IP AMaths Notes (Upper Sec, Year 3-4): 17) Integration Essentials

Study guide08 Nov 2025, 00:00 ZUpdated 17 Jul 2026

Reverse power rule, substitution, and definite integral basics for IP AMaths.

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For Integrated Programme students: Your current school materials, teacher instructions, and assessment scope take precedence because IP topic sequence and depth vary by school. This is an Eclat IP guide, not the O-Level / SEC G3 exam-track guide.

How this chapter applies

Eclat core: reverse differentiation, standard algebraic, trigonometric, and exponential forms, linear inner functions, definite integrals, and signed area form the main route.
School-sensitive extension: substitution beyond simple linear inner functions and area between two curves may be school-sensitive depth.
2027 national comparison: K341 Topic C1 covers rational-power, sine, cosine, secant-squared, and exponential integrals, linear inner functions, definite integrals, areas bounded by a curve and lines, and regions below the x-axis.
Check your school: confirm whether area between two curves or broader substitution is assessed.
Exam-track route: use the separate O-Level and SEC G3 Additional Mathematics notes for K341 topic ownership.

Q: What does IP AMaths Notes (Upper Sec, Year 3-4): 17) Integration Essentials cover?
A: Reverse power rule, substitution, and definite integral basics for IP AMaths.

Integration is the inverse of differentiation. Start with antiderivatives and practice evaluating definite integrals.

Keep the full topic roadmap handy via our IP Maths tuition hub so you can jump into related drills, quizzes, or diagnostics as you move through these notes.

New to the Integrated Programme? Start with What is IP? | Browse all free IP notes.

The core idea is simple: Integration reverses differentiation and accumulates area or displacement.

Use it as a working check: Pick the simplest antiderivative rule first, keep the constant for indefinite integrals, and substitute both bounds for definite integrals.

Then go one layer deeper: Example: integrate velocity to get displacement, then use the initial condition to find the constant before answering the motion question.

Choosing the integration route

Before integrating, decide whether the question needs an antiderivative, a number from bounds, or a model with an initial condition.

Question cue

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Question cue	First integration move	How to use the condition	Common trap
Given $\dfrac{dy}{dx}$ and one point $(a,b)$	Integrate to get $y = F(x) + C$ .	Substitute $x=a$ , $y=b$ , then solve for $C$ .	Treating $b$ as the final answer instead of the value that finds $C$ .
Given velocity $v(t)$ and initial position	Integrate velocity to get position $s(t)$ .	Substitute the initial time and position to find $C$ .	Reporting displacement when the question asks for position.
Given acceleration $a(t)$ and initial velocity	Integrate acceleration to get velocity first.	Use the initial velocity before integrating again for position.	Jumping straight from acceleration to position and losing one constant.

Question wording	What the integral gives	Extra step	Common trap
Evaluate $\int_a^b f(x) \space dx$	Signed area between the curve and the $x$ -axis.	Use upper bound minus lower bound.	Changing every negative part to positive without being asked.
Find the total area enclosed	Positive geometric area.	Split at roots where the curve crosses the $x$ -axis, then add magnitudes.	Integrating across a crossing and letting positive and negative regions cancel.
Find displacement from velocity	Signed change in position.	Integrate velocity over the time interval.	Calling this total distance travelled.
Find total distance from velocity	Total path length.	Split at times when velocity changes sign, then add absolute displacements.	Using one integral when the object reverses direction.

Inner expression	Derivative to look for	If the integrand has...	First rewrite
$x^{2} + 1$	$2x$	$6x$	$6x \space dx = 3 \space du$
$3x + 5$	$3$	no extra factor	$dx = \dfrac{1}{3} \space du$
$1 - 4x$	$-4$	no extra factor	$dx = -\dfrac{1}{4} \space du$
$x^{2} + 4$	$2x$	$x$	$x \space dx = \dfrac{1}{2} \space du$

IP AMaths Notes (Upper Sec, Year 3-4): 17) Integration Essentials

How this chapter applies

Choosing the integration route

Sources

Initial-condition checkpoint

1 Fundamental rules

Bounded-region checkpoint

Reverse-power exception checkpoint

2 Worked example - Definite integral

Signed area checkpoint

3 Worked example - Substitution

Substitution multiplier checkpoint

4 Worked example - Log form via substitution

5 Practice Quiz

6 Try this

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