← Integration methods: past papers

AS 91579 · 2025 paper · substitution-type, constants, kinematics

Question 2(b) Achieved

Practises Differential equations

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 rate of change of a particular function, p, is given by dp/dt = 5 cos(4t). Find the function, given that p = 8 when t = π/24.

The 2025 paper, page 6 ↗ Marking schedule, page 3 ↗

Work it out

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

  1. 1 Integrate 5 cos(4t). What does the 4 inside the cosine do?

    p = (5/4) sin(4t) + c. Dividing by 4 undoes the 4 the chain rule would bring out.

  2. 2 Use p = 8 when t = π/24. What is 4t then?

    4 × π/24 = π/6, and sinπ/6 = 1/2.
    8 = 5/4 × 1/2 + c = 5/8 + c, so c = 59/8 = 7.375.
Answer p = (5/4) sin(4t) + 59/8, that is p = 1.25 sin 4t + 7.375.

What the marker wanted

  • Achieved The whole function, with c found and the integration shown.

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

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Common mistake

Finding sinπ/6 with the calculator in degrees. NZQA’s 2025 report says many students made calculator errors putting radian values into trig functions. In calculus an angle is in radians unless the question says otherwise.

How did it go?

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