Secondary 3 Additional Mathematics | Singapore G3
Integration

How Can You Reconstruct a Function From Its Rate of Change?

If you know that a function’s derivative is 6x, you still do not know the original function completely.

It could be:

• 3x²;
• 3x²+5;
• 3x²−100;
• 3x²+π.

Every one of them differentiates to 6x.

So integration is not merely “differentiate backwards.” It asks a deeper reconstruction question:

What whole family of functions could have produced this rate of change, and what extra information is needed to identify one member of the family?

That is why the constant of integration is not an optional symbol added because a textbook says so. It records information that differentiation erased.

The 2027 Singapore G3 Additional Mathematics syllabus includes integration as the reverse of differentiation; integration of rational powers, sin x, cos x, sec²x and eˣ; corresponding forms involving ax+b; and later definite integrals, area and straight-line kinematics. This page owns the indefinite-integration machinery. Definite area and kinematics have separate owners so the concepts do not blur together.

SEAB 2027 G3 Additional Mathematics syllabus (K341) →

The Topic Job

Recover a function from a known derivative, preserve the family of possible originals, and use additional conditions when one specific original function is required.

If a learner can only apply an integration formula, the method is fragile. The durable model is:

rate of change → reverse the derivative → restore lost constant information → differentiate back to verify.

What Differentiation Loses

Differentiate:

y=4x³+7.

We obtain:

dy/dx=12x².

The +7 vanished.

Differentiate y=4x³−12 or y=4x³+1000. The derivative is still 12x².

Therefore, when we reverse the process:

∫12x² dx = 4x³ + C.

The symbol C represents an arbitrary constant.

Why +C Is Mathematically Necessary

If you wrote only 4x³, you would be claiming there is exactly one antiderivative.

But:

d/dx(4x³+C)=12x²

for every constant C.

The constant of integration is a receipt for information destroyed by differentiation.

Integration Notation

The expression:

∫f(x) dx

asks for a function whose derivative is f(x).

• ∫ is the integral sign;
• f(x) is the integrand;
• dx identifies x as the variable of integration;
• +C is required for an indefinite integral.

The dx is not decorative. It tells you which variable the reversal is with respect to.

The Reverse Power Rule

Differentiation says:

d/dx(xⁿ⁺¹)=(n+1)xⁿ.

Reverse it:

∫xⁿ dx = xⁿ⁺¹/(n+1)+C, provided n≠−1.

The routine is:

add 1 to the power → divide by the new power → add C.

But the routine is only trustworthy if you can differentiate your answer back.

Worked Example 1 — Rational Powers

Integrate:

6x²−4x⁻³+5√x.

Rewrite √x as x¹ᐟ²:

∫(6x²−4x⁻³+5x¹ᐟ²)dx.

Integrate term by term:

2x³ + 2x⁻² + (10/3)x³ᐟ² + C.

Now differentiate it:

• d/dx(2x³)=6x²;
• d/dx(2x⁻²)=−4x⁻³;
• d/dx[(10/3)x³ᐟ²]=5x¹ᐟ².

The original integrand returns. That is the strongest routine check available in this topic.

The Exceptional Power n = −1

The reverse power rule would require division by n+1=0 when n=−1, so it breaks.

Instead:

∫1/x dx = ln|x| + C

on intervals where x≠0.

The current K341 syllabus explicitly lists ln x among derivative functions but its stated integration list focuses on xⁿ, sin x, cos x, sec²x and eˣ plus linear-composite forms. Teachers may connect 1/x to ln|x| for structural understanding, but students should follow their school’s exact examinable scope. The important reasoning boundary is: do not force the power rule through n=−1.

Standard Integrals You Should Be Able to Reverse

Derivative fact	Reverse integral
d/dx(sin x)=cos x	∫cos x dx = sin x + C
d/dx(cos x)=−sin x	∫sin x dx = −cos x + C
d/dx(tan x)=sec²x	∫sec²x dx = tan x + C
d/dx(eˣ)=eˣ	∫eˣ dx = eˣ + C

Notice the sign in ∫sin x dx. A quick differentiation check prevents the common error +cos x.

Integration Is Linear

Constants can be taken outside and sums can be integrated term by term:

∫[af(x)+bg(x)]dx = a∫f(x)dx + b∫g(x)dx.

This is why long polynomial integrals can be decomposed into simple pieces.

Linear Composite Forms: The Hidden Chain Rule

Consider:

∫(3x+2)⁵ dx.

If we guessed:

(3x+2)⁶/6,

differentiating gives:

3(3x+2)⁵,

which is three times too large.

So divide by the derivative of the inner linear expression:

∫(3x+2)⁵ dx = (3x+2)⁶/18 + C.

Differentiate back:

6(3x+2)⁵·3/18=(3x+2)⁵.

Reverse chain rule for ax+b: integrate the outer pattern, then compensate for the inner derivative a.

General Linear-Power Form

For n≠−1:

∫(ax+b)ⁿ dx = (ax+b)ⁿ⁺¹/[a(n+1)] + C.

The denominator contains two pieces:

• n+1 reverses the power rule;
• a compensates for differentiating ax+b.

Linear Trigonometric Forms

Because:

d/dx[sin(ax+b)] = a cos(ax+b),

we have:

∫cos(ax+b)dx = (1/a)sin(ax+b)+C.

Similarly:

∫sin(ax+b)dx = −(1/a)cos(ax+b)+C.

Linear Exponential Forms

Since:

d/dx[e^(ax+b)] = a e^(ax+b),

the reverse is:

∫e^(ax+b)dx = (1/a)e^(ax+b)+C.

The same compensation idea appears again.

Worked Example 2 — Mixed Standard Forms

Integrate:

4(2x−1)³ + 6cos(3x) − 5e^(5x).

Term 1:

∫4(2x−1)³dx = 4(2x−1)⁴/[2·4] = (2x−1)⁴/2.

Term 2:

∫6cos(3x)dx = 2sin(3x).

Term 3:

∫−5e^(5x)dx = −e^(5x).

Therefore:

(2x−1)⁴/2 + 2sin(3x) − e^(5x) + C.

Differentiate the whole expression and every coefficient should reconstruct the integrand exactly.

Finding the Constant From a Point

Suppose:

dy/dx = 6x−4

and the curve passes through (2,7).

Integrate:

y=3x²−4x+C.

Use the point:

7=3(2²)−4(2)+C
7=12−8+C
C=3.

So the specific curve is:

y=3x²−4x+3.

The point did not help us integrate. It selected one curve from the family after integration.

Initial Conditions Are Information-Recovery Conditions

In geometry, an initial condition may be a point on a curve.

In kinematics, it may be initial displacement or velocity.

In every case, the logic is the same:

differentiation erased a constant → integration restores C → one known state determines C.

Integration Is Not Algebraic Inversion

If dy/dx=3x², we do not “move d x across” and divide symbols until y appears.

The notation is powerful, but the mathematical operation is finding antiderivatives.

At later levels, differential notation supports deeper substitution and differential-equation techniques, but at Sec 3/4 the safest conceptual model is function reversal, not symbolic fraction manipulation.

How to Choose the Method

Integrand structure	First thought
sum of powers of x	reverse power rule term by term
(ax+b)ⁿ	reverse power + compensate by 1/a
sin(ax+b), cos(ax+b)	reverse standard trig derivative + 1/a
e^(ax+b)	same exponential form + 1/a
known derivative plus a point	integrate first, then use point to find C

The Earliest Weak Link

What you see	Likely earliest weak link
forgets +C	does not understand information loss under differentiation
adds 1 to power but forgets division	reverse power rule memorised incompletely
gets ∫sin x = cos x	does not differentiate answer back
integrates (3x+1)⁴ without 1/3 factor	chain-rule compensation not recognised
substitutes point before integrating	does not distinguish derivative equation from original function
uses power rule on x⁻¹	exception condition n≠−1 ignored

Common Mistakes to Repair

• No +C on an indefinite integral.
• Differentiate instead of integrate because the symbols look familiar.
• Divide by the old power instead of the new power.
• Forget the inner linear coefficient.
• Use +cos x for ∫sin x dx.
• Force the power rule through n=−1.
• Find C but forget to write the final function.

Retrieval Check

1. Why is +C needed in an indefinite integral?
2. Integrate 5x⁴.
3. Integrate 3x⁻².
4. Integrate cos x.
5. Integrate sin x.
6. Integrate e^(4x).
7. Integrate (2x+3)⁵.
8. What is the strongest routine check for an indefinite integral?

Transfer Set

1. Find ∫(8x³−6x¹ᐟ²+4)dx.
2. Find ∫[5cos(2x)+3e^(3x)]dx.
3. Find ∫6(4x−1)²dx.
4. Given dy/dx=4x³−2x and y=5 when x=1, find y in terms of x.
5. Find a function f if f′(x)=3/√x and f(4)=10.
6. A student writes ∫(5x+2)³dx=(5x+2)⁴/4+C. Diagnose the exact missing factor and prove your diagnosis by differentiation.

Answer outline — open only after attempting

1. (8/4)x⁴−6/(3/2)x³ᐟ²+4x+C = 2x⁴−4x³ᐟ²+4x+C.
2. (5/2)sin(2x)+e^(3x)+C.
3. (4x−1)³/2+C.
4. y=x⁴−x²+C; 5=1−1+C, so C=5; y=x⁴−x²+5.
5. 3x⁻¹ᐟ² integrates to 6√x+C; 10=12+C, so C=−2; f(x)=6√x−2.
6. It is missing division by the inner derivative 5. Correct answer=(5x+2)⁴/20+C; differentiating gives (4/20)(5x+2)³·5=(5x+2)³.

Unfamiliar Transfer — Reconstruct Before You Calculate

Suppose a curve has gradient:

dy/dx = 2e^(2x) − 3sin(3x)

and passes through the origin.

Before calculating, name the structure:

• e^(2x) needs compensation by 1/2;
• sin(3x) integrates to −(1/3)cos(3x);
• a point will determine C after the antiderivative is found.

The calculation is now an execution of a model already chosen.

A Strong Answer Has Receipts

• Derivative receipt: differentiate your answer and recover the integrand.
• Condition receipt: substitute the given point/initial state and verify it.
• Domain receipt: check that rational powers or logs are used where defined.
• Structure receipt: confirm every ax+b form contains the compensating factor 1/a.

The check should be capable of disagreeing with your working. Otherwise it is not really a check.

Why This Is Worth Learning

Integration answers a recurring mathematical question: if we observe how something changes locally, can we reconstruct the accumulated whole?

That idea later appears in:

• recovering displacement from velocity;
• recovering velocity from acceleration;
• finding accumulated area from infinitesimal strips;
• probability densities;
• work done by changing forces;
• population and finance models;
• differential equations across science and engineering.

The Sec 3/4 rules are therefore not an isolated examination ritual. They are the first controlled version of reconstruction from change.

For Parents and Tutors — What This Topic Is Really Testing

Integration exposes whether differentiation was actually understood.

If the student cannot differentiate the proposed answer back, do not assign more integration questions yet. Repair the derivative map first.

Use this teaching sequence:

1. Ask what differentiation would erase.
2. Reconstruct one antiderivative.
3. Add +C and explain why.
4. Differentiate back.
5. Only then introduce ax+b compensation.
6. Use a point to choose one member of the family.
7. Mix polynomial, trig and exponential forms so method choice becomes necessary.

Listen for a student who can say:

“I know what this must differentiate back to.”

That sentence is a much stronger sign of readiness than speed alone.

Where This Connects Next

• Differentiation | How a Curve Can Have a Gradient at One Exact Point
• Definite Integrals & Area — signed accumulation and geometric area
• Kinematics — displacement, velocity and acceleration
• Secondary 3 Additional Mathematics: Complete A-Math Map
• Additional Mathematics Directory

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Bukit Timah Tutor Mathematics
Integration is not the opposite button to differentiation. It is the disciplined reconstruction of what differentiation can no longer tell you by itself.