
Total distance: \(\int_a^b |v(t)| \, dt\) (use absolute value)
Displacement: \(\int_a^b v(t) \, dt\) (can be negative)

Average Value of a Function

Average Value Formula:

\(f_{\text{avg}} = \frac{1}{b-a}\int_a^b f(x) \, dx\)

Average height of function over interval \([a, b]\)

Rates of Change

Accumulation from Rates:

If \(f'(x)\) represents rate of change, then:

\(\int_a^b f'(x) \, dx = f(b) - f(a)\)

Total change = integral of rate of change

📋 Integral Calculus Quick Reference

Concept	Formula	Notes
Power Rule	\(\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\)	except \(n = -1\)
Fundamental Theorem	\(\int_a^b f(x) \, dx = F(b) - F(a)\)	Upper minus lower
Area Under Curve	\(A = \int_a^b f(x) \, dx\)	If \(f(x) \geq 0\)
Area Between Curves	\(A = \int_a^b [f(x) - g(x)] \, dx\)	Upper minus lower
Volume (x-axis)	\(V = \pi\int_a^b y^2 \, dx\)	Disk method

🎯 IB Exam Strategy

Common Question Types:

"Find ∫f(x)dx": Identify rule, apply, add +C for indefinite integrals
"Evaluate definite integral": Integrate then substitute limits (upper minus lower)
"Find area": Set up integral, determine limits from intersections
"Area between curves": Upper minus lower, find intersection points
"Find volume": Use \(\pi\int y^2 dx\), square the function first
"Given v(t), find s(t)": Integrate velocity, use initial conditions for C

Key Reminders:

Indefinite integrals: always include +C
Definite integrals: no +C needed, evaluate at limits
Area: sketch region first to identify limits and which curve is on top
Volume: square the function before integrating, multiply by π
Use GDC to verify numerical answers
Check signs: negative area means curve is below x-axis

🎉 Master Integral Calculus!

Integral calculus completes the calculus toolkit by enabling accumulation calculations and area/volume computations. From finding total distance traveled to calculating volumes of complex solids, integration provides essential tools for engineering, physics, economics, and all quantitative sciences. Master integration rules, area calculations, and volume formulas to excel in IB exams and prepare for university mathematics!

Key Success Factors:

✓ Power rule: \(\int x^n dx = \frac{x^{n+1}}{n+1} + C\) (except \(n = -1\))
✓ Definite integral: \(\int_a^b f(x)dx = F(b) - F(a)\) (no +C)
✓ Area between curves: \(\int_a^b [\text{upper} - \text{lower}] dx\)
✓ Volume about x-axis: \(V = \pi\int_a^b y^2 dx\)
✓ Always sketch regions before setting up integrals
✓ Find intersection points for limits of integration

Integrate Correctly • Evaluate Carefully • Verify Results

Master integral calculus and excel in IB Mathematics! 🚀