Integration Techniques — SL 5.6 Techniques and standard integrals (basic) - IB Mathematics Analysis and Approaches

SECTION 1

Historical Context & Motivation

For thousands of years, mathematicians struggled with a deceptively simple question: how do you find the exact area of a shape with curved boundaries? Ancient Greek mathematicians could handle rectangles and triangles, but circles and parabolas required something entirely new. The quest to solve this problem eventually led to one of the most powerful tools in all of mathematics — integration. Today, integration is essential in physics, engineering, economics, and biology, wherever quantities accumulate over time or space.

~250 BCE • Archimedes and Exhaustion

Archimedes approximated the area under a parabola by filling it with ever-thinner triangles — an early precursor to integration that relied on geometric reasoning rather than algebra.

1665–1676 • Newton & Leibniz Invent Calculus

Working independently, Isaac Newton and Gottfried Wilhelm Leibniz discovered that differentiation and integration are inverse operations. Leibniz introduced the ∫ symbol we still use today, modeled after an elongated letter S for "summa" (Latin for sum).

1823 • Cauchy Formalizes the Integral

Augustin-Louis Cauchy provided rigorous definitions for limits and integrals, giving the subject a firm logical foundation and enabling mathematicians to prove results rather than rely on intuition.

1854 • Riemann's Definite Integral

Bernhard Riemann defined the definite integral as the limit of sums of rectangles, the framework that modern calculus courses still teach as Riemann sums.

In this lesson, you will learn the standard integral formulas that appear on the IB SL 5.6 syllabus. These formulas let you reverse the process of differentiation — finding a function whose derivative is the expression you started with. Think of differentiation as breaking something down and integration as building it back up.

SECTION 2

Core Principles & Definitions

Before diving into formulas, you need a solid grasp of what integration actually means and the vocabulary that surrounds it. The core idea is beautifully simple: if differentiation answers "what is the rate of change?", then integration answers "what quantity produced that rate of change?" The following foundational ideas will guide everything in this lesson.

Antiderivative

An antiderivative of f(x) is any function F(x) whose derivative equals f(x). For example, since d/dx(x³) = 3x², an antiderivative of 3x² is x³.

Indefinite Integral

The indefinite integral ∫ f(x) dx represents the family of all antiderivatives of f(x). It always includes a constant of integration, C, because adding any constant to F(x) still gives the same derivative.

Constant of Integration (C)

Since the derivative of any constant is zero, we can never know the exact constant from the derivative alone. The constant of integration C captures all possible vertical shifts of the antiderivative.

Integrand & Differential

In the expression ∫ f(x) dx, the function f(x) is called the integrand and dx is the differential, telling you which variable to integrate with respect to.

Linearity of Integration

Integration is linear: you can pull constants out and split sums. That is, ∫ [af(x) + bg(x)] dx = a∫ f(x) dx + b∫ g(x) dx.

✦

KEY TAKEAWAY

Think of integration like reversing a recipe. Differentiation "cooks" a function into its rate of change; integration "un-cooks" it back into the original function. Just as a finished dish doesn't tell you exactly how much salt the chef used, you always need a constant of integration C because some information is lost during differentiation.

SECTION 3

Visual Explanation — Antiderivatives as Area

One of the most intuitive ways to understand integration is through area. The definite integral of a function from a to b gives the signed area between the curve and the x-axis. The diagram below shows the function f(x) = x² and the shaded region representing ∫₀² x² dx. Each thin vertical strip has width dx and height x², so the integral sums all those tiny rectangles to get the exact area.

The shaded region shows thin rectangular strips of width dx and height x². As we make infinitely many infinitely thin strips, the sum of their areas converges to the exact integral value of 8/3 ≈ 2.67.

Notice how the rectangular strips get taller as x increases, since x² grows. The indefinite integral gives you the general antiderivative (x³/3 + C), while a definite integral evaluates that antiderivative at the boundaries to produce a single number. In this course, you'll work with both, but the SL 5.6 topic focuses primarily on finding antiderivatives using standard formulas.

SECTION 4

Mathematical Framework — Standard Integral Formulas

The IB SL 5.6 syllabus requires you to know a set of standard integrals. Each formula is simply the reverse of a familiar differentiation rule. Let's build them up one at a time, starting with the most important.

POWER RULE FOR INTEGRATION

∫ xⁿ dx = xⁿ⁺¹ / (n + 1) + C, n ≠ −1

Raise the exponent by 1, then divide by the new exponent. This works for any real number n except n = −1 (which would cause division by zero). For n = −1, see the natural log rule below.

NATURAL LOGARITHM RULE

∫ (1/x) dx = ln|x| + C, x ≠ 0

This handles the gap in the power rule. The absolute value is needed because ln is only defined for positive inputs, and 1/x can be negative.

EXPONENTIAL RULE

∫ eˣ dx = eˣ + C

The exponential function eˣ is its own antiderivative — one of the most elegant facts in calculus. The constant e ≈ 2.718.

TRIGONOMETRIC INTEGRALS

∫ sin x dx = −cos x + C ∫ cos x dx = sin x + C

Notice the negative sign in the sine integral. You can verify: d/dx(−cos x) = sin x ✓. Similarly, d/dx(sin x) = cos x ✓. Always double-check by differentiating your answer.

💡

IB Formula Booklet Tip

All of these standard integrals are provided in the IB formula booklet during your exam. You don't need to memorize them, but you do need to know when and how to apply each one. Fluency with these formulas will save you valuable time under exam conditions.

SECTION 5

Detailed Breakdown — Table of Standard Integrals

The table below collects all the basic standard integrals you need for SL 5.6 in one place. For each integral, the third column shows how you can verify the result by differentiating the answer. Building the habit of checking your work this way is one of the most effective strategies in calculus.

Integrand f(x)	Integral ∫ f(x) dx	Verification: d/dx [result]
xⁿ (n ≠ −1)	xⁿ⁺¹/(n + 1) + C	(n+1) · xⁿ/(n+1) = xⁿ ✓
1/x (x ≠ 0)	ln\|x\| + C	1/x ✓
eˣ	eˣ + C	eˣ ✓
sin x	−cos x + C	−(−sin x) = sin x ✓
cos x	sin x + C	cos x ✓
k (constant)	kx + C	k ✓
sec²x	tan x + C	sec²x ✓

Standard integrals required for IB SL 5.6

This diagram shows three function pairs. Following the cyan arrow (differentiate) goes from F(x) to f(x). Following the violet arrow (integrate) goes back. Integration is the reverse of differentiation.

When you integrate a sum of terms, use the linearity property: integrate each term separately and add the results. For instance, ∫ (3x² + 2x − 5) dx = ∫ 3x² dx + ∫ 2x dx − ∫ 5 dx. Then apply the power rule to each piece. Constants can always be pulled out in front of the integral sign.

SECTION 6

Worked Example — Integrating a Polynomial Expression

Let's work through a complete example that combines several standard integrals and demonstrates the step-by-step process you should use on exams.

Find ∫ (4x³ − 6x + 3eˣ + 2/x) dx

1
Step 1 — Split Using Linearity

Apply the linearity property to separate the integral into four simpler integrals: ∫ 4x³ dx − ∫ 6x dx + ∫ 3eˣ dx + ∫ (2/x) dx. Notice that we pull the constant multipliers out as well.

4∫ x³ dx − 6∫ x dx + 3∫ eˣ dx + 2∫ (1/x) dx
2
Step 2 — Apply the Power Rule (first two terms)

For ∫ x³ dx, use the power rule with n = 3: raise the exponent to 4 and divide by 4, giving x⁴/4. For ∫ x dx (which is x¹), raise the exponent to 2 and divide by 2, giving x²/2.

4 · x⁴/4 − 6 · x²/2 = x⁴ − 3x²
3
Step 3 — Apply the Exponential Rule (third term)

Since the antiderivative of eˣ is simply eˣ, we get 3eˣ.

3eˣ
4
Step 4 — Apply the Logarithm Rule (fourth term)

For ∫ (1/x) dx, use the natural log rule: the antiderivative is ln|x|. With the constant 2 out front, we get 2 ln|x|.

2 ln|x|
5
Step 5 — Combine and Add C

Combine all four results into one expression, and don't forget the constant of integration C. Even though each separate integral technically has its own constant, they all merge into a single C.

∫ (4x³ − 6x + 3eˣ + 2/x) dx = x⁴ − 3x² + 3eˣ + 2 ln|x| + C
6
Step 6 — Verify by Differentiating

Check: d/dx(x⁴ − 3x² + 3eˣ + 2 ln|x| + C) = 4x³ − 6x + 3eˣ + 2/x ✓. The derivative of C is 0, confirming our answer matches the original integrand.

Verified ✓

💡

Pro Tip: Always Verify

The beauty of integration is that you can always check your answer by differentiating it. If differentiating your result gives back the original integrand, you know you're correct. This is the single most reliable error-checking method in calculus.

SECTION 7

Common Mistakes & How to Avoid Them

Integration is often called "the harder twin" of differentiation because there are more ways to go wrong. Understanding the most common mistakes will help you avoid losing marks on exams. The table below compares frequent errors with the correct approach.

Common Mistake	Why It's Wrong	Correct Approach
Forgetting + C on indefinite integrals	The antiderivative is a family of functions, not a single function. Without C, you've only found one member.	Always write + C for indefinite integrals. Only definite integrals (with limits) don't need C.
Writing ∫ x⁻¹ dx = x⁰/0	The power rule explicitly excludes n = −1 because division by zero is undefined.	Use the logarithm rule: ∫ x⁻¹ dx = ln\|x\| + C.
Dropping the negative sign in ∫ sin x dx	Students often write cos x + C instead of −cos x + C. Differentiating cos x gives −sin x, not sin x.	Remember: ∫ sin x dx = −cos x + C. Verify by differentiating.
Treating constants inside the integrand incorrectly	Constants multiply the integrand; they don't change which rule applies.	Pull constants out: ∫ 5eˣ dx = 5∫ eˣ dx = 5eˣ + C.
Trying to use the power rule on eˣ	eˣ is not a power function of x — the variable is in the exponent, not the base.	Use the exponential rule: ∫ eˣ dx = eˣ + C.

Five most common integration errors at SL level

✦

KEY TAKEAWAY

Think of the + C as your safety net. When you jump off the differentiation cliff, information about the starting height (the constant) is lost. The constant of integration reminds you that infinitely many curves share the same slope pattern — they're just shifted up or down. Forgetting it on an IB exam typically costs you 1 mark per question, which adds up quickly.

SECTION 8

Connection to Advanced Integration

The standard integrals you've learned in this lesson are your foundation, but more complex functions require additional techniques. In the IB HL course and in university mathematics, you'll encounter methods that extend what the basic rules can handle. Understanding where SL 5.6 fits in the bigger picture will help you see why these foundational skills matter so much.

Feature	SL 5.6 (This Lesson)	HL & Beyond
Types of functions	Powers of x, eˣ, sin x, cos x, 1/x, sec²x	Composite functions, products, fractions, inverse trig
Techniques needed	Direct application of standard formulas + linearity	Substitution, integration by parts, partial fractions
Difficulty level	Recognize the form, apply the rule, simplify	Choose a strategy, transform the integral, then apply rules
Example integral	∫ (3x² + eˣ) dx	∫ x · eˣ dx or ∫ 1/(x² + 1) dx

How SL 5.6 compares to advanced integration

Even the most advanced integration techniques eventually reduce to applying the standard integrals from this lesson. When you learn integration by substitution, for example, you'll transform a complex integral into a standard form, then use the power rule or exponential rule to finish. Mastering SL 5.6 is like learning your scales before playing a concerto — every advanced move builds on these basics.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

Explain in your own words why every indefinite integral includes a constant of integration, C. Give a specific example of two different functions that have the same derivative.

Show answer

The derivative of any constant is zero, so if F(x) is an antiderivative of f(x), then F(x) + 7, F(x) − 3, or F(x) + any number are also antiderivatives. For example, x² + 5 and x² − 10 both have derivative 2x. The constant C accounts for all these vertically shifted versions of the antiderivative. Without C, you would be claiming there is only one antiderivative, which is false.

Question 2: BASIC CALCULATION

Find ∫ (5x⁴ + 3) dx.

Show answer

Apply linearity to separate into ∫ 5x⁴ dx + ∫ 3 dx. For the first term, use the power rule with n = 4: 5 × x⁵/5 = x⁵. For the second term, the integral of a constant k is kx, so ∫ 3 dx = 3x. Combining: ∫ (5x⁴ + 3) dx = x⁵ + 3x + C. You can verify by differentiating: d/dx(x⁵ + 3x + C) = 5x⁴ + 3 ✓.

Question 3: INTERMEDIATE

Find ∫ (2x⁻³ + 4cos x − 1/x) dx. Write x⁻³ in the final answer with positive exponents only.

Show answer

Separate into three integrals. For ∫ 2x⁻³ dx, use the power rule with n = −3: 2 × x⁻²/(−2) = −x⁻² = −1/x². For ∫ 4cos x dx, the standard integral gives 4sin x. For ∫ (1/x) dx, the log rule gives ln|x|, and with the negative sign: −ln|x|. Combining: −1/x² + 4sin x − ln|x| + C. Verify: d/dx(−x⁻² + 4sin x − ln|x| + C) = 2x⁻³ + 4cos x − 1/x ✓.

Question 4: APPLIED

A car's velocity is given by v(t) = 6t² + 4 m/s. If the car's position at time t = 0 is s(0) = 10 metres, find the position function s(t) and determine the car's position at t = 3 seconds.

Show answer

Position is the integral of velocity: s(t) = ∫ (6t² + 4) dt = 6 × t³/3 + 4t + C = 2t³ + 4t + C. Use the initial condition s(0) = 10: 2(0)³ + 4(0) + C = 10, so C = 10. Therefore s(t) = 2t³ + 4t + 10. At t = 3: s(3) = 2(27) + 4(3) + 10 = 54 + 12 + 10 = 76 metres. This problem shows how the constant C is determined by real-world initial conditions.

Question 5: CRITICAL THINKING

A student claims that ∫ (eˣ + sin x) dx = eˣ + cos x + C. Without evaluating the integral yourself first, explain how you can determine whether the student is correct or incorrect, and then identify any error.

Show answer

To check any antiderivative, differentiate the proposed answer and see if it equals the integrand. Differentiating eˣ + cos x + C gives eˣ − sin x + 0 = eˣ − sin x. This does NOT equal the original integrand eˣ + sin x, because of the sign on sin x. The error is in the sin x term: the student wrote +cos x instead of −cos x. The correct integral is eˣ − cos x + C, since d/dx(−cos x) = sin x. This verification method — differentiating to check — is the most powerful self-checking tool in integration.

SUMMARY

Lesson Summary

Summary

Integration is the reverse process of differentiation. The power rule ∫ xⁿ dx = xⁿ⁺¹/(n + 1) + C works for all n ≠ −1, while the special case n = −1 uses the natural logarithm rule ∫ (1/x) dx = ln|x| + C. The exponential rule ∫ eˣ dx = eˣ + C and the trigonometric rules ∫ sin x dx = −cos x + C and ∫ cos x dx = sin x + C complete the core toolkit for SL 5.6.

Always use the linearity property to break complex integrals into simpler pieces, apply the matching standard formula to each term, and recombine the results with a single constant of integration C. The most reliable way to check your work is to differentiate your answer — if you recover the original integrand, you know you're correct. These basic techniques form the essential foundation for all more advanced integration methods you will encounter later.