A definite integral gives a number. It measures net accumulation over an interval and, when interpreted geometrically, gives signed area between a graph and the horizontal axis.
The limits are part of the question. They tell you where accumulation begins and ends, and they determine whether regions above and below the axis reinforce or cancel one another.
1. Definite-integral notation
The expression
∫_a^b f(x)dx
means accumulate f(x) with respect to x from x=a to x=b.
If F′(x)=f(x), then the Fundamental Theorem of Calculus gives
∫_a^b f(x)dx=F(b)−F(a).
2. Worked basic example
Evaluate ∫_0^2 (3x²+1)dx.
An antiderivative is x³+x.
[x³+x]_0^2=(8+2)−0=10.
No +C remains in the final value because any constant would cancel when F(b)−F(a) is formed.
3. Why order of limits matters
Reversing the limits changes the sign:
∫_b^a f(x)dx=−∫_a^b f(x)dx.
This reflects the orientation of accumulation. The same geometric region is being traversed in the opposite direction.
4. Signed area above the axis
If f(x)≥0 throughout [a,b], then the definite integral equals the ordinary geometric area under the curve and above the horizontal axis.
Example: ∫_0^3 2x dx=[x²]_0^3=9. The graph is above the axis, so the area is 9 square units.
5. Signed area below the axis
If f(x)≤0 throughout [a,b], the definite integral is negative.
For f(x)=−2 on 1≤x≤4,
∫_1^4 −2dx=−6.
The geometric rectangle has area 6, but the signed integral is −6 because the graph lies below the axis.
6. Total area requires splitting
If a curve crosses the horizontal axis, first find the roots inside the interval. Split the integral at those crossing points and treat below-axis contributions by magnitude when total geometric area is required.
7. Worked crossing-axis example
Find the total area between y=x−1 and the horizontal axis from x=0 to x=3.
The root is x=1. On [0,1], the graph is below the axis; on [1,3], it is above.
Signed integral from 0 to 1:
∫_0^1(x−1)dx=[x²/2−x]_0^1=−1/2.
Area contribution is 1/2.
From 1 to 3:
[x²/2−x]_1^3=(9/2−3)−(1/2−1)=2.
Total area=1/2+2=5/2.
8. Net accumulation can cancel
For an odd function over a symmetric interval, positive and negative contributions may cancel. For example,
∫_−2^2 x³dx=0.
This does not mean the graph encloses no geometric area with the axis. It means the signed accumulations balance.
9. Symmetry as a checking tool
If f is odd, ∫_−a^a f(x)dx=0. If f is even, ∫_−a^a f(x)dx=2∫_0^a f(x)dx.
These symmetry facts can reduce work and provide a useful independent check.
10. Integral of a rate
If v(t) is velocity, ∫_a^b v(t)dt gives displacement change, not necessarily total distance travelled. Negative velocity contributes negatively.
To find total distance, split where velocity changes sign and add the magnitudes of the displacement contributions.
11. Worked motion interpretation
Suppose v(t)=t−2 for 0≤t≤5. The velocity changes sign at t=2.
Net displacement:
∫_0^5(t−2)dt=[t²/2−2t]_0^5=25/2−10=5/2.
Total distance requires splitting at 2 and adding absolute contributions.
12. Integral properties
- ∫_a^a f(x)dx=0.
- ∫_a^b [f(x)+g(x)]dx=∫_a^b f(x)dx+∫_a^b g(x)dx.
- ∫_a^b kf(x)dx=k∫_a^b f(x)dx for constant k.
- ∫_a^c f(x)dx=∫_a^b f(x)dx+∫_b^c f(x)dx.
These properties reflect how accumulation behaves when intervals or quantities are combined.
13. Units of a definite integral
Integral units come from vertical units multiplied by horizontal units. A rate of litres/minute integrated over minutes gives litres. Velocity in m/s integrated over seconds gives metres.
In pure coordinate geometry, an area interpretation may use square units.
14. Exact limits and values
Keep exact values such as π, √2 or logarithms where appropriate. Rounding a root before using it as a splitting point can introduce unnecessary area error.
15. Graph before calculus
A quick sketch can reveal whether the curve crosses the axis, where a split may be needed, whether the final sign is plausible and whether symmetry can help.
This is especially important in area problems, where correct integration with incorrect region interpretation still produces the wrong answer.
16. Common mistakes
- Adding +C to a final definite-integral value.
- Substituting limits into the original function rather than the antiderivative.
- Calling a negative definite integral a negative geometric area.
- Failing to split when the graph crosses the axis.
- Confusing displacement with total distance.
- Reversing limits without changing the sign.
17. Independent practice
- Evaluate ∫_1^3 4x dx.
- Evaluate ∫_−1^1 x²dx.
- Evaluate ∫_−2^2 x³dx using symmetry.
- Find the total area between y=x−2 and the axis from x=0 to x=5.
- If velocity is v=t−1 on 0≤t≤3, find net displacement.
- Explain why total distance may differ from displacement.
18. Answers
- [2x²]_1^3=18−2=16.
- Even symmetry: 2∫_0^1 x²dx=2/3.
- 0.
- Root x=2. Area on [0,2] is 2; area on [2,5] is 9/2. Total 13/2.
- ∫_0^3(t−1)dt=[t²/2−t]_0^3=3/2.
- Displacement is signed net change; total distance adds magnitudes across changes of direction.
19. What mastery looks like
A student has definite-integration control when they can evaluate limits correctly, distinguish signed accumulation from total area, split intervals at sign changes, use symmetry, preserve units and interpret the result in context.
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