← Integration methods: past papers

AS 91579 · 2023 paper · an exponential integral, a rate constant, a chocolate factory

Question 2(c) Merit

Practises Standard integrals

An exponential integral, a negative-power differential equation, an upper limit that appears on both sides, a trig quotient whose top is a multiple of the derivative of its bottom, and a proportional-to-the-square model built from scratch.

Find the value of k, given that int2k ((6x − 3)/(2x − 3)) dx = 3k.

The 2023 paper, page 7 ↗ Marking schedule, page 3 ↗

Work it out

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

  1. 1 The top has the same degree as the bottom. Divide first.

    6x − 3 = 3(2x − 3) + 6, so (6x − 3)/(2x − 3) = 3 + 6/(2x − 3).

  2. 2 Integrate both pieces.

    int (3 + 6/(2x − 3))dx = 3x + 3 ln |2x − 3|

  3. 3 Put in the limits 2 and k. What is 2x − 3 at x = 2?

    It is 1, and ln 1 = 0.
    (3k + 3 ln(2k − 3)) − (6 + 0) = 3k

  4. 4 Solve for k.

    3 ln(2k − 3) = 6, so ln(2k − 3) = 2 and 2k − 3 = e2.
    k = (e2 + 3)/2 ≈ 5.1945
Answer k = (e2 + 3)/2 ≈ 5.19

What the marker wanted

  • Achieved The integral correct.
  • Merit The value of k, 5.1945, from correct integration.

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

✕

Common mistake

Writing ln|2x − 3| straight away because there is a bracket on the bottom. NZQA’s 2023 report lists believing every fraction integrates to a log of its bottom among the Not Achieved answers.

How did it go?

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