The definite integral and area: WACE Year 12 Mathematics Methods Unit 3

Syllabus placement. In the SCSA Mathematics Methods ATAR Year 12 syllabus, integrals are Unit 3 (Topic 3.2), not Unit 4. This page sits in our Unit 4 folder for historical reasons; the content is Unit 3.

What this dot point is asking

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 $\Delta x$, and approximate the area under $y=f(x)$ by the sum of rectangle areas $\sum f(x_i)\Delta x$. 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 $f(x_i)$ 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.

• $\displaystyle\int_a^a f(x)\,dx = 0$ (zero width).
• $\displaystyle\int_b^a f(x)\,dx = -\int_a^b f(x)\,dx$ (reversing the limits negates).
• $\displaystyle\int_a^c f(x)\,dx + \int_c^b f(x)\,dx = \int_a^b f(x)\,dx$ (additivity over adjacent intervals).
• $\displaystyle\int_a^b k\,f(x)\,dx = k\int_a^b f(x)\,dx$ (constants factor out).
• $\displaystyle\int_a^b \big(f(x)\pm g(x)\big)\,dx = \int_a^b f(x)\,dx \pm \int_a^b g(x)\,dx$ (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.