Question 2(e) Merit to Excellence

A power-of-a-linear integral, a rate of change integrated back to its function, a constant found from a definite integral, a kinematics chain (acceleration → velocity → displacement), and an odd-power trig area.

The graph shows y = sin³ x cos³ x, shaded between the curve and the x-axis from the origin to (π/2, 0).

Find the shaded area under the curve between x = 0 and x = π/2.

The 2025 paper, page 9 ↗ Marking schedule, page 4 ↗

Work it out

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

1. 1 Both powers are odd. Keep one cos x aside and turn the cosines that are left into sines.

   sin³ x cos³ x = sin³ x cos² x cos x = sin³ x(1 − sin² x) cos x

2. 2 Which substitution does the spare cos x invite? Change the limits too.

   u = sin x, so du = cos x dx.
   x = 0 gives u = 0; x = π/2 gives u = 1.

3. 3 Integrate in u.

   int₀¹ (u³ − u⁵)du = Bigl[u⁴/4 − (u⁶/6)Bigr]₀¹ = 1/4 − 1/6 = 1/12

4. 4 Is that the area, or could part of the curve be below the axis?

   Between 0 and π/2 both sin x and cos x are positive, so the curve stays above the axis and the integral is the area.

What the marker wanted

• Merit A correct integration, with the method set out clearly. Keeping one sin x aside and turning the rest into cosines works just as well.
• Excellence The area, 1/12 = 0.0833 units², fully justified. That is E8; this part has no E7.

The schedule gives this part no Achieved cell: its first credit is Merit, for the integration.

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

How did it go?

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