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.
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.
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.
The rule fails for x−1, which uses a logarithm instead.
A fraction shaped like f(x)f′(x)
Try the log form or a substitution.
Keep the multiplier from f′(x) consistent.
Definite integral with bounds
Find the antiderivative, then substitute upper bound minus lower bound.
Do not add C in the final definite integral answer.
Rate or velocity context
Integrate the rate first, then use the given condition.
The constant matters because the question usually asks for the original quantity.
Common trap: an indefinite integral needs +C, but a definite integral does not. The bounds remove the constant when the upper and lower values are subtracted.
Initial-condition checkpoint
When integration is used to recover an original function from a rate, the antiderivative still contains C. Use the given value only after integrating, then substitute it back before answering the question.
Question cue
First integration move
How to use the condition
Common trap
Given dxdy 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.
Worked check: if dxdy=6x+2 and y=5 when x=1, then y=3x2+2x+C. Substituting x=1, y=5 gives 5=3+2+C, so C=0 and y=3x2+2x.
Misconception check: the constant is not optional just because the question gives a value. The given value is the tool for finding the constant.
For a linear inner function, divide by its gradient. For example, ∫cos(3x+1),dx=31sin(3x+1)+C. The same adjustment applies to sin(ax+b), sec2(ax+b), and eax+b.
Bounded-region checkpoint
For an area bounded by a curve, the x-axis, and vertical lines, integrate the curve between the given bounds. If the curve lies below the x-axis, the definite integral is negative but geometric area is positive, so take its magnitude. If a curve and a line bound the region, integrate "upper function minus lower function" after finding their intersection points.
Reverse-power exception checkpoint
Before using the reverse power rule, check whether the term has power −1. That one term uses the logarithm rule, not 0x0.
| Term to integrate | Rule to use | First answer line | Common trap |
| ------------------------- | ----------------------------------- | ---------------------------- | ----------------------------------------------- | ------ | ----------------------------------------------------------- |
| x4 | Reverse power rule | 5x5+C | Forgetting to divide by the new power. |
| 3x−2 | Reverse power rule | −3x−1+C | Dividing by −2 instead of adding 1 first. |
| x1 | Log rule | ln∣x∣+C | Treating it as x−1 and trying to divide by zero. |
| 2x+15 | Log rule with multiplier adjustment | 25ln∣2x+1∣+C | Missing the derivative 2 of the denominator. |
Worked check: for ∫(4x3+x2)dx, integrate the terms separately. The first term gives x4, while the second gives 2ln∣x∣, so the answer is x4+2ln∣x∣+C.
Misconception check: x−1 is still a power when you rewrite the expression, but it is the exception to the reverse power rule because the new exponent would be zero.
2 Worked example - Definite integral
Evaluate ∫13(4x2−6x+5)dx.
Antiderivative: 34x3−3x2+5x.
Substitute bounds: [34x3−3x2+5x]13.
Compute: at x=3, value is 34×27−27+15=36−27+15=24
At x=1, value is 34−3+5=34+2=310
Difference: 24−310=372−310=362
Signed area checkpoint
Before interpreting a definite integral, check whether the question wants a signed quantity or total area.
Question wording
What the integral gives
Extra step
Common trap
Evaluate ∫abf(x)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.
Worked check: if v(t) is positive from 0 to 2 and negative from 2 to 5, then ∫05v(t)dt gives displacement. Total distance needs ∫02v(t)dt+∫25v(t)dt.
3 Worked example - Substitution
Evaluate ∫016x(x2+1)2dx.
Let u=x2+1; then dxdu=2x⟹xdx=21du.
When x=0, u=1; when x=1, u=2.
Integral becomes ∫126×21u2du=3∫12u2du
Evaluate: 3[31u3]12=[u3]12=8−1=7
Substitution multiplier checkpoint
After choosing u, compare dxdu with the factor already in the integrand. Any missing multiplier becomes a constant outside the new integral.
Inner expression
Derivative to look for
If the integrand has...
First rewrite
x2+1
2x
6x
6xdx=3du
3x+5
3
no extra factor
dx=31du
1−4x
−4
no extra factor
dx=−41du
x2+4
2x
x
xdx=21du
Worked check: for ∫10x(x2+3)4dx, let u=x2+3. Since du=2xdx, rewrite 10xdx as 5du, not 10du.
Misconception check: substitution does not make constants disappear. Any factor that is not part of du stays outside the new integral.
4 Worked example - Log form via substitution
Evaluate ∫x2+4xdx.
Let u=x2+4, so dxdu=2x and xdx=21du.
Substitute: ∫x2+4xdx=21∫u1du
Integrate: 21ln∣u∣+C=21ln(x2+4)+C
Answer: ∫x2+4xdx=21ln(x2+4)+C.
5 Practice Quiz
Get comfortable with antiderivatives, substitutions, and definite integrals so area interpretations and kinematics tie-ins stay accurate.
6 Try this
Integrate ∫(5x4−3x+2)dx and state the constant of integration explicitly.
