QCE Mathematical Methods - Unit 4 - Further integration

Applications of Integration | QCE Mathematical Methods

Revise QCE Mathematical Methods applications of integration, including area under curves, area between curves, trapezoidal rule and motion.

Updated 2026-05-18 - 4 min read

Integration becomes useful when a question asks for accumulation: total area, total change, displacement, distance, or a quantity built from a changing rate.

Area under a curve

If $f(x)>0$ on $[a,b]$, the area between $y=f(x)$ and the $x$-axis is:

$ \int_a^b f(x)\,dx $

If the curve is below the axis, the integral is negative, but area is positive. That is why you may need an absolute value or a split interval.

If a curve crosses the $x$-axis, find the intercepts first. For example, if the curve is positive on one part of the interval and negative on another, total area is found by splitting:

$ \text{Area}=\left|\int_a^c f(x)\,dx\right|+\left|\int_c^b f(x)\,dx\right|. $

This is different from net signed area:

$ \int_a^b f(x)\,dx. $

The wording of the question decides which one is wanted.

Area between curves

If the upper curve is $f(x)$ and the lower curve is $g(x)$, then:

$ \text{Area}=\int_a^b [f(x)-g(x)]\,dx $

The hard part is often deciding which curve is on top and finding the intersection points that form the boundaries.

If no bounds are given, solve the intersection points:

$ f(x)=g(x). $

Those $x$-values usually become the terminals. If the curves switch order between intersections, split the integral. A graph from technology is useful, but the final setup should still show the top-minus-bottom reasoning.

For example, between $y=x^2-1$ and $y=-x^2+7$, the intersections satisfy:

$ x^2-1=-x^2+7. $

So $x=\pm2$, and the upper curve is $-x^2+7$. The area is:

$ \int_{-2}^{2}\left[(-x^2+7)-(x^2-1)\right]\,dx. $

Trapezoidal rule

The trapezoidal rule estimates area using trapezia instead of rectangles:

$ \int_a^b f(x)\,dx \approx \frac{w}{2}\left[f(x_0)+2(f(x_1)+\cdots+f(x_{n-1}))+f(x_n)\right] $

where:

$ w=\frac{b-a}{n} $

The end values are counted once. The interior values are counted twice.

You can always derive the rule by adding trapezium areas:

$ \text{area of one trapezium}=\frac{w}{2}(h_1+h_2). $

When adjacent trapezia are added, interior heights appear twice. That is why the coefficient pattern becomes:

$ 1,\ 2,\ 2,\ldots,\ 2,\ 1. $

The trapezoidal rule is an approximation. It is most useful when an exact anti-derivative is unavailable, when a table of values is given, or when a technology-active question asks for a numerical estimate.

For table questions, lay the values out before substituting:

| $x$ | $x_0$ | $x_1$ | $\cdots$ | $x_n$ | | --- | --- | --- | --- | --- | | $f(x)$ | first height | interior height | $\cdots$ | last height | | coefficient | $1$ | $2$ | $\cdots$ | $1$ |

Then multiply height by coefficient, add, and finally multiply by $\frac{w}{2}$. Keeping the coefficient row visible prevents the common error of doubling the endpoints.

Motion and total change

If $v(t)$ is velocity, then:

$ \int_a^b v(t)\,dt $

gives displacement. If the velocity changes sign, displacement and distance are different. Distance travelled requires total area under the speed graph, or splitting where $v(t)=0$ and adding absolute values.

Integrating an instantaneous rate gives total change. If:

$ \frac{dV}{dt}=4t-32 $

models water flow in litres per second, then:

$ V(t)=\int(4t-32)\,dt=2t^2-32t+c. $

An initial condition, such as $V(0)=200$, gives $c=200$. If the question asks when the water stops flowing, set the rate equal to zero:

$ 4t-32=0. $

This distinction is important: integrate to recover the accumulated quantity, but differentiate or set the derivative to zero to answer rate questions.

Units are a strong check on integration applications. If velocity is in metres per second and time is in seconds, then:

$ \int v(t)\,dt $

has units of metres. If a marginal cost is dollars per item and you integrate with respect to number of items, the result is dollars. When your final units do not match the accumulated quantity, the setup probably needs revisiting.

Worked example

Quick check

Sources

• QCAA Mathematical Methods subject page
• QCAA Mathematical Methods 2025 syllabus
• OpenStax Calculus Volume 1: The Definite Integral
• OpenStax Calculus Volume 1: Integration Formulas and the Net Change Theorem