2024 HSC Mathematics Extension 1
Question 13(d):
Antidifferentiation
Using the substitution , and considering , find .
How to recognise this question
Question type: antidifferentiation
- The command and notation point to antidifferentiation.
- The requested response is a worked solution.
How to handle it: choose a substitution or identity that simplifies the integrand, then reverse-differentiate to check.
Watch out: Do not select a formula until its domain, interval, sign and units match the question.
Step-by-step answer
- Let . Then so .
- Also .
- Rewrite the integrand by factoring in numerator and denominator so that the numerator becomes a multiple of and the denominator becomes (NESA algebra: denominator after rearrangement).
- The integral reduces to .
- Back-substitute: .
- NESA 13(d): 3 marks correct; 2 marks for transforming the integrand algebraically; 1 mark for obtaining .
Common mistake: choosing a familiar formula before checking that its conditions, signs, interval and units match this question.
The paper records 3 total marks for this group. Where the official marking guide provides useful partial-credit criteria, those criteria are stated in the steps above.
Try your hand at it again by answering these questions
Question 1
Using , find .
Question 2
If , compute and show that the exam integrand becomes . Hence state the antiderivative in .
Question 3
Using , find (hint: ).