Integration is the mathematics of accumulation. It allows us to reconstruct quantities from rates of change, find total change from a varying rate, and measure signed area under a graph.
Students often meet integration after differentiation and assume it is simply “doing differentiation backwards”. That is an important part of the story, but it is not the whole story. Integration connects three ideas: antiderivatives, accumulation and area.
1. Reverse differentiation
If d/dx(x³)=3x², then an antiderivative of 3x² is x³. More generally, if F′(x)=f(x), then F is an antiderivative of f.
This gives the basic indefinite-integral notation
∫f(x) dx = F(x)+C.
The constant C matters because every vertical shift F(x)+C has the same derivative.
2. Why the constant of integration appears
Differentiate x³+5, x³−7 and x³+100. Each gives 3x². Differentiation removes constants, so reversing the process cannot recover one unique constant unless extra information is supplied.
Therefore ∫3x² dx=x³+C, not just x³.
3. The power rule for integration
For n≠−1,
∫x^n dx = x^(n+1)/(n+1)+C.
Integration raises the power by one and divides by the new power. Differentiate the result to verify it.
4. Worked algebraic example
Find ∫(6x²−4x+5) dx.
Integrate term by term:
∫6x² dx=2x³, ∫−4x dx=−2x², and ∫5 dx=5x.
Therefore
∫(6x²−4x+5)dx=2x³−2x²+5x+C.
Differentiate the answer: 6x²−4x+5. The reverse check confirms the integration.
5. Accumulation from a rate
If v(t) is velocity, then integrating velocity over time gives change in displacement. If a(t) is acceleration, integrating acceleration gives change in velocity.
This is a broader principle: integrating a rate accumulates the total change produced by that rate.
6. Definite integrals measure net accumulation
A definite integral has limits:
∫_a^b f(x) dx.
If F′=f, then
∫_a^b f(x)dx=F(b)−F(a).
The result is a number representing net accumulation over the interval.
7. Worked definite-integral example
Evaluate ∫_1^3 2x dx.
An antiderivative is x². Therefore
x²|_1^3=9−1=8.
Because y=2x is above the horizontal axis on this interval, the result also equals the geometric area under the graph.
8. Signed area, not always ordinary area
If f(x) is below the horizontal axis, the definite integral contributes negatively. Therefore a definite integral represents signed area or net accumulation, not automatically total geometric area.
To find total area between a curve and the axis, split at roots and make below-axis contributions positive.
9. Worked signed-area example
Evaluate ∫_−1^1 x dx.
An antiderivative is x²/2. Thus
[x²/2]_−1^1=1/2−1/2=0.
The geometric regions are not zero; the negative contribution on the left cancels the positive contribution on the right.
10. Integration reconstructs a family of curves
If dy/dx=2x, then y=x²+C. Different values of C produce vertically shifted parabolas with the same gradient function.
If the curve passes through (2,7), then 7=4+C, so C=3 and y=x²+3. A condition selects one member from the antiderivative family.
11. Graph interpretation
If an accumulation function A(x)=∫_a^x f(t)dt is introduced, then A′(x)=f(x). Where f is positive, A increases; where f is negative, A decreases.
This connects integration and differentiation at a deeper level: the original rate controls the slope of the accumulated-total function.
12. Units matter
If velocity is measured in metres per second and you integrate with respect to seconds, the result has units metres. If a graph shows a rate in litres per minute against time in minutes, the area under the graph has units litres.
Multiplying vertical units by horizontal units is a useful interpretation check.
13. Common mistakes
- Forgetting +C in an indefinite integral.
- Using +C in a final definite-integral value.
- Increasing the power but forgetting to divide by the new power.
- Calling every definite integral “area” without checking sign.
- Substituting limits into the integrand instead of the antiderivative.
14. Independent practice
- Find ∫4x³ dx.
- Find ∫(3x²+2x−6)dx.
- Evaluate ∫_0^2 3x² dx.
- If dy/dx=6x and y=5 when x=1, find y in terms of x.
- Explain why ∫_−2^2 x dx=0 does not mean there is no geometric area between y=x and the axis.
- State the units of the integral of a speed in km/h with respect to time in hours.
15. Answers
- x⁴+C.
- x³+x²−6x+C.
- [x³]_0^2=8.
- y=3x²+C; 5=3+C, so y=3x²+2.
- The below-axis and above-axis signed contributions cancel; total geometric area would add their magnitudes.
- Kilometres.
16. What mastery looks like
A student understands integration when they can explain it as reverse differentiation and accumulation, distinguish indefinite and definite integrals, interpret signed area, preserve units and verify an antiderivative by differentiating it.
Continue with the Additional Mathematics Calculus Guide or return to the Additional Mathematics Hub.