Level 3 Integration Walkthrough

2017 NCEA Level 3 Integration Question 2(d)

2017 Paper

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Question

Part of the graph of \(y=\sin(3x)\cos(2x)\) is shown below.

Area under y equals sine three x cosine two x The curve passes through the origin and is positive until x equals pi over four. That first positive region above the x-axis is shaded. The curve is then slightly negative until its next zero at pi over three. x y π/4 π/3 y = sin(3x) cos(2x)

Find the area enclosed between the curve \(y=\sin(3x)\cos(2x)\) and the lines \(y=0\), \(x=0\), and \(x=\frac{\pi}{4}\).

You must use calculus and show the results of any integration needed to solve the problem.

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First walkthrough idea

Focus to try first

Convert the product of sine and cosine into a sum, then integrate over the first positive region.

Step 1

Use the product-to-sum identity

Apply \(\sin A\cos B=\frac12[\sin(A+B)+\sin(A-B)]\).

Show the first step’s working
\[ \sin(3x)\cos(2x) =\frac12\bigl(\sin(5x)+\sin x\bigr). \]

Walkthrough overview

What this question practises

This 2017 walkthrough is part of AS91579 — Apply integration methods in solving problems.

Method: Product-to-sum before integrating a trigonometric area.

This is Question 2(d) from the 2017 NCEA Level 3 Integration paper for AS91579 — Apply integration methods in solving problems. Use the guided hints to practise the method before revealing the full working.

Calc.nz is an independent learning resource. Compare questions, diagrams, and assessment information with the official NZQA resources for AS91579.

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Learning summary

Review the method, not only the answer

This walkthrough helps you practise product-to-sum before integrating a trigonometric area. Use the hints to plan the method, then repeat the question without hints and check each step.

Common mistake to avoid

Check the antiderivative by differentiating it, and handle constants and bounds explicitly.

Continue practising

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