AS 91579 · 2025 paper · trapezium rule, areas, centroid
Question 3(d) Merit
Practises Areas under & between curves
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 graph shows part of the curve y = (x2 + 6)/x4, where x > 0, shaded between the curve and the x-axis from x = p to x = 2p. The area of the shaded region is 9/4 units².
Prove that 9p3 − 2p2 − 7 = 0.
Work it out
Try it on paper first. Stuck? Take one hint at a time.
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1 Split the fraction into two powers of x.
(x2 + 6)/x4 = x−2 + 6x−4 -
2 Integrate.
int (x−2 + 6x−4)dx = −x−1 − 2x−3 = −1/x − 2/x3 -
3 Put in the limits p and 2p, and set the area equal to 9/4.
Bigl(−1/2p − (2/8p3)Bigr) − Bigl(−1/p − (2/p3)Bigr) = 9/4
1/2p + 7/4p3 = 9/4 -
4 Clear the fractions: multiply every term by 4p3.
2p2 + 7 = 9p3, so 9p3 − 2p2 − 7 = 0, as required.
What the marker wanted
- Achieved The integration correct.
- Merit The proof finished, with the integration shown.
In our words, from NZQA’s 2025 marking schedule.
✕Common mistake
Integrating and substituting correctly, then losing a term or a sign while clearing the fractions. NZQA’s 2025 report names this part, with Two (c) and Two (d), as one where many students got that far and did not finish.How did it go?
Got it moves it further out. Review brings it back tomorrow.