AS 91579 · 2025 paper · trapezium rule, areas, centroid

Question 3(a) Achieved

Opens with a numerical estimate from measured depths, then a log integral left in terms of a limit, a separable differential equation, an area that must be turned into a cubic, and a balance point found from a ratio of two integrals.

The diagram shows the cross-section of a hole dug in the ground, 20 m across the top, with straight sides. The depth of the hole is measured at both edges and every 5 metres in between.

Using the trapezium rule, find an estimate for the area of the cross-section of the hole.

strip width 5 m

depths, left to right: 2.12, 2.32, 2.65, 2.54, 1.88 m

The 2025 paper, page 10 ↗ Marking schedule, page 5 ↗

See it move: Rectangles, and the integral they become

Work it out

Try it on paper first. Stuck? Take one hint at a time.

1 What is the strip width h, and how many strips are there?
h = 5 m. Five depths, at both edges and every 5 m between, make four strips across 20 m.
2 Trapezium rule: the two end depths once, every depth in between twice, all times h/2.
A ≈ (h/2)[y₀ + y₄ + 2(y₁ + y₂ + y₃)]
= (5/2)[2.12 + 1.88 + 2(2.32 + 2.65 + 2.54)]
3 Work it out.
= (5/2)[4.00 + 2 × 7.51] = 5/2 × 19.02 = 47.55

Answer About 47.55 m² (47.6 m²).

What the marker wanted

Achieved The estimate, 47.55 m², from the trapezium rule.

In our words, from NZQA’s 2025 marking schedule.

Common mistake

Using the wrong h. NZQA’s 2025 report lists not recognising the value of h among the Not Achieved answers. It is the gap between measurements, 5 m, not the number of strips or the full 20 m.

How did it go?

Got it moves it further out. Review brings it back tomorrow.