AQA Paper 3 2023 June — Question 8 7 marks

Exam BoardAQA
ModulePaper 3 (Paper 3)
Year2023
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicIntegration by Substitution
TypeShow definite integral equals specific value (algebraic/exponential substitution)
DifficultyStandard +0.3 This is a straightforward substitution question with clear guidance. Students must find du = 5x^4 dx, rewrite x^9 as x^4 · x^5, change limits (u=2 to u=3), and integrate u^-3. While it requires careful algebraic manipulation across multiple steps, the method is explicitly given and follows a standard A-level integration technique with no conceptual surprises.
Spec1.08h Integration by substitution

Use the substitution \(u = x^5 + 2\) to show that $$\int_0^1 \frac{x^9}{(x^5 + 2)^3} \, dx = \frac{1}{180}$$ [7 marks]

Question 8:
AnswerMarks
8Obtains 5x4
4
1 −
PI by (u−2) 5
AnswerMarks Guidance
51.1b B1
du
=5x4
dx
9
x 1
∫ du
3 4
u 5x
1 3u−2
∫ du
5 2 u3
= 1 ∫ 3 u −2 −2u −3du
5 2
1 3
−u−1+u−2
 
5 2
11 1 1 1
= − − −
   
59 3 4 2
1
=
180
Substitutes for denominator and
dx operator
PI by fully correct substitution
Condone any limits or missing
integral sign or du
AnswerMarks Guidance
Condone dx in place du1.1a M1
Substitutes x 5 =u−2
x=(u−2 ) 1
or 5 in at least one
AnswerMarks Guidance
place1.1a M1
1 u−2
Obtains ∫ du
5 u 3
1
Condone missing or incorrect
5
or any limits
AnswerMarks Guidance
Must have du1.1b A1
Integrates u−2 or u−3correctly1.1a M1
1
−u−1+u−2
Obtains
 
5
AnswerMarks Guidance
Condone any limits1.1b A1
Completes reasoned argument
by substituting correct limits
consistent with their variable to
show the given result
AG
R1 could be scored if du is
AnswerMarks Guidance
missing throughout2.1 R1
Question 8 Total7
QMarking instructions AO
Question 8:
8 | Obtains 5x4
4
1 −
PI by (u−2) 5
5 | 1.1b | B1 | u = x 5 +2
du
=5x4
dx
9
x 1
∫ du
3 4
u 5x
1 3u−2
∫ du
5 2 u3
= 1 ∫ 3 u −2 −2u −3du
5 2
1 3
−u−1+u−2
 
5 2
11 1 1 1
= − − −
   
59 3 4 2
1
=
180
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 | 1.1a | M1
Substitutes x 5 =u−2
x=(u−2 ) 1
or 5 in at least one
place | 1.1a | M1
1 u−2
Obtains ∫ du
5 u 3
1
Condone missing or incorrect
5
or any limits
Must have du | 1.1b | A1
Integrates u−2 or u−3correctly | 1.1a | M1
1
−u−1+u−2
Obtains
 
5
Condone any limits | 1.1b | A1
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 | 2.1 | R1
Question 8 Total | 7
Q | Marking instructions | AO | Marks | Typical solution
Use the substitution $u = x^5 + 2$ to show that
$$\int_0^1 \frac{x^9}{(x^5 + 2)^3} \, dx = \frac{1}{180}$$

[7 marks]

\hfill \mbox{\textit{AQA Paper 3 2023 Q8 [7]}}