Question 1(d) Merit

Five independent parts, no shared scenario: a standard integral, a differential equation with a boundary condition, a pair of definite integrals that pin down two constants, a trig area, and a differential equation you have to rearrange before you can separate it.

The graph shows part of y = 4 sin(5x) cos(3x), shaded between the curve and the x-axis from the origin to x = π/6, where the curve comes back down to the axis.

Find the shaded area under the curve between x = 0 and x = π/6.

The 2025 paper, page 4 ↗ Marking schedule, page 1 ↗

Work it out

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

1. 1 A sine times a cosine has no standard integral. Which identity turns the product into a sum?

   2 sin A cos B = sin(A + B) + sin(A − B). With A = 5x and B = 3x:
   4 sin 5x cos 3x = 2(sin 8x + sin 2x) = 2 sin 8x + 2 sin 2x

2. 2 Is the curve above the axis the whole way from 0 to π/6?

   Yes. Across that interval 5x runs from 0 to 5π/6, where sine is not negative, and 3x from 0 to π/2, where cosine is not negative. So the integral is the area.

3. 3 Integrate each sine, dividing by the number in front of x.

   int (2 sin 8x + 2 sin 2x) dx = −(2 cos 8x)/8 − (2 cos 2x)/2 = −(1/4) cos 8x − cos 2x

4. 4 Put in the limits. What are cos4π/3 and cosπ/3?

   At π/6: −(1/4) cos4π/3 − cosπ/3 = −(1/4)(−1/2) − 1/2 = −3/8
   At 0: −1/4 − 1 = −5/4
   Area = −3/8 − (−5/4) = 7/8

What the marker wanted

• Achieved The integral done correctly.
• Merit The area, 0.875 units², with correct integration shown.

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

How did it go?

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