Integration is the part of Additional Mathematics that rebuilds a quantity from its rate of change. Differentiation asks how a function is changing at an instant. Integration asks what total quantity could have produced that change, or how much accumulated effect has been created across an interval.
This Secondary 4 Additional Mathematics Learning Guide develops integration as a connected reasoning system: antiderivatives, constants of integration, definite integrals, area under and between curves, kinematics, modelling, verification and examination control. It belongs to the Additional Mathematics Learning Hub.
Differentiation takes a function apart into its local rate. Integration reconstructs a total from that rate.
Integration reverses differentiation
If differentiating x³ gives 3x², then integrating 3x² should return x³—but not only x³. The functions x³+1, x³−7 and x³+100 all have derivative 3x². That is why indefinite integration includes an arbitrary constant C.
∫3x² dx = x³ + C.
| Process | Question |
|---|---|
| Differentiation | How is this quantity changing now? |
| Indefinite integration | What family of functions could have produced this rate? |
| Definite integration | What net accumulation occurs across this interval? |
The integration power rule
For n≠−1, ∫xndx=xn+1/(n+1)+C. Increase the exponent by one and divide by the new exponent.
- ∫x⁴dx=x⁵/5+C
- ∫6x²dx=2x³+C
- ∫x⁻²dx=−x⁻¹+C
- ∫5dx=5x+C
A strong checking habit is to differentiate the proposed antiderivative. If it does not recover the original integrand, the integration is wrong.
Worked Example 1 | Integrate a polynomial
∫(4x³−6x+5)dx=x⁴−3x²+5x+C.
Rewrite awkward forms before integrating
Roots and denominators often become easier as indices. For example, √x=x1/2 and 1/x³=x−3.
Worked Example 2 | Fractional and negative powers
∫(2x1/2+3x−3)dx=(4/3)x3/2−(3/2)x−2+C.
The special 1/x case
The power rule fails at n=−1 because the new exponent would be zero and division by zero would occur. Instead:
∫(1/x)dx=ln|x|+C.
Reverse chain rule
Look for an inside function and its derivative. If differentiation of (2x+1)⁵ produces 10(2x+1)⁴, then integrating 10(2x+1)⁴ returns (2x+1)⁵+C.
Worked Example 3 | Reverse chain structure
∫6(3x+2)⁴dx=(2/5)(3x+2)⁵+C. Differentiate the answer to verify the factor.
Recovering the constant from a point
An indefinite integral produces a family of curves. A known point selects one member of that family.
Worked Example 4 | Find the original function
Given dy/dx=6x−4 and the curve passes through (3,10), integrate to get y=3x²−4x+C. Substitution gives 10=27−12+C, so C=−5. Therefore y=3x²−4x−5.
Definite integrals
If F′(x)=f(x), then ∫abf(x)dx=F(b)−F(a). The arbitrary constant cancels, so a definite integral produces a numerical accumulation rather than a family of functions.
Worked Example 5 | Evaluate a definite integral
∫13(2x+1)dx=[x²+x]13=(12)−(2)=10.
Area under a curve
When f(x) lies above the x-axis from a to b, the definite integral equals the geometric area bounded by the curve, the x-axis and the vertical boundaries.
Worked Example 6 | Area under a parabola
Area under y=x² from x=0 to 2 is ∫02x²dx=[x³/3]02=8/3 square units.
Signed integral versus geometric area
When a curve lies below the x-axis, its definite integral contribution is negative. Geometric area is always non-negative. If a question asks for total area, split the interval at crossings and add magnitudes.
Integral = signed accumulation. Area = positive geometric size.
Worked Example 7 | Crossing the x-axis
For y=x−1 on 0≤x≤3, the definite integral is 3/2, but total geometric area is 5/2 because the small region below the axis must be added positively.
Area between two curves
If f(x) lies above g(x), the vertical height is f(x)−g(x), so area=∫[f(x)−g(x)]dx across the correct boundaries.
Worked Example 8 | Line and parabola
For y=2x and y=x², intersections satisfy x²=2x, so x=0,2. Between them, 2x is above x². The area is ∫02(2x−x²)dx=4/3 square units.
When the top curve changes
- Find every relevant intersection.
- Order the boundaries.
- Determine the upper curve in each subinterval.
- Integrate top minus bottom on each piece.
- Add the positive areas.
Accumulated change beyond area
If v(t) is velocity, integrating velocity across time gives displacement. If a(t) is acceleration, integrating acceleration gives change in velocity.
Rate × a very small interval → a very small change. Integration adds all those changes.
Kinematics chain
s(t) differentiates to v(t), which differentiates to a(t). Integration reverses the direction. Constants are determined from initial or known conditions.
Worked Example 9 | Recover velocity from acceleration
If a=6t−4 and v=3 when t=0, then v=3t²−4t+C and C=3. Hence v=3t²−4t+3.
Worked Example 10 | Recover displacement
If v=4t−t² and s=5 when t=0, then s=2t²−t³/3+5.
Distance is not displacement
Integrating velocity gives signed displacement. If velocity changes sign, the particle reverses direction. Total distance requires splitting at the times when v=0 and adding the magnitudes of the separate displacements.
Worked Example 11 | Displacement versus total distance
For v=t−2 on 0≤t≤5, displacement is 5/2 units. Velocity changes sign at t=2, so total distance is 2+9/2=13/2 units.
Exponential and trigonometric integration
- ∫exdx=ex+C
- ∫ekxdx=(1/k)ekx+C
- ∫cos x dx=sin x+C
- ∫sin x dx=−cos x+C
Worked Example 12 | Exponential integration
∫4e2xdx=2e2x+C.
Worked Example 13 | Trigonometric integration
∫3cos(2x)dx=(3/2)sin(2x)+C.
Graphical plausibility
Before integrating, inspect whether the integrand is mostly positive or negative, where it crosses an axis, and whether the result is an integral value, area, displacement or distance. A positive function lying below y=4 from x=0 to 2 cannot enclose more than 8 square units with the x-axis.
Common integration failure modes
| Error | Cause | Repair |
|---|---|---|
| Power increased but not divided | Differentiation rule copied forward | Differentiate the answer |
| +C omitted | Family of antiderivatives not understood | Ask what constants do under differentiation |
| Negative area reported | Integral confused with geometric area | Split and add magnitudes |
| Area between curves negative | Bottom minus top | Sketch or test a point |
| Wrong limits | Intersections not found first | Solve boundaries before integrating |
| Distance equals displacement | Direction reversal ignored | Find where v=0 |
| Condition unused | C not resolved | Substitute the known point |
| 1/x forced through power rule | Logarithmic exception missed | Use ln|x| |
Integration decision tree
- What is being accumulated?
- Is the integral indefinite or definite?
- Can the integrand be rewritten?
- Is there a reverse-chain structure?
- Are boundaries given or must they be found?
- Does sign carry meaning?
- How will the answer be checked?
Examination control
In a mixed paper, the integral itself may be short. The hard part may be solving an intersection, factorising, identifying a region or applying a condition. Do not rush to the integration symbol before the correct integrand and interval are established.
Build the right integrand. Build the right limits. Then calculate.
Independent practice
- Integrate 6x²−4x+3.
- Integrate 2x⁻³+4x1/2.
- Given dy/dx=8x−3 and y=7 when x=2, find y.
- Evaluate ∫02(3x²+1)dx.
- Find the area under y=4x−x² from x=0 to 4.
- Find the area enclosed by y=x and y=x².
- Given a=4t+2 and v=5 when t=0, find v.
- Given v=6−2t and s=3 when t=0, find s.
- For v=t−3 on 0≤t≤6, find displacement and total distance.
- Integrate 5e5x.
- Integrate 4cos(2x).
- Integrate 3/x.
Explained answers
1. 2x³−2x²+3x+C. 2. −x⁻²+(8/3)x3/2+C. 3. y=4x²−3x−3. 4. 10. 5. 32/3 square units. 6. 1/6 square unit. 7. v=2t²+2t+5. 8. s=6t−t²+3. 9. displacement 0, total distance 9. 10. e5x+C. 11. 2sin(2x)+C. 12. 3ln|x|+C.
Teaching sequence
Begin by linking antiderivatives directly to differentiation. Check early answers by differentiating back. Add rewritten powers, the logarithmic exception, reverse-chain structures and conditions for C. Then introduce definite integrals as accumulation, followed by area, signed-area control, area between curves and kinematics. Finally remove the chapter label and mix integration with algebra, functions, trigonometry and exponentials.
How this guide connects to Secondary 4 A-Math
- Differentiation, tangents, normals and stationary points
- Trigonometric functions, identities and equations
- Exponential and logarithmic functions and equations
Final thought
Find the quantity being accumulated. Find the correct boundaries. Integrate. Then return the answer to meaning.
Return to the Additional Mathematics Learning Hub.
