Integration Basics — SL 5.5 Anti-derivatives and definite integrals; area under curves - IB Mathematics Analysis and Approaches

SECTION 1

Historical Context & Motivation

Long before calculus had a name, ancient mathematicians wrestled with a deceptively simple question: how do you measure the exact area of a curved region? Straight-sided shapes like rectangles and triangles were easy, but curves demanded something new. The quest to answer this question spanned thousands of years and eventually gave rise to one of the most powerful tools in all of mathematics — integration.

~250 BCE • Archimedes and the Method of Exhaustion

The Greek mathematician Archimedes approximated the area of a parabolic segment by filling it with ever-smaller triangles. This early technique foreshadowed modern integration by using infinitely many slices to pin down a curved area.

1665 • Newton's Fluxions

Isaac Newton developed his method of "fluxions," showing that finding areas under curves was the reverse process of finding slopes. This insight connected two seemingly unrelated problems — rates of change and accumulated quantities.

1675 • Leibniz Introduces ∫ Notation

Gottfried Wilhelm Leibniz independently discovered calculus and introduced the elongated "S" symbol ∫ for integration — short for the Latin word "summa" (sum). His notation is still the standard used worldwide today.

1823 • Cauchy Formalizes the Definite Integral

Augustin-Louis Cauchy gave the definite integral a rigorous definition using limits of sums, placing integration on a firm logical foundation that mathematicians still rely on.

The central question that integration answers is this: if you know the rate at which something changes (its derivative), can you work backwards to find the original quantity? And can you use that idea to compute the exact area trapped between a curve, the x-axis, and two vertical boundaries? In IB SL 5.5, you will learn that the answer to both questions is yes — and the tool that makes it possible is the Fundamental Theorem of Calculus.

SECTION 2

Core Principles & Definitions

Integration rests on a handful of foundational ideas. Before you start computing integrals, it is essential to understand what each term means and how the pieces fit together. Think of differentiation as taking something apart to see how fast it changes; integration is the process of putting it back together to see how much has accumulated.

Anti-derivative

A function F(x) is an anti-derivative of f(x) if F′(x) = f(x). In other words, differentiating F gives you back f. Because any constant disappears when you differentiate, anti-derivatives always include a "+ C" (the constant of integration).

Indefinite Integral

The notation ∫ f(x) dx represents the indefinite integral — the general family of all anti-derivatives of f(x). The result is a function (plus C), not a single number.

Definite Integral

The definite integral ∫ from a to b of f(x) dx gives a single numerical value. It measures the net signed area between the curve y = f(x) and the x-axis from x = a to x = b.

Fundamental Theorem of Calculus

This theorem links anti-derivatives to definite integrals: ∫ from a to b of f(x) dx = F(b) − F(a), where F is any anti-derivative of f. It means you can evaluate areas exactly without summing infinitely many rectangles.

The Constant of Integration (C)

Because the derivative of any constant is zero, every indefinite integral includes "+ C." This represents a whole family of curves, each shifted vertically. When you compute a definite integral, the C cancels out.

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KEY TAKEAWAY

Think of integration like an odometer in a car. The speedometer tells you your speed at each instant (that's the derivative). The odometer adds up all those tiny speeds over time to tell you the total distance you've traveled — that's the integral. Finding the anti-derivative is like figuring out the odometer reading from the speedometer.

SECTION 3

Visualizing Area Under a Curve

The most intuitive way to understand integration is to see it. The diagram below shows the function y = x² from x = 0 to x = 3. The shaded region between the curve and the x-axis is exactly the area that the definite integral ∫ from 0 to 3 of x² dx calculates. Notice how the region is not a simple rectangle or triangle — its top boundary is curved, which is precisely why we need integration rather than basic geometry formulas.

The shaded cyan region represents ∫ from 0 to 3 of x² dx. The curve y = x² rises steeply, and the dashed pink line marks the upper bound x = 3. The exact area equals 9 square units.

In the diagram, the cyan shaded region is bounded on the left by x = 0, on the right by the pink dashed line at x = 3, below by the x-axis, and above by the parabola y = x². This region has a curved top edge, so you cannot simply use length × width. Instead, you can imagine slicing the region into infinitely thin vertical strips, each with width dx and height x². Adding them all up — that infinite sum — is precisely what the definite integral computes.

SECTION 4

Mathematical Framework

The IB SL 5.5 syllabus requires you to master two key integration rules (the power rule and simple polynomial integration) and to apply the Fundamental Theorem of Calculus to evaluate definite integrals. Below are the essential formulas you need.

POWER RULE FOR INTEGRATION

∫ xⁿ dx = xⁿ⁺¹ / (n + 1) + C, where n ≠ −1

This is the reverse of the power rule for differentiation. Increase the exponent by 1, then divide by the new exponent. The constant C accounts for any vertical shift.

CONSTANT MULTIPLE RULE

∫ k · f(x) dx = k · ∫ f(x) dx

A constant factor k can be pulled out in front of the integral. This works because constants do not depend on x.

SUM / DIFFERENCE RULE

∫ [f(x) ± g(x)] dx = ∫ f(x) dx ± ∫ g(x) dx

You can integrate term by term. This means polynomials are integrated by handling each term separately.

FUNDAMENTAL THEOREM OF CALCULUS

∫ₐᵇ f(x) dx = F(b) − F(a)

To evaluate a definite integral, find any anti-derivative F(x) of f(x), substitute the upper limit b and the lower limit a, then subtract. The constant C cancels, so you do not need it for definite integrals.

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IB Formula Booklet Tip

On the IB exam, the formula booklet lists the standard integrals including ∫ xⁿ dx. However, you are expected to apply these rules fluently. Make sure you can integrate basic polynomials, constants, and simple powers without reaching for the booklet every time.

SECTION 5

Standard Integrals & the Role of Signed Area

The table below collects the most important anti-derivatives you need for SL 5.5. Memorizing this short list will let you integrate any polynomial quickly. After the table, we explore a subtlety that catches many students off guard: the concept of signed area.

f(x)	∫ f(x) dx	Example
k (constant)	kx + C	∫ 5 dx = 5x + C
x	x² / 2 + C	∫ x dx = x² / 2 + C
x²	x³ / 3 + C	∫ x² dx = x³ / 3 + C
xⁿ (n ≠ −1)	xⁿ⁺¹ / (n + 1) + C	∫ x⁴ dx = x⁵ / 5 + C
ax² + bx + c	ax³ / 3 + bx² / 2 + cx + C	∫ (3x² + 2x − 1) dx = x³ + x² − x + C

Standard anti-derivatives for polynomial functions

When a curve dips below the x-axis, that portion of the definite integral contributes a negative value. The green regions above the x-axis contribute positive values. The definite integral gives the net signed area — the positive parts minus the negative parts.

A common mistake is to assume the definite integral always gives the "total" area. It does not — it gives the net signed area. When the IB asks for the total area enclosed between the curve and the x-axis, you must split the integral at any x-intercepts, take the absolute value of each piece, and then add them together. This distinction between signed area and total area is tested frequently on IB exams.

SECTION 6

Worked Example: Area Under a Polynomial

Let's work through a complete IB-style problem step by step. We will find the area enclosed between the curve y = 6x − x² and the x-axis.

Find the area enclosed between y = 6x − x² and the x-axis.

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Step 1 — Find the x-intercepts

Set y = 0: 6x − x² = 0. Factor out x: x(6 − x) = 0. So x = 0 or x = 6. These are the boundaries of the enclosed region.

Limits of integration: a = 0, b = 6
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Step 2 — Check whether the curve is above or below the x-axis

Pick a test point between 0 and 6, say x = 3. Then y = 6(3) − 3² = 18 − 9 = 9 > 0. Since the curve is above the x-axis throughout [0, 6], the definite integral will give the actual area directly (no absolute value splitting needed).

Curve is above x-axis on (0, 6)
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Step 3 — Find the anti-derivative

Integrate term by term using the power rule. ∫ (6x − x²) dx = 6 × (x²/2) − x³/3 = 3x² − x³/3. We do not need + C because this is a definite integral.

F(x) = 3x² − x³/3
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Step 4 — Apply the Fundamental Theorem of Calculus

Evaluate F(b) − F(a): F(6) − F(0) = [3(6)² − (6)³/3] − [3(0)² − (0)³/3] = [3 × 36 − 216/3] − [0] = [108 − 72] − 0 = 36.

Area = 36 square units

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Common Mistake Alert

Don't forget to check whether the curve goes below the x-axis within your limits. If it does, you must split the integral at each x-intercept. In this example, the curve stayed above the axis, so the definite integral directly equaled the area.

SECTION 7

Differentiation vs. Integration — Strengths & Pitfalls

Because integration is the reverse of differentiation, students sometimes assume the two processes are mirror images with identical difficulty. In reality, integration has its own set of challenges. The comparison below highlights what each process does well and where integration requires extra care.

Feature	Differentiation	Integration
Purpose	Finds the instantaneous rate of change (slope)	Finds the accumulated total (area)
Output	A single, unique derivative	A family of functions (+ C) or a number
Power rule direction	Bring exponent down, decrease by 1	Increase exponent by 1, divide by new exponent
Constant concern	Constants vanish (derivative of C is 0)	Must add + C in indefinite integrals
Sign area issue	Not applicable	Regions below x-axis count as negative

Comparing differentiation and integration at the SL level

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KEY TAKEAWAY

Differentiation and integration are like tearing a building down versus building one up. Tearing down (differentiation) is fairly mechanical — apply a rule, get a unique answer. Building up (integration) requires you to figure out what the original structure looked like, and there's always some ambiguity (the + C). The definite integral resolves that ambiguity by locking in specific starting and ending points.

SECTION 8

Connection to Advanced Integration Topics

The skills you build in SL 5.5 form the foundation for everything that comes next in calculus. At the HL level and in university courses, integration extends far beyond polynomials. The table below previews how the basic ideas you've learned connect to more advanced techniques.

SL 5.5 Concept	Advanced Extension	Where You'll See It
Power rule for xⁿ	Integration of eˣ, sin x, cos x, 1/x	IB HL 5.11 and university calculus
Area under one curve	Area between two curves	IB HL 5.12 and AP Calculus
Definite integral as a number	Volumes of revolution (rotating areas around an axis)	IB HL 5.12 and Calculus 2
Fundamental Theorem of Calculus	Differential equations (solving rates of change)	IB HL 5.18 and engineering/physics

How SL 5.5 concepts extend into advanced topics

For now, focus on mastering the polynomial anti-derivatives and the mechanics of definite integrals. Once these become second nature, you will find the jump to trigonometric and exponential integrals much smoother. The core logic — find the anti-derivative, apply the limits, subtract — never changes.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

Explain in your own words why the indefinite integral ∫ 3x² dx produces a family of functions rather than a single function. Why is the "+ C" necessary?

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The derivative of any constant is zero, so if F(x) = x³ is one anti-derivative of 3x², then F(x) = x³ + 7, F(x) = x³ − 100, and F(x) = x³ + C (for any real number C) are all also anti-derivatives. We write "+ C" to represent this entire family. There is no way to determine the specific constant without additional information, such as a point the curve passes through. When evaluating a definite integral, the C cancels in the subtraction F(b) − F(a), which is why you do not need it for definite integrals.

Question 2: BASIC CALCULATION

Evaluate the indefinite integral: ∫ (4x³ − 6x + 5) dx.

Show answer

Apply the power rule term by term. For 4x³: increase the exponent to 4 and divide by 4 → 4x⁴/4 = x⁴. For −6x: increase the exponent to 2 and divide by 2 → −6x²/2 = −3x². For 5 (a constant): integrate to 5x. Combining and adding the constant of integration: ∫ (4x³ − 6x + 5) dx = x⁴ − 3x² + 5x + C.

Question 3: INTERMEDIATE

Evaluate the definite integral ∫ from 1 to 4 of (2x + 3) dx and interpret the result geometrically.

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First find the anti-derivative: F(x) = x² + 3x. Then apply the Fundamental Theorem: F(4) − F(1) = (16 + 12) − (1 + 3) = 28 − 4 = 24. Geometrically, y = 2x + 3 is a straight line with positive slope, so the region from x = 1 to x = 4 is a trapezoid above the x-axis. The parallel sides have heights y(1) = 5 and y(4) = 11, and the width is 3. The trapezoid area formula gives (5 + 11)/2 × 3 = 24, confirming the integral. The definite integral equals the area of this trapezoid: 24 square units.

Question 4: APPLIED

A particle moves along a straight line with velocity v(t) = t² − 4t + 3 m/s. Find the total distance traveled by the particle from t = 0 to t = 4 seconds.

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First find where the velocity is zero: t² − 4t + 3 = (t − 1)(t − 3) = 0, so t = 1 and t = 3. Check signs: v(0) = 3 > 0, v(2) = 4 − 8 + 3 = −1 < 0, v(4) = 16 − 16 + 3 = 3 > 0. Since total distance requires absolute values, split into three intervals. ∫ from 0 to 1 of (t² − 4t + 3) dt = [t³/3 − 2t² + 3t] from 0 to 1 = (1/3 − 2 + 3) − 0 = 4/3. ∫ from 1 to 3 = [t³/3 − 2t² + 3t] from 1 to 3 = (9 − 18 + 9) − (4/3) = 0 − 4/3 = −4/3. ∫ from 3 to 4 = [t³/3 − 2t² + 3t] from 3 to 4 = (64/3 − 32 + 12) − (0) = 64/3 − 20 = 4/3. Total distance = |4/3| + |−4/3| + |4/3| = 4/3 + 4/3 + 4/3 = 4 metres.

Question 5: CRITICAL THINKING

The curve y = kx − x² encloses a region with the x-axis that has an area of 36 square units. Find the positive value of k.

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The x-intercepts are found by setting kx − x² = 0, giving x(k − x) = 0, so x = 0 and x = k. Since k > 0, the parabola opens downward and stays above the x-axis on (0, k). The area is ∫ from 0 to k of (kx − x²) dx = [kx²/2 − x³/3] from 0 to k = k³/2 − k³/3 = k³(1/2 − 1/3) = k³/6. Setting this equal to 36: k³/6 = 36, so k³ = 216, therefore k = 6. This problem shows how integration can be used "in reverse" — you know the area and work backwards to find a parameter in the function.

SUMMARY

Lesson Summary

Summary

Integration is the reverse of differentiation. An anti-derivative F(x) of a function f(x) satisfies F′(x) = f(x), and the indefinite integral ∫ f(x) dx = F(x) + C represents the entire family of anti-derivatives. The power rule for integration — ∫ xⁿ dx = xⁿ⁺¹/(n + 1) + C — is the workhorse formula for integrating polynomials. You can integrate term by term, pulling out constant factors as needed.

The definite integral ∫ from a to b of f(x) dx yields a single number using the Fundamental Theorem of Calculus: evaluate F(b) − F(a). This number equals the net signed area between the curve and the x-axis. Regions above the axis contribute positively and regions below contribute negatively. To find the total area, split the integral at the x-intercepts and sum the absolute values of each piece.