2020HSCShort answer

Q16 (b) (iii)

3 marks

Question

Let
$J_n=\int_0^1x^n(1-x)^n\,dx,\qquad n=0,1,2,\ldots.$

Using the result of part (ii), or otherwise, show that
$J_n=\frac{(n!)^2}{(2n+1)!}.$

Try the question before revealing the answer.

Solution

Criteria	Marks
Provides correct solution	3
Obtains an integral of $\sin^{2n+1}(2\theta)$ , or equivalent merit	2
Attempts the substitution $x=\sin^2\theta$ , or equivalent merit	1

Sample answer:

J_n=\int_0^1x^n(1-x)^n\,dx

Put $x=\sin^2\theta$ .

\frac{dx}{d\theta}=2\sin\theta\cos\theta

\therefore J_n=\int_0^{\pi/2}2\sin^{2n+1}\theta\cos^{2n+1}\theta\,d\theta

=\int_0^{\pi/2}\frac1{2^{2n}}(2\sin\theta\cos\theta)^{2n+1}\,d\theta

=\frac1{2^{2n}}\int_0^{\pi/2}\sin^{2n+1}(2\theta)\,d\theta

=\frac1{2^{2n}}I_n

=\frac{(n!)^2}{(2n+1)!}

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Syllabus alignment

CalculusWorking Mathematically

Further IntegrationApply substitution, partial fractions, integration by parts and recurrence relationships to theoretical and practical problems.

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