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

Question 2(c) Merit

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.

Find the value of the constant k, given that int₀^k (1/(√(4x + 1))) dx = 1.

The 2025 paper, page 7 ↗ Marking schedule, page 3 ↗

Work it out

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

1 Write the integrand as a power of a linear bracket.
1/(√(4x + 1)) = (4x + 1)^−1/2
2 Integrate it: add one to the power, then divide by the new power and by the 4 inside.
int (4x + 1)^−1/2 dx = (4x + 1)^1/2/(1/2 × 4) = (1/2)(4x + 1)^1/2
3 Put in the limits 0 and k, and set the result equal to 1. The lower limit does not give zero.
(1/2)√(4k + 1) − (1/2)√1 = 1
√(4k + 1) − 1 = 2
√(4k + 1) = 3
4 Square both sides and solve.
4k + 1 = 9, so k = 2.

Answer k = 2

What the marker wanted

Achieved The integration correct.
Merit The integration correct, and k = 2 found from it.

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

Common mistake

Getting the integral and the limits right, then slipping in the algebra. NZQA’s 2025 report names this part, with Two (d) and Three (d), as one where many students reached that point and went no further, which kept them out of Merit.

How did it go?

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