Question 3(c) Merit

Opens with Simpson’s Rule on five measured offsets from a garden path, then a root quotient, a product-of-sines area, an exponential-quotient acceleration integrated to a velocity, and a differential equation that has to be factorised before it will separate.

The graph shows y = 5 sin(3x) sin(x), shaded between the curve and the x-axis from the origin to x = π/3, where it meets the axis.

Find the shaded area under y = 5 sin(3x) sin(x).

The 2023 paper, page 11 ↗ Marking schedule, page 5 ↗

Work it out

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

1. 1 A sine times a sine of different angles. Which identity makes it a sum?

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

2. 2 Where does the shaded region end?

   At x = π/3, where sin 3x = sinπ = 0. Between 0 and π/3 both sines are positive, so the curve is above the axis.

3. 3 Integrate and put in the limits.

   (5/2)Bigl[(1/2) sin 2x − (1/4) sin 4xBigr]₀^π/3 = (5/2)(1/2 · √3/2 − 1/4 · (−√3/2)) = 5/2 · 3√3/8 = 15√3/16

What the marker wanted

• Achieved The integral correct.
• Merit The area, 1.6238 (15√3/16), from correct integration.

In our words, from NZQA’s 2023 marking schedule.

How did it go?

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