June 2023 Paper 3 Q8

AQACurrent spec7 marksIntegration

8 Use the substitution \(u = x^5 + 2\) to show that

\[\int_0^1 \frac{x^9}{(x^5 + 2)^3}\,\mathrm{d}x = \frac{1}{180}\]

[7 marks]

Mark scheme

Scheme	Marks	AO
Obtains \(5x^4\) PI by \(\dfrac{1}{5}(u - 2)^{-\frac{4}{5}}\)	B1	1.1b
Substitutes for denominator and dx operator PI by fully correct substitution Condone any limits or missing integral sign or \(du\) Condone \(dx\) in place \(du\)	M1	1.1a
Substitutes \(x^5 = u - 2\) or \(x = (u - 2)^{\frac{1}{5}}\) in at least one place	M1	1.1a
Obtains \(\dfrac{1}{5}\displaystyle\int \frac{u - 2}{u^3}\,du\) Condone missing or incorrect \(\dfrac{1}{5}\) or any limits Must have \(du\)	A1	1.1b
Integrates \(u^{-2}\) or \(u^{-3}\) correctly	M1	1.1a
Obtains \(\dfrac{1}{5}\left[-u^{-1} + u^{-2}\right]\) Condone any limits	A1	1.1b
Completes reasoned argument by substituting correct limits consistent with their variable to show the given result AG R1 could be scored if \(du\) is missing throughout	R1	2.1
	(7 marks)

Typical solution

\[u = x^5 + 2\]\[\frac{du}{dx} = 5x^4\]\[\int \frac{x^9}{u^3}\,\frac{1}{5x^4}\,du\]\[\frac{1}{5}\int_2^3 \frac{u - 2}{u^3}\,du\]\[= \frac{1}{5}\int_2^3 u^{-2} - 2u^{-3}\,du\]\[\frac{1}{5}\left[-u^{-1} + u^{-2}\right]_2^3\]\[= \frac{1}{5}\left(\left(\frac{1}{9} - \frac{1}{3}\right) - \left(\frac{1}{4} - \frac{1}{2}\right)\right)\]\[= \frac{1}{180}\]