Question 18 - A-Level Maths

AQA Paper 1 2024 June — Question 18 11 marks

Exam Board	AQA
Module	Paper 1 (Paper 1)
Year	2024
Session	June
Marks	11
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Integration by Substitution
Type	Show substitution transforms integral, then evaluate
Difficulty	Standard +0.3 This is a straightforward integration by substitution question with standard techniques. Part (a) requires a routine substitution u = 2x + 1, part (b) is direct integration of powers, and part (c) tests basic understanding of numerical integration bounds. The algebra is simple, and all steps follow standard A-level procedures with no novel insight required. Slightly easier than average due to the guided structure and explicit hints.
Spec	1.08h Integration by substitution

Use a suitable substitution to show that $$\int_0^1 (4x + 1)(2x + 1)^{\frac{1}{2}} dx$$ can be written as $$\frac{1}{2}\int_a^9 (2u^{\frac{1}{2}} - u^{\frac{1}{2}}) du$$ where $a$ is a constant to be found. [5 marks]
Hence, or otherwise, show that $$\int_0^1 (4x + 1)(2x + 1)^{\frac{1}{2}} dx = \frac{1322}{15}$$ [4 marks]
A graph has the equation $$y = (4x + 1)\sqrt{2x + 1}$$ A student uses four rectangles to approximate the area under the graph between the lines $x = 0$ and $x = 4$ The rectangles are all the same width. All the rectangles are drawn under the curve as shown in the diagram below. \includegraphics{figure_18c} The total area of the four rectangles is $A$ The student decides to improve their approximation by increasing the number of rectangles used. Explain why the value of the student's improved approximation will be greater than $A$, but less than $\frac{1322}{15}$ [2 marks]

Show mark scheme Show mark scheme source

Question 18:

Answer	Marks
18(a)	Selects the substitution

u = 2 x + 1 and differentiates or

uses it to replace 2 x + 1 in the

Answer	Marks	Guidance
integrand.	3.1a	B1

d u 1

 = 2 d x = d u

d x 2

4 x u + = 1 2 − 1

1  4 ( ) x + 4 1 ( x + 2 1 ) 2 d x

0 = 9 (  u − 2 ) ( ) u 1 1 2 1 d u

1 2

 3 1 

1 9

=  u  2 u − 2 2 d u

 

1 2



Differentiates their substitution

and uses the result to replace

Answer	Marks	Guidance
d x in the integral.	1.1a	M1

Makes a complete substitution

to write the integrand in terms of

u

leading to an integrand of the

1 A( 2u−k)u2

form

Or

FT their substitution

1 u =( 2x+1 ) 2 or u2 =2x+1

leading to an integrand of the

( )

Answer	Marks	Guidance
form A 2u2 −1 u2	3.1a	M1

Obtains correct lower limit for

Answer	Marks	Guidance
their substitution	1.1a	M1

Completes a reasoned

argument to show the required

Answer	Marks	Guidance
result with a = 1	2.1	R1
Subtotal	5
Q	Marking instructions	AO

Answer	Marks
18(b)	Integrates to obtain

5 5 3

1 4u2 4u2 1 2 u 2

or or − or

2 5 5 2 3

3 2u2

Answer	Marks	Guidance
3	1.1a	M1

0  3 12 

1 9

=  u 2 2 − u d u

2 1

9  52 3 

1 4 u 2 u 2

= −

2 5 3

1  5 3   5 3 

1492 292  412 212 

= − − −

   

2  5 3   5 3 

   

1322

=

15  5 3 

1 4 u 2 2 u 2

Obtains −

2 5 3

5 3

4u2 2u2

or −

Answer	Marks	Guidance
5 3	1.1b	A1

Substitutes limits explicitly into

their integrated expression of

5 3

the form A u 2 − B u 2

Where A and B are both positive

FT their non-zero a

Condone omission of powers on

Answer	Marks	Guidance
substitution of 1	1.1a	M1

Completes argument to show

the given result with no

Answer	Marks	Guidance
unrecovered slips.	2.1	R1
Subtotal	4
Q	Marking instructions	AO

Answer	Marks
18(c)	Explains that increasing the

number of rectangles will lead to

an increased value so the

(improved) approximation will be

Answer	Marks	Guidance
greater than A	2.4	E1

underestimate, increasing their

number will give an

increased total > A

No matter how many rectangles are

used their total will be less than the

1322

exact area so <

15 Gives a reason why the

approximation will be an under-

estimate so will be less than

1322

15

Answer	Marks	Guidance
AG	2.4	E1
Subtotal	2
Question 18 Total	11
Q	Marking instructions	AO

Question 18:
--- 18(a) ---
18(a) | Selects the substitution
u = 2 x + 1 and differentiates or
uses it to replace 2 x + 1 in the
integrand. | 3.1a | B1 | L u x = e t 2 + 1
d u 1
 = 2 d x = d u
d x 2
4 x u + = 1 2 − 1
1
 4 ( ) x + 4 1 ( x + 2 1 ) 2 d x
0
= 9 (  u − 2 ) ( ) u 1 1 2 1 d u
1 2
 3 1 
1 9
=  u  2 u − 2 2 d u
 
1 2

Differentiates their substitution
and uses the result to replace
d x in the integral. | 1.1a | M1
Makes a complete substitution
to write the integrand in terms of
u
leading to an integrand of the
1
A( 2u−k)u2
form
Or
FT their substitution
1
u =( 2x+1 ) 2 or u2 =2x+1
leading to an integrand of the
( )
form A 2u2 −1 u2 | 3.1a | M1
Obtains correct lower limit for
their substitution | 1.1a | M1
Completes a reasoned
argument to show the required
result with a = 1 | 2.1 | R1
Subtotal | 5
Q | Marking instructions | AO | Marks | Typical solution
--- 18(b) ---
18(b) | Integrates to obtain
5 5 3
1 4u2 4u2 1 2 u 2
or or − or
2 5 5 2 3
3
2u2
3 | 1.1a | M1 |  4 ( 4 x + ) ( 1 2 x + 1 ) 12 d x
0
 3 12 
1 9
=  u 2 2 − u d u
2 1
9
 52 3 
1 4 u 2 u 2
= −
2 5 3
1
 5 3   5 3 
1492 292  412 212 
= − − −
   
2  5 3   5 3 
   
1322
=
15
 5 3 
1 4 u 2 2 u 2
Obtains −
2 5 3
5 3
4u2 2u2
or −
5 3 | 1.1b | A1
Substitutes limits explicitly into
their integrated expression of
5 3
the form A u 2 − B u 2
Where A and B are both positive
FT their non-zero a
Condone omission of powers on
substitution of 1 | 1.1a | M1
Completes argument to show
the given result with no
unrecovered slips. | 2.1 | R1
Subtotal | 4
Q | Marking instructions | AO | Marks | Typical solution
--- 18(c) ---
18(c) | Explains that increasing the
number of rectangles will lead to
an increased value so the
(improved) approximation will be
greater than A | 2.4 | E1 | Since the area of the rectangles is an
underestimate, increasing their
number will give an
increased total > A
No matter how many rectangles are
used their total will be less than the
1322
exact area so <
15
Gives a reason why the
approximation will be an under-
estimate so will be less than
1322
15
AG | 2.4 | E1
Subtotal | 2
Question 18 Total | 11
Q | Marking instructions | AO | Marks | Typical solution

Show LaTeX source

\begin{enumerate}[label=(\alph*)]
\item Use a suitable substitution to show that
$$\int_0^1 (4x + 1)(2x + 1)^{\frac{1}{2}} dx$$
can be written as
$$\frac{1}{2}\int_a^9 (2u^{\frac{1}{2}} - u^{\frac{1}{2}}) du$$
where $a$ is a constant to be found.
[5 marks]

\item Hence, or otherwise, show that
$$\int_0^1 (4x + 1)(2x + 1)^{\frac{1}{2}} dx = \frac{1322}{15}$$
[4 marks]

\item A graph has the equation
$$y = (4x + 1)\sqrt{2x + 1}$$

A student uses four rectangles to approximate the area under the graph between the lines $x = 0$ and $x = 4$

The rectangles are all the same width.

All the rectangles are drawn under the curve as shown in the diagram below.

\includegraphics{figure_18c}

The total area of the four rectangles is $A$

The student decides to improve their approximation by increasing the number of rectangles used.

Explain why the value of the student's improved approximation will be

greater than $A$, but less than $\frac{1322}{15}$
[2 marks]
\end{enumerate}

\hfill \mbox{\textit{AQA Paper 1 2024 Q18 [11]}}

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