Integration: Reconstructing a Function from Its Rate of Change
Differentiation asks how a function changes. Integration reverses that process: from a rate or gradient, rebuild the accumulated function.
Integration is introduced in Additional Mathematics as the reverse of differentiation, but that phrase deserves to be taken seriously. If differentiation removes information about the original vertical position, integration must restore a family of possible functions using an arbitrary constant. If a derivative describes rate, integration accumulates that rate. If a graph lies above or below the x-axis, a definite integral combines signed contributions over an interval.
This guide develops integration through three linked ideas: antiderivatives, definite integrals and area. The aim is not to memorise a table of reversed derivatives. The aim is to understand why the constant of integration appears, how limits convert a family of antiderivatives into one numerical result, and why geometric area sometimes requires more than simply reporting a signed integral.
AI Extraction Box: The Core Map
- Integration as reverse differentiation: if F′(x)=f(x), then ∫f(x)dx=F(x)+C.
- Power rule: ∫xⁿdx=xⁿ⁺¹/(n+1)+C, for n≠−1.
- ∫1/x dx=ln|x|+C where appropriate in the real domain; syllabus use of ln x often appears over positive domains.
- ∫sin x dx=−cos x+C.
- ∫cos x dx=sin x+C.
- ∫sec²x dx=tan x+C.
- ∫eˣ dx=eˣ+C.
- Linear inside: ∫(ax+b)ⁿdx=(ax+b)ⁿ⁺¹/[a(n+1)]+C for n≠−1.
- ∫sin(ax+b)dx=−cos(ax+b)/a+C.
- ∫cos(ax+b)dx=sin(ax+b)/a+C.
- ∫e^(ax+b)dx=e^(ax+b)/a+C.
- Definite integral: ∫ₐᵇf(x)dx=F(b)−F(a).
- Signed area: contributions below the x-axis are negative in the definite integral.
Why the Constant of Integration Exists
Differentiate:
x²+3, x²−7 and x²+100.
All three derivatives are 2x. Differentiation removes constant vertical shifts because constants have derivative zero. Therefore, if we are told only that dy/dx=2x, we cannot know which vertical shift was originally present.
So:
∫2x dx=x²+C.
The arbitrary constant C represents the information differentiation erased.
+C is not a ritual. It stores the missing vertical position.
The Power Rule for Integration
Since differentiating xⁿ⁺¹/(n+1) gives xⁿ, we obtain:
∫xⁿdx=xⁿ⁺¹/(n+1)+C, n≠−1.
The routine is:
- increase the power by 1;
- divide by the new power;
- include the constant of integration for an indefinite integral.
Examples:
- ∫x⁴dx=x⁵/5+C;
- ∫x⁻²dx=−x⁻¹+C=−1/x+C;
- ∫√x dx=∫x¹ᐟ²dx=(2/3)x³ᐟ²+C.
The exception n=−1 matters because increasing −1 by 1 gives zero, so the power formula would require division by zero. That case connects instead to the logarithm.
Worked Example 1: Integrate a Polynomial
Find:
∫(6x²−4x+5)dx.
Integrate term by term:
6(x³/3)−4(x²/2)+5x+C
=2x³−2x²+5x+C.
Differentiate the result to check:
6x²−4x+5.
This reverse check is one of the best habits in integration because differentiation is usually quicker than redoing the whole integral.
Standard Trigonometric and Exponential Integrals
From the standard derivative table:
∫sin x dx=−cos x+C
∫cos x dx=sin x+C
∫sec²x dx=tan x+C
∫eˣ dx=eˣ+C.
The negative sign in ∫sin x dx is a common error point. Differentiate −cos x: the derivative is +sin x. That quick derivative check is often enough to restore the correct sign.
Linear Inner Functions: Reverse the Chain Rule
Consider:
∫(3x+2)⁵dx.
If we guessed (3x+2)⁶/6 and differentiated, the Chain Rule would introduce an extra factor 3. To compensate, divide by that inner derivative:
∫(3x+2)⁵dx=(3x+2)⁶/(18)+C.
In general:
∫(ax+b)ⁿdx=(ax+b)ⁿ⁺¹/[a(n+1)]+C, n≠−1.
This is a controlled reverse-Chain-Rule pattern.
Worked Example 2: Linear-Inside Trigonometric and Exponential Forms
Integrate cos(4x−1).
∫cos(4x−1)dx=(1/4)sin(4x−1)+C.
Integrate e^(5x+2).
∫e^(5x+2)dx=(1/5)e^(5x+2)+C.
Integrate sin(2x+3).
∫sin(2x+3)dx=−(1/2)cos(2x+3)+C.
Each answer can be checked by differentiation. The inner coefficient should cancel with the dividing factor.
Recovering a Particular Function from a Derivative
If dy/dx is given together with one point on the curve, integration finds the general family and the point determines C.
Example: dy/dx=6x−4 and the curve passes through (2,3).
Integrate:
y=3x²−4x+C.
Use (2,3):
3=3(4)−8+C=4+C,
so C=−1.
Therefore:
y=3x²−4x−1.
The point restores the information differentiation lost.
Definite Integrals: Accumulation Between Two Bounds
A definite integral has limits:
∫ₐᵇ f(x)dx.
If F′(x)=f(x), then:
∫ₐᵇf(x)dx=F(b)−F(a).
The arbitrary constant disappears because:
[F(b)+C]−[F(a)+C]=F(b)−F(a).
This is why definite integrals do not end with +C.
Worked Example 3: Evaluate a Definite Integral
Evaluate:
∫₀²(3x²+1)dx.
An antiderivative is x³+x. Therefore:
[x³+x]₀²=(8+2)−0=10.
The result is a number because the interval has been accumulated completely.
Area Under a Curve
When f(x)≥0 on [a,b], the area between y=f(x), the x-axis and the vertical lines x=a and x=b is:
Area=∫ₐᵇ f(x)dx.
But the word area must be interpreted geometrically. A definite integral is signed. If the graph lies below the x-axis, the integral contribution is negative even though geometric area is positive.
Integral measures signed accumulation; geometric area measures positive size.
Worked Example 4: Area Above the x-Axis
Find the area under y=x²+1 from x=0 to x=2.
Area=∫₀²(x²+1)dx
=[x³/3+x]₀²
=8/3+2
=14/3 square units.
The function is positive over the interval, so the definite integral equals the geometric area directly.
Regions Below the x-Axis
If f(x)<0 on an interval, then ∫f(x)dx over that interval is negative. To obtain geometric area, take the magnitude of that contribution.
For a curve that crosses the x-axis, split the interval at the roots. Evaluate each section separately and convert negative contributions to positive area.
This is why finding x-intercepts often comes before an area calculation. The roots tell us where the sign of the function can change.
Worked Example 5: Area with a Sign Change
Consider y=x−1 from x=0 to x=3. The line crosses the x-axis at x=1.
From 0 to 1, the graph is below the axis:
∫₀¹(x−1)dx=[x²/2−x]₀¹=−1/2.
Geometric area there is 1/2.
From 1 to 3:
∫₁³(x−1)dx=[x²/2−x]₁³=2.
Total geometric area:
1/2+2=5/2 square units.
If we had integrated from 0 to 3 in one step, the signed integral would be 3/2, which is not the total geometric area.
Area Between a Curve and a Line
Within the syllabus scope, a region can be bounded by a curve and line(s). If the line lies above the curve on [a,b], the vertical gap is:
line − curve.
Then:
Area=∫ₐᵇ[(upper function)−(lower function)]dx.
The intersection points usually determine the limits. Solving the line-curve equation therefore comes before integration.
The 2027 G3 syllabus excludes area of a region between two curves, so this guide keeps examples within curve-and-line boundaries.
Worked Example 6: Region Between a Curve and a Line
Find the area between y=x² and y=2x from x=0 to x=2.
On this interval, 2x≥x². Therefore:
Area=∫₀²(2x−x²)dx
=[x²−x³/3]₀²
=4−8/3
=4/3 square units.
A sketch is useful because it confirms which function is upper and prevents reversing the subtraction.
Integration as Accumulation
The area interpretation is one example of accumulation. More generally, if a rate is integrated over time, the result is accumulated change. This prepares the way for motion:
- integrating velocity over time gives displacement change;
- integrating acceleration gives velocity change;
- initial conditions determine the constant of integration.
That motion system is developed further in the next batch. The key foundation here is that integration accumulates a rate across an interval.
The Integration Decision Tree
- Indefinite integral? Find antiderivative and include +C.
- Derivative given with a point? Integrate first, then use the point to find C.
- Definite integral? Find F(b)−F(a); no +C needed in final evaluation.
- Area above x-axis? Definite integral can equal area directly.
- Curve crosses x-axis? Find roots, split the interval and convert negative contributions to positive area.
- Area between line and curve? Find intersections, identify upper−lower, then integrate.
- Linear inner expression? Reverse the Chain Rule by compensating for the inner derivative.
Checking an Integral by Differentiation
After finding an indefinite integral F(x)+C, differentiate F(x). If the original integrand does not return exactly, the integration contains an error.
This check is especially useful for:
- linear-inside forms such as (ax+b)ⁿ;
- signs in sine/cosine integrals;
- missing inner-coefficient divisors;
- fractional or negative powers.
Differentiation is the audit trail of integration.
Common Failure Modes
| Visible error | Underlying issue | Repair |
|---|---|---|
| +C omitted from indefinite integral | Lost-information meaning forgotten | Remember differentiation removes constants |
| Power decreased instead of increased | Differentiation rule used in reverse incorrectly | Add 1 to exponent, divide by new exponent |
| ∫sin x dx=cos x | Sign not checked | Differentiate proposed answer |
| Inner coefficient not divided out | Reverse Chain Rule incomplete | Differentiate answer and compensate |
| +C included after evaluating definite limits | Indefinite and definite forms confused | C cancels in F(b)−F(a) |
| Negative definite integral reported as negative area | Signed accumulation confused with geometric size | Split at roots and use positive magnitudes for area |
| Upper/lower functions reversed | No sketch or comparison | Identify vertical gap before integrating |
Transfer Set
Question A
Find ∫(4x³−6x)dx.
Answer: x⁴−3x²+C.
Question B
Find ∫e^(3x)dx.
Answer: (1/3)e^(3x)+C.
Question C
Evaluate ∫₁³2x dx.
Answer: [x²]₁³=9−1=8.
Question D
If dy/dx=3x² and y=5 when x=1, find y.
Answer: y=x³+C. 5=1+C, so C=4. Hence y=x³+4.
Question E
Why must an area calculation sometimes be split at an x-intercept?
Answer: the definite integral is signed; a region below the x-axis contributes negatively, whereas geometric area is positive.
A 50-Minute Integration Repair Session
- 7 minutes: retrieve standard integrals and the power rule.
- 8 minutes: integrate six powers, including negative and fractional exponents.
- 8 minutes: integrate four linear-inside trig/exponential/power expressions and differentiate to check.
- 7 minutes: recover two functions from derivatives plus initial conditions.
- 8 minutes: evaluate three definite integrals.
- 8 minutes: calculate one area crossing the x-axis and one area between a line and curve.
- 4 minutes: classify every error as rule, constant, sign, limit, boundary or interpretation.
What Mastery Looks Like
- The learner explains +C as information lost under differentiation.
- The learner integrates rational powers and standard trig/exponential forms accurately.
- The learner recognises and compensates for linear inner derivatives.
- The learner recovers a particular function from a derivative and one point.
- The learner distinguishes indefinite and definite integrals.
- The learner interprets a definite integral as signed accumulation.
- The learner splits area problems at sign changes and identifies upper−lower for curve-line regions.
- The learner checks indefinite integrals by differentiation.
Syllabus Alignment
This guide aligns with the integration component of the 2027 Singapore-Cambridge SEC G3 Additional Mathematics syllabus C1: integration as reverse differentiation; integration of rational powers, sin x, cos x, sec²x and eˣ with constants, sums and differences; integration of (ax+b)ⁿ, sin(ax+b), cos(ax+b) and e^(ax+b); definite integrals; area under a curve; regions bounded by a curve and line(s), excluding the area between two curves; and regions below the x-axis. Motion applications are continued separately in the next batch.
Official SEAB 2027 G3 syllabus index
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Batch 03 now connects proof and the main calculus engine: differentiation rules, applications of differentiation and integration.
