Additional Mathematics Classroom | Chapter 13: Integration — Reverse Differentiation, Constants, Linear-Factor Integrals and G3 Trigonometric/Exponential/Logarithmic Extension | SEC G2 K232 / G3 K341

The 2027 SEC crosswalk

G2 K232 requires integration as reverse differentiation, integration of xⁿ for rational n excluding n = −1, integration of (ax+b)ⁿ for rational n excluding n = −1, together with constant multiples, sums and differences. The syllabus then continues into definite integrals and area, which this classroom series will treat in the next chapter.

G3 K341 contains the same reverse-differentiation architecture but widens the integration family. It includes xⁿ for rational n, sin x, cos x, sec²x and e^x, plus corresponding linear-argument forms such as (ax+b)ⁿ, sin(ax+b), cos(ax+b) and e^ax+b. This naturally brings the logarithmic antiderivative of 1/x and 1/(ax+b) into the G3 route when n = −1.

Official references: SEAB 2027 G2 Additional Mathematics K232 · SEAB 2027 G3 Additional Mathematics K341.

Integration is easiest to remember when every rule is treated as a derivative running backward.

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What this chapter is really teaching

Students often learn integration as “add one to the power and divide”. That shortcut works for one important family and fails as a complete model of the topic. Integration has four deeper ideas.

1. Inverse structure: integration asks which function differentiates to the expression we have.
2. Lost information: indefinite integration returns a family because differentiation erased constants.
3. Reverse Chain Rule recognition: a linear inside factor changes the coefficient required in the antiderivative.
4. Verification: differentiating the proposed antiderivative should return the original integrand.

This verification loop is unusually powerful. A differentiation answer can be difficult to check from first principles. An indefinite integration answer can be checked almost immediately: differentiate it. If the original integrand does not return, the antiderivative is wrong.

Chapter map

1. Integration as reverse differentiation
2. Antiderivatives and families of curves
3. Why +C is necessary
4. Power integration for rational powers
5. Constant-multiple and sum/difference rules
6. Simplifying before integrating
7. Functions involving a linear factor
8. Finding a particular function from an initial condition
9. G3: 1/x and logarithmic antiderivatives
10. G3: sine, cosine and sec² integration
11. G3: trigonometric linear arguments
12. G3: exponential integration
13. Reverse Chain Rule reasoning
14. Mixed algebraic preparation
15. Verification by differentiation
16. Common integration errors
17. G2/G3 route separation
18. Original guided practice
19. Worked answers and mastery checkpoint

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1 · Integration as reverse differentiation

Suppose

d/dx(x³) = 3x².

Read the statement backward:

an antiderivative of 3x² is x³.

We write

∫3x² dx = x³ + C.

The elongated ∫ symbol asks for accumulation/antidifferentiation. In an indefinite integral, dx identifies the variable with respect to which we are integrating.

Derivative language and integral language

If F′(x) = f(x), then

∫f(x) dx = F(x) + C.

F is called an antiderivative or primitive of f. Because F(x)+C has the same derivative for every constant C, the indefinite integral represents an entire family.

Worked reverse reading

• d/dx(5x⁴) = 20x³ → ∫20x³ dx = 5x⁴ + C.
• d/dx(−3/x²) = 6/x³ → ∫6x⁻³ dx = −3x⁻² + C.
• d/dx(4√x) = 2/√x → ∫2x^−1/2 dx = 4x^1/2 + C.

The cleanest memory is not “integration has a strange separate rule”. It is “what would differentiate to this?”

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2 · Why the constant of integration cannot be optional

Differentiate each of these:

• F₁(x)=x²;
• F₂(x)=x²+7;
• F₃(x)=x²−100.

All three derivatives are 2x.

If we are given only f(x)=2x and asked to integrate, we cannot know which constant the original function carried. Therefore

∫2x dx = x² + C.

C is not a decorative symbol added because the teacher expects it. It records information destroyed by differentiation.

Geometric meaning

All functions x²+C have the same shape and same gradient at corresponding x-values. They are vertical translations of one another. The derivative sees slope but not vertical position. Integration restores the family of possible vertical positions.

When C disappears

In a definite integral, the same constant appears at both limits and cancels. That is why +C belongs to indefinite integration but is not carried through the final evaluation of a definite integral. We will make that distinction explicit in the next chapter.

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3 · Power integration: reverse the Power Rule

For rational n with n ≠ −1:

∫xⁿ dx = xⁿ⁺¹/(n+1) + C.

Why add one and divide? Differentiate the proposed result:

d/dx[xⁿ⁺¹/(n+1)] = [(n+1)/(n+1)]xⁿ = xⁿ.

The integration rule is the differentiation Power Rule reversed exactly.

Worked route A · positive power

∫6x⁵ dx

= 6[x⁶/6] + C

= x⁶ + C.

Worked route B · negative power

∫4x⁻³ dx

= 4[x⁻²/(−2)] + C

= −2x⁻² + C

= −2/x² + C.

Worked route C · fractional power

∫3x^1/2 dx

= 3[x^3/2/(3/2)] + C

= 2x^3/2 + C.

Worked route D · reciprocal square root

∫5/√x dx = ∫5x^−1/2 dx.

= 5[x^1/2/(1/2)] + C

= 10√x + C.

Why n = −1 breaks the formula

If n = −1, then n+1 = 0 and the denominator in the power-antiderivative formula would be zero. That is not a small technical inconvenience. It signals a different antiderivative family: logarithms. G2 explicitly excludes this case. G3 includes the broader integration system in which the logarithmic antiderivative appears.

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4 · Constant multiples, sums and differences

Integration distributes across constant multiples, sums and differences:

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

Be careful with the word constant. A scalar coefficient can be taken outside the integral. A standalone constant term must itself be integrated.

Worked polynomial

Integrate

12x³ − 6x + 5.

∫(12x³ − 6x + 5)dx

= 3x⁴ − 3x² + 5x + C.

Answer: 3x⁴ − 3x² + 5x + C.

Notice the constant 5 became 5x. It did not “stay 5” and it did not disappear. Constants disappear under differentiation; they become linear terms under integration.

A common constant error

∫k dx = kx + C, where k is constant. Do not confuse this with the rule that a constant multiplier may be moved outside the integral.

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5 · Simplify before integrating

Just as differentiation becomes easier after a useful algebraic rewrite, integration often rewards preparation.

Example A · split a quotient

Integrate

(3x²+2x)/x.

For x ≠ 0, simplify:

3x + 2.

Then

∫(3x+2)dx = (3/2)x²+2x+C.

Example B · expand a simple product

Integrate x(x²+4).

Expand:

x³+4x.

Antiderivative:

x⁴/4+2x²+C.

Integration does not have a Product Rule inverse

Differentiation has Product Rule. There is no school-level rule saying “integrate each factor and multiply”. If an integrand is a product, first ask whether it can be expanded, simplified or recognised as a listed reverse-Chain form. More advanced integration techniques exist beyond the present SEC core, but they should not be invented from differentiation rules.

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6 · Integrating a function involving a linear factor

Both G2 and G3 require integration of

(ax+b)ⁿ

for the relevant rational n. For n ≠ −1:

∫(ax+b)ⁿdx = (ax+b)ⁿ⁺¹/[a(n+1)] + C.

The factor 1/a is the reverse of the Chain Rule. Differentiating (ax+b)ⁿ⁺¹ produces an extra factor a, so integration must compensate by dividing by a.

Worked route A

Integrate (3x+2)⁴.

Increase power to 5 and divide by 3×5:

∫(3x+2)⁴dx = (3x+2)⁵/15 + C.

Check:

d/dx[(3x+2)⁵/15] = (5/15)(3x+2)⁴·3 = (3x+2)⁴.

Worked route B · negative power

Integrate (2x−5)⁻³.

New power = −2.

Divide by 2(−2)=−4:

∫(2x−5)⁻³dx = −(1/4)(2x−5)⁻² + C.

Worked route C · fractional power

Integrate √(4x+1) = (4x+1)^1/2.

New power = 3/2.

Coefficient = 1/[4(3/2)] = 1/6.

∫√(4x+1)dx = (4x+1)^3/2/6 + C.

Reverse Chain Rule language

Differentiation:

outside derivative × inside derivative.

Integration of a linear inner factor:

reverse the outside derivative and compensate for the inside derivative.

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7 · Coefficients can already contain the inner derivative

Sometimes the integrand includes a coefficient that matches the derivative of the inner function. Then the reverse Chain Rule becomes especially visible.

Example A

Integrate 6(3x+1)⁴.

Since derivative of 3x+1 is 3, the coefficient 6 is twice the inner derivative. Direct rule:

∫6(3x+1)⁴dx = 6·(3x+1)⁵/15 + C

= (2/5)(3x+1)⁵ + C.

Example B

Integrate 10x(5x²+2)³.

This is a useful structural extension of reverse-Chain thinking. The inner derivative of 5x²+2 is 10x, which is already present. Therefore

∫10x(5x²+2)³dx = (5x²+2)⁴/4 + C.

This exact substitution-style pattern may appear as a recognition exercise when it reduces immediately to known antiderivatives. Do not turn it into a catalogue of advanced substitution techniques beyond the current syllabus; keep the reasoning anchored to reverse differentiation.

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8 · Finding one particular function from a condition

An indefinite integral gives F(x)+C. A condition such as “the curve passes through (2,5)” determines C and selects one member of the family.

Worked route A

Given

dy/dx = 6x² − 4x + 3

and y = 5 when x = 1, find y in terms of x.

Integrate:

y = 2x³ − 2x² + 3x + C.

Use x=1,y=5:

5 = 2 − 2 + 3 + C

5 = 3 + C

C = 2.

Therefore

y = 2x³ − 2x² + 3x + 2.

Worked route B · gradient information and point

A curve has gradient function

dy/dx = 4x − 6

and passes through (3,10). Find its equation.

Integrate:

y = 2x² − 6x + C.

At (3,10):

10 = 18 − 18 + C.

C = 10.

y = 2x² − 6x + 10.

The condition is not optional decoration

Without the point condition, there are infinitely many vertically translated curves with the same gradient function. The condition restores the missing information.

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9 · G3 extension: the logarithmic antiderivative

The power integration formula fails at n=−1 because division by n+1 would require division by zero. G3 has the function that fills this gap:

d/dx[ln|x|] = 1/x

on intervals that do not cross zero. Hence

∫(1/x)dx = ln|x| + C.

When a school question is restricted to x>0, this may be written ln x + C. The absolute-value form records the wider real antiderivative on positive or negative intervals separately.

Linear denominator

Since

d/dx[ln|ax+b|] = a/(ax+b),

we have

∫1/(ax+b) dx = (1/a)ln|ax+b| + C.

Worked route A

∫1/(3x+2) dx

= (1/3)ln|3x+2| + C.

Worked route B · numerator matches derivative

∫6x/(3x²+5) dx.

The derivative of 3x²+5 is 6x, so

∫6x/(3x²+5)dx = ln(3x²+5)+C.

The inside expression is always positive, so absolute-value notation is unnecessary here.

G2 boundary

G2 K232 explicitly excludes n=−1 from its rational-power integration requirements. Do not allow an older textbook’s broader chapter to silently turn logarithmic antiderivatives into G2 core content.

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10 · G3 extension: integrating sine, cosine and sec²

Reverse the derivative facts from Chapter 12:

• d(sin x)/dx = cos x → ∫cos x dx = sin x + C;
• d(cos x)/dx = −sin x → ∫sin x dx = −cos x + C;
• d(tan x)/dx = sec²x → ∫sec²x dx = tan x + C.

These standard trigonometric antiderivatives use radian measure, matching the derivative formulas from which they are reversed.

Worked route A

∫(4cos x − 3sin x)dx

= 4sin x + 3cos x + C.

Answer: 4sin x + 3cos x + C.

The sign on the sine antiderivative is easy to lose. Differentiate the answer: derivative of +3cos x is −3sin x, which returns the integrand.

Worked route B

∫5sec²x dx = 5tan x + C.

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11 · G3 extension: trigonometric functions with a linear argument

Reverse Chain Rule again.

• ∫cos(ax+b)dx = (1/a)sin(ax+b)+C;
• ∫sin(ax+b)dx = −(1/a)cos(ax+b)+C;
• ∫sec²(ax+b)dx = (1/a)tan(ax+b)+C.

The 1/a compensates for the factor a created when the inner linear expression is differentiated.

Worked route A

∫cos(4x−1)dx

= (1/4)sin(4x−1)+C.

Worked route B

∫sin(3−2x)dx.

Here a=−2:

−1/a = −1/(−2)=1/2.

Therefore

∫sin(3−2x)dx = (1/2)cos(3−2x)+C.

Check: derivative of (1/2)cos(3−2x) = (1/2)[−sin(3−2x)](−2)=sin(3−2x).

Worked route C

∫6sec²(3x+5)dx

= 6·(1/3)tan(3x+5)+C

= 2tan(3x+5)+C.

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12 · G3 extension: exponential integration

Since d(e^x)/dx=e^x:

∫e^xdx = e^x+C.

For a linear exponent:

∫e^ax+bdx = (1/a)e^ax+b+C.

Worked route A

∫e^5x−2dx

= (1/5)e^5x−2+C.

Worked route B

∫7e^−2xdx

= 7(−1/2)e^−2x+C

= −(7/2)e^−2x+C.

The self-reproducing function still needs inner compensation

e^x integrates to itself. e^3x does not integrate to e^3x unchanged because differentiating e^3x creates a factor 3. The antiderivative must contain 1/3.

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13 · Reverse Chain Rule is the unifying idea

The linear-factor power, trig and exponential rules look like separate formulae until their derivative origin is compared.

Integrand	Antiderivative idea
(ax+b)n	Reverse power, divide by a
cos(ax+b)	Reverse sine derivative, divide by a
sin(ax+b)	Reverse cosine derivative with minus sign, divide by a
sec²(ax+b)	Reverse tangent derivative, divide by a
eax+b	Keep exponential form, divide by a
1/(ax+b)	Reverse log derivative, divide by a

One principle controls all six rows: differentiating the antiderivative would create an inner factor a, so integration compensates by dividing by a.

Do not memorise six unrelated 1/a rules. Recognise one reversed Chain Rule.

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14 · Mixed examples: simplify until the integrand belongs to the library

The old textbook’s later integration examples are valuable because they show that difficult-looking integrands often become routine after algebraic or trigonometric simplification. The modern boundary is simple: use identities and algebra when they reduce the expression to integration forms actually in the syllabus.

Shared example · algebraic expansion

Integrate (x+1)(x²−x+3).

Expand:

x³+2x+3.

Therefore

∫(x³+2x+3)dx = x⁴/4+x²+3x+C.

G3 example · trigonometric identity first

Integrate sin²x+cos²x.

Use the identity:

sin²x+cos²x=1.

Hence

∫1 dx = x+C.

G3 example · sec² identity

Integrate 1+tan²x.

Since 1+tan²x=sec²x:

∫(1+tan²x)dx = tan x+C.

The integrand did not initially look like one of the memorised rows. The identity transformed it into a listed derivative pair.

G3 supporting example · logarithmic pattern

Integrate

(2x+3)/(x²+3x+7).

The denominator derivative is 2x+3, exactly the numerator. Therefore

∫(2x+3)/(x²+3x+7) dx = ln(x²+3x+7)+C

because x²+3x+7 is always positive. This is reverse-Chain recognition, not a new integration technique.

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15 · What is not the current integration core

It is useful to know the boundary so revision time stays focused.

• Formal integration by parts is not listed in the 2027 SEC Additional Mathematics core.
• General substitution methods beyond direct reverse-Chain recognition are not stated as a separate required technique.
• Advanced partial-fraction integration should not be allowed to dominate unless the expression reduces transparently to listed antiderivative forms.
• Area between two curves is explicitly excluded from the stated area requirement; the next chapter will respect that boundary.

An older textbook may contain useful examples beyond a current boundary. Keep them as support or enrichment only when they strengthen the listed mathematics without crowding out the actual examination route.

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16 · Verification by differentiation

Every indefinite integration answer comes with a built-in unit test:

Differentiate the answer. Do you recover the integrand?

Check A

Proposed answer:

∫(4x+1)³dx = (4x+1)⁴/16+C.

Differentiate:

(4/16)(4x+1)³·4 = (4x+1)³.

Correct.

Check B

Proposed answer:

∫cos(5x)dx = sin(5x)+C.

Differentiate:

5cos(5x).

We get five times the required integrand. The antiderivative is too large by factor 5. Correct answer:

(1/5)sin(5x)+C.

Check C

Proposed answer:

∫sin x dx = cos x+C.

Differentiate cos x: −sin x. Wrong sign. Correct answer is −cos x+C.

This verification habit should become automatic. It catches coefficient, sign and power errors before they survive to the final answer.

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17 · The integration error map

• Derivative-direction error: reducing the power instead of increasing it.
• Division error: adding one to n but forgetting to divide by n+1.
• n=−1 error: trying to use the power formula and dividing by zero.
• Constant-of-integration omission: giving one antiderivative when a family was requested.
• Constant-term error: ∫5dx written as 5 instead of 5x.
• Linear-inner-factor error: forgetting the 1/a compensation in (ax+b)ⁿ.
• Trig-sign error: ∫sin x dx written as +cos x.
• Exponential-factor error: ∫e^3xdx written as e^3x.
• Log-domain error: logarithmic antiderivative used without respecting zero/argument restrictions.
• Initial-condition error: integrating correctly but never using the point to determine C.
• Product fantasy: integrating a product by multiplying two antiderivatives.
• Expansion avoidance: leaving a simple polynomial product in a form that hides the easy integration route.
• Level-boundary error: teaching logarithmic/trig/exp antiderivatives as G2 core despite the K232 specification.

First weak-link diagnostic

1. Integrate x⁵.
2. Integrate x^−3/2.
3. Integrate (3x+1)⁴.
4. Recover y from dy/dx plus one point.
5. G3: integrate sin x, cos x, sec²x, e^x, 1/x.
6. G3: integrate cos(4x), e^−2x, 1/(3x+1).

Repair the first failure. If +C is the only problem, do not reteach powers. If the 1/a factor is missing repeatedly, return to reverse Chain Rule.

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18 · G2 and G3: one inverse idea, two integration libraries

The conceptual idea is shared: integration reverses differentiation. The difference is the library of functions whose antiderivatives students are required to know.

• G2 K232: xⁿ and (ax+b)ⁿ for rational n excluding −1, with constants, sums and differences.
• G3 K341: the wider rational-power family together with logarithmic n=−1 behaviour, sin, cos, sec² and e^x, including the stated linear-argument forms.

This is exactly why the old textbook remains useful but cannot be followed blindly. Its broader Chapter 14 can teach the whole mathematical landscape, while the current SEC codes decide which lane each learner must master.

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19 · Original guided practice

The following questions are original to this classroom guide. Questions marked G3 use the broader K341 integration library.

A · Shared power integration

1. Integrate 8x⁷.
2. Integrate 6x²−4x+3.
3. Integrate 5x⁻³.
4. Integrate 4x^1/2.
5. Integrate 3/√x.
6. Integrate 2x^5/3−7x^−1/2.
7. Integrate (3x²+6x)/x after simplifying.
8. Integrate x(x²−5x+2) after expanding.

B · Shared linear-factor powers

9. Integrate (2x+1)⁵.
10. Integrate (5x−3)⁻².
11. Integrate √(6x+1).
12. Integrate 4(2x+5)³.
13. Integrate 3/(4x+1)².
14. Integrate (7−3x)^1/2.

C · Particular functions

15. Given dy/dx=4x³−6x+2 and y=5 when x=1, find y.
16. A curve has gradient dy/dx=6x²+4 and passes through (0,−3). Find its equation.
17. Given f′(x)=3(x+2)² and f(0)=10, find f(x).
18. A particle has velocity v=6t²−4t+3 and displacement s=2 when t=0. Find s in terms of t. Treat this as reverse differentiation; full kinematics interpretation comes later.

D · G3 logarithmic integration

19. G3: Integrate 1/x.
20. G3: Integrate 1/(5x+2).
21. G3: Integrate 8/(4x−3).
22. G3: Integrate 2x/(x²+5).
23. G3: Integrate (6x+1)/(3x²+x+7).
24. G3: Explain why the ordinary power formula cannot be used when n=−1.

E · G3 trigonometric integration

25. G3: Integrate 5cos x.
26. G3: Integrate 4sin x.
27. G3: Integrate 3sec²x.
28. G3: Integrate cos(6x).
29. G3: Integrate sin(4x−1).
30. G3: Integrate sec²(3x+2).
31. G3: Integrate 2cos x−5sin x+4sec²x.
32. G3: Integrate sin²x+cos²x after simplifying.
33. G3: Integrate 1+tan²x after simplifying.

F · G3 exponential integration

34. G3: Integrate 7e^x.
35. G3: Integrate e^3x.
36. G3: Integrate 5e^−2x.
37. G3: Integrate 4e^1−5x.
38. G3: Given dy/dx=3e^2x and y=1 when x=0, find y.

G · Mixed reasoning

39. A student integrates x⁴ as 4x³+C. Diagnose the direction error.
40. A student integrates 5 as 5+C. Diagnose the constant-term error.
41. A student integrates (3x+1)⁴ as (3x+1)⁵/5+C. Diagnose the missing factor.
42. G3: A student integrates sin x as cos x+C. Diagnose the sign error.
43. G3: A student integrates e^4x as e^4x+C. Diagnose the missing factor.
44. Why is differentiating an indefinite-integral answer such a strong check?
45. Explain why there is no direct inverse Product Rule for elementary integration.
46. Explain the role of an initial condition after indefinite integration.
47. State the integration-library difference between G2 K232 and G3 K341.

20 · Worked answers and reasoning checkpoints

1. x⁸+C.
2. 2x³−2x²+3x+C.
3. 5x⁻³ integrates to 5x⁻²/(−2): −(5/2)x⁻²+C.
4. 4x^1/2 → 4x^3/2/(3/2): (8/3)x^3/2+C.
5. 3x^−1/2 → 6√x+C.
6. 2·x^8/3/(8/3) − 7·x^1/2/(1/2) = (3/4)x^8/3−14√x+C.
7. Simplify to 3x+6. Integral (3/2)x²+6x+C.
8. Expand x³−5x²+2x. Integral x⁴/4−(5/3)x³+x²+C.
9. (2x+1)⁶/12+C.
10. New exponent −1; denominator 5(−1)=−5: −1/[5(5x−3)]+C.
11. (6x+1)^3/2/9+C.
12. 4·(2x+5)⁴/[2·4] = (2x+5)⁴/2+C.
13. 3(4x+1)⁻² integrates to 3(4x+1)⁻¹/[4(−1)] = −3/[4(4x+1)]+C.
14. a=−3, n=1/2: coefficient 1/[−3(3/2)]=−2/9. −(2/9)(7−3x)^3/2+C.
15. Integrate: y=x⁴−3x²+2x+C. At x=1,y=5: 5=1−3+2+C=C, so y=x⁴−3x²+2x+5.
16. y=2x³+4x+C. At x=0,y=−3 gives C=−3. y=2x³+4x−3.
17. ∫3(x+2)²dx=(x+2)³+C. f(0)=8+C=10 → C=2. f(x)=(x+2)³+2.
18. s=2t³−2t²+3t+C. s(0)=2 gives C=2. s=2t³−2t²+3t+2.
19. ln|x|+C.
20. (1/5)ln|5x+2|+C.
21. 8·(1/4)ln|4x−3| = 2ln|4x−3|+C.
22. Inner derivative 2x matches numerator: ln(x²+5)+C.
23. Denominator derivative =6x+1: ln(3x²+x+7)+C.
24. Because the reverse-power formula divides by n+1. At n=−1 that denominator is zero; the antiderivative is logarithmic instead.
25. 5sin x+C.
26. −4cos x+C.
27. 3tan x+C.
28. (1/6)sin6x+C.
29. ∫sin(4x−1)dx = −(1/4)cos(4x−1)+C.
30. (1/3)tan(3x+2)+C.
31. 2sin x+5cos x+4tan x+C.
32. Identity reduces integrand to 1: x+C.
33. 1+tan²x=sec²x: tan x+C.
34. 7e^x+C.
35. (1/3)e^3x+C.
36. −(5/2)e^−2x+C.
37. a=−5: −(4/5)e^1−5x+C.
38. y=(3/2)e^2x+C. At x=0: 1=3/2+C → C=−1/2. y=(3/2)e^2x−1/2.
39. That is differentiation, not integration. Integration reverses the power rule: increase exponent to 5 and divide by 5, giving x⁵/5+C.
40. A constant term integrates to a linear term: ∫5dx=5x+C.
41. Differentiating (3x+1)⁵/5 creates an extra factor 3, giving 3(3x+1)⁴. Divide by 3 as well: correct (3x+1)⁵/15+C.
42. d(cos x)/dx=−sin x, so ∫sin x dx must be −cos x+C.
43. Differentiating e^4x gives 4e^4x. Correct antiderivative is (1/4)e^4x+C.
44. Because integration and differentiation are inverse operations. Differentiating the answer should reproduce the original integrand exactly, exposing wrong powers, signs and coefficients.
45. Product Rule describes how a product differentiates, but reversing a sum u′v+uv′ does not allow an arbitrary product uv to be integrated factor-by-factor. Integration requires recognition/simplification or more advanced techniques not created by simply reversing Product Rule.
46. The initial condition determines C and selects one particular function from the family of antiderivatives.
47. G2 integrates rational powers and linear-factor powers excluding n=−1. G3 has the wider library including logarithmic n=−1 behaviour, sine, cosine, sec² and exponential functions with the stated linear-argument forms.

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21 · A two-week integration architecture

1. Day 1: reverse Power Rule and +C.
2. Day 2: negative/fractional powers and algebraic rewriting.
3. Day 4: (ax+b)ⁿ and reverse Chain Rule.
4. Day 6: particular functions from initial conditions.
5. Day 8: G3 logarithmic antiderivatives; G2 continues mixed rational-power retrieval.
6. Day 10: G3 trig and exponential antiderivatives with linear arguments.
7. Day 12: mixed simplification and verification by differentiation.
8. Day 14: unlabelled integration set preparing for definite integrals and area.

Six-minute retrieval test

• State ∫xⁿdx for n≠−1.
• Explain +C.
• Integrate (3x+1)⁴.
• Recover a function from dy/dx plus one point.
• G3: state ∫1/x, ∫sin x, ∫cos x, ∫sec²x, ∫e^x.
• Explain the 1/a factor for a linear inner function.

22 · For teachers: always close the loop with differentiation

Integration lessons can become fragile because students are asked to trust an unfamiliar reverse rule. Make verification part of the standard solution, especially while a new family is being learned.

After integrating (4x+1)³, differentiate the answer aloud. After integrating sin x, check the sign by differentiating −cos x. After integrating e^5x, show why the factor 1/5 is exactly what cancels the inner derivative.

The goal is not to double the length of every exam solution. It is to build an internal error-detection mechanism until the inverse pairs become automatic.

Teach n=−1 as a boundary, not an exception to memorise

Let students try the power rule symbolically. They will reach division by zero. That failure creates a meaningful reason for the logarithmic antiderivative and also makes the G2/G3 syllabus boundary easy to understand.

Do not rush to area too early

Indefinite integration should stabilise first. Area questions add a new conceptual layer: limits, signed accumulation and region geometry. The old textbook was right to separate Integration from Applications of Integration; this series keeps that separation.

23 · For parents: what real progress sounds like

A student relying on a slogan may say, “Integration means add one and divide.” A student gaining structural control begins to say:

• “I can check this by differentiating my answer.”
• “I need +C because the derivative lost the vertical constant.”
• “The inside is 3x+1, so I need to compensate for the factor 3.”
• “n=−1 is not a normal power-rule case; in G3 it becomes a logarithm.”
• “This product expands into ordinary powers, so I should simplify first.”
• “The point condition is what tells me which value of C belongs to this curve.”
• “This trig integral is G3 extension content, not part of my G2 core.”

That language shows integration has become reverse reasoning rather than a memorised motion of the exponent.

24 · What carries forward

The old textbook’s next chapter is Applications of Integration: definite integrals and area under a curve. SEC 2027 keeps that progression. Both G2 and G3 require definite integrals, evaluation and area of a region bounded by a curve and line(s), with area between two curves excluded. G3 additionally requires areas of regions below the x-axis.

The next classroom chapter will therefore move from “which family of functions differentiates to this?” to a second meaning of integration: how much signed accumulation occurs between two boundaries? That is where limits, F(b)−F(a), area, axis crossings and geometric interpretation become central.

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Official syllabus references

• Singapore Examinations and Assessment Board · 2027 SEC G2 Additional Mathematics K232
• Singapore Examinations and Assessment Board · 2027 SEC G3 Additional Mathematics K341

Curriculum and assessment requirements can change. The official SEAB syllabuses remain the controlling sources for current subject codes, examinable content and examination structure.

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