How is the definite integral defined as a limit of sums, and why does it give signed area under a curve?
Interpret the definite integral as a limit of Riemann sums and as the signed area between a curve and the x-axis
WACE Year 12 Mathematics Methods Unit 4 the definite integral: the limit of Riemann sums, signed area above and below the axis, properties of the definite integral, and total area, with worked SCSA-style examples.
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SCSA Unit 4 defines the definite integral as a limit of sums before connecting it to antidifferentiation. This dot point asks you to interpret the integral as accumulated signed area and to use its basic properties, supporting questions in both examination sections.
The limit of Riemann sums
Divide [a,b] into n thin strips of width Δx, and approximate the area under y=f(x) by the sum of rectangle areas ∑f(xi)Δx. As the strips become infinitely thin, this sum approaches the exact area.
This construction explains why the integral measures accumulated area: each rectangle contributes its height times its width, and the limit removes the approximation error. A left-endpoint sum, a right-endpoint sum and a midpoint sum all converge to the same value as the strips shrink, which is why the definite integral is well defined regardless of how the sample point in each strip is chosen. In practice we never evaluate the limit directly; the Fundamental Theorem of Calculus turns it into a simple subtraction of antiderivative values, but the limit-of-sums definition is what gives the integral its meaning as accumulated area.
Signed area
The height f(xi) is negative wherever the curve dips below the axis, so those strips contribute negative amounts. The definite integral therefore gives the signed area: positive above the axis, negative below.
Properties of the definite integral
These properties follow directly from the limit definition and are used to simplify calculations.
∫aaf(x)dx=0 (zero width).
∫baf(x)dx=−∫abf(x)dx (reversing the limits negates).
∫acf(x)dx+∫cbf(x)dx=∫abf(x)dx (additivity over adjacent intervals).
∫abkf(x)dx=k∫abf(x)dx (constants factor out).
∫ab(f(x)±g(x))dx=∫abf(x)dx±∫abg(x)dx (the integral of a sum is the sum of the integrals).
The additivity property is especially useful in exam questions that give you integrals over part of an interval and ask for the rest, since you can add or subtract known integrals rather than recomputing. The constant-multiple and sum properties let you break a complicated integrand into manageable standard pieces.
Exam-style practice questions
Practice questions written in the style of SCSA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.
WACE 20215 marksCalculator-free. (a) Evaluate ∫13(2x+1)dx. (b) Given ∫04f(x)dx=10 and ∫02f(x)dx=3, find ∫24f(x)dx.
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A definite integral and a use of the additivity property.
(a) ∫13(2x+1)dx=[x2+x]13=(9+3)−(1+1)=12−2=10.
(b) By additivity, ∫04f=∫02f+∫24f, so 10=3+∫24f, giving ∫24f(x)dx=7.
Markers reward the antiderivative with both limits in (a) and the additivity property in (b).
WACE 20235 marksCalculator-assumed. Consider f(x)=x2−4 on the interval [0,3]. (a) Evaluate ∫03f(x)dx and state what it represents. (b) Find the total area between the curve and the x-axis on [0,3].
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A signed-versus-total area question.
(a) ∫03(x2−4)dx=[3x3−4x]03=(9−12)−0=−3. This is the signed area: the region below the axis outweighs the region above.
(b) The curve crosses the axis at x=2. On [0,2], ∫02(x2−4)dx=38−8=−316 (area 316). On [2,3], ∫23(x2−4)dx=(9−12)−(38−8)=−3+316=37. Total area =316+37=323 square units.
Markers reward the signed value, splitting at x=2, and summing absolute areas.