AQA Paper 1 2019 June — Question 14 10 marks

Exam BoardAQA
ModulePaper 1 (Paper 1)
Year2019
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicNumerical integration
TypeTrapezium rule then exact integration comparison
DifficultyStandard +0.3 This is a standard A-level integration question combining trapezium rule (routine application with n=4) and exact integration using substitution (u = x² + 1). Part (b) requires recognizing the substitution method but follows a predictable pattern for rational functions. The limit concept in part (c) is straightforward. Slightly easier than average due to the structured guidance and standard techniques.
Spec1.08d Evaluate definite integrals: between limits1.08e Area between curve and x-axis: using definite integrals1.09f Trapezium rule: numerical integration

The graph of \(y = \frac{2x^3}{x^2 + 1}\) is shown for \(0 \leq x \leq 4\)
Caroline is attempting to approximate the shaded area, A, under the curve using the trapezium rule by splitting the area into \(n\) trapezia.
  1. When \(n = 4\)
    1. State the number of ordinates that Caroline uses. [1 mark]
    2. Calculate the area that Caroline should obtain using this method. Give your answer correct to two decimal places. [3 marks]
  2. Show that the exact area of \(A\) is $$16 - \ln 17$$ Fully justify your answer. [5 marks]
  3. Explain what would happen to Caroline's answer to part (a)(ii) as \(n \to \infty\) [1 mark]

Question 14:

(a)(i) ---
14
AnswerMarks Guidance
(a)(i)States correct number of
ordinates1.2 B1
14
(a)(ii)
AnswerMarks
14(b)Obtains at least 4 correct y
values (condone 7.5… for y )
4
AnswerMarks Guidance
and correct h1.1b B1
Area= 1 ×1×( 0+7.529+2(1+3.2+5.4))
2
=13.36
Substitutes their y values into
trapezium rule with correct
number of strips.
Condone missing 0
May see working on graph
0.5+2.1+4.3+6.4647.
AnswerMarks Guidance
(Exact value 1136/85)1.1a M1
Obtains correct area
NMS correct answer award full
AnswerMarks Guidance
marks CAO1.1b A1
Selects substitution u = x2 +1or
du
u = x2 and obtains =2x
dx
or writes the integrand in the
Bx
form Ax+
AnswerMarks Guidance
x2 +13.1a M1
u= x2+1⇒ =2x
dx
∫ 172x3 1
× du
u 2x
1
17 x2
= du
u
1
17u−1
= du
u
1
∫ 17 1
= 1− du
u
1
=[ u−lnu ]17
1
=(17−ln17)−(1−ln1)
=16−ln17
17 1
Obtains 1− du OE
u
1
Or
4 2x
∫ 2x− dx
0 x2 +1
AnswerMarks Guidance
Ignore limits1.1b A1
Integrates their expression
AnswerMarks Guidance
obtaining an ln term correctly.1.1a M1
Obtains fully correct integral
( )
u−lnuor x2 −ln x2 +1
AnswerMarks Guidance
Condone missing limits.1.1b A1
Completes fully correct argument
Substituting correct limits for their
method to show the correct
required result with correct
AnswerMarks Guidance
notation with AG2.1 R1

AnswerMarks
14(c)Explains that as n increases the
approximation found will tend to
4 2x3
the value of dx
x2 +1
0
AnswerMarks Guidance
OE2.4 E1
Total10
QMarking Instructions AO
15
(a)(i)
15
AnswerMarks
(a)(ii)Uses model with t=0 to find correct
value of h
AnswerMarks Guidance
AWRT 5.9AO3.4 B1
Uses v=0 to set up quadratic
AnswerMarks Guidance
equation for tAO3.4 M1
2𝑡𝑡
= 4 - 3 – 2
t = 6
8 am
Obtains t = 6
AnswerMarks Guidance
PI correct answerAO1.1b A1
Interprets their lowest positive
solution correctly
AnswerMarks Guidance
NMS can score 3AO3.2a A1F
Question 14:
--- 14
(a)(i) ---
14
(a)(i) | States correct number of
ordinates | 1.2 | B1 | 5
14
(a)(ii)
14(b) | Obtains at least 4 correct y
values (condone 7.5… for y )
4
and correct h | 1.1b | B1
Area= 1 ×1×( 0+7.529+2(1+3.2+5.4))
2
=13.36
Substitutes their y values into
trapezium rule with correct
number of strips.
Condone missing 0
May see working on graph
0.5+2.1+4.3+6.4647.
(Exact value 1136/85) | 1.1a | M1
Obtains correct area
NMS correct answer award full
marks CAO | 1.1b | A1
Selects substitution u = x2 +1or
du
u = x2 and obtains =2x
dx
or writes the integrand in the
Bx
form Ax+
x2 +1 | 3.1a | M1 | du
u= x2+1⇒ =2x
dx
∫ 172x3 1
× du
u 2x
1
17 x2
∫
= du
u
1
17u−1
∫
= du
u
1
∫ 17 1
= 1− du
u
1
=[ u−lnu ]17
1
=(17−ln17)−(1−ln1)
=16−ln17
17 1
∫
Obtains 1− du OE
u
1
Or
4 2x
∫ 2x− dx
0 x2 +1
Ignore limits | 1.1b | A1
Integrates their expression
obtaining an ln term correctly. | 1.1a | M1
Obtains fully correct integral
( )
u−lnuor x2 −ln x2 +1
Condone missing limits. | 1.1b | A1
Completes fully correct argument
Substituting correct limits for their
method to show the correct
required result with correct
notation with AG | 2.1 | R1
--- 14(c) ---
14(c) | Explains that as n increases the
approximation found will tend to
4 2x3
∫
the value of dx
x2 +1
0
OE | 2.4 | E1 | Area →16−ln17
Total | 10
Q | Marking Instructions | AO | Mark | Typical Solution
15
(a)(i)
15
(a)(ii) | Uses model with t=0 to find correct
value of h
AWRT 5.9 | AO3.4 | B1 | 5.88metres
Uses v=0 to set up quadratic
equation for t | AO3.4 | M1 | 0 ( )2
2𝑡𝑡
= 4 - 3 – 2
t = 6
8 am
Obtains t = 6
PI correct answer | AO1.1b | A1
Interprets their lowest positive
solution correctly
NMS can score 3 | AO3.2a | A1F
The graph of $y = \frac{2x^3}{x^2 + 1}$ is shown for $0 \leq x \leq 4$

\begin{tikzpicture}[>=latex]

    % Define the mathematical function
    \def\myfunc{2*pow(\x,3) / (pow(\x,2) + 1)}

    % 1. Fill the shaded region 'A'
    \fill[gray!35] (0,0) -- plot[domain=0:4, samples=100] (\x, {\myfunc}) -- (4,0) -- cycle;

    % 2. Draw the grid
    % Vertical grid lines (every 0.2 units = 1/5)
    \foreach \i in {0,1,...,25} {
        \draw[gray!50, very thin] (\i/5, 0) -- (\i/5, 9);
    }
    % Horizontal grid lines (every 0.2 units = 1/5)
    \foreach \j in {0,1,...,45} {
        \draw[gray!50, very thin] (0, \j/5) -- (5, \j/5);
    }

    % 3. Draw the function curve
    \draw[line width=1.2pt] plot[domain=0:4, samples=100] (\x, {\myfunc});

    % 4. Draw the coordinate axes
    \draw[line width=0.8pt, ->] (0,0) -- (5.5,0) node[below] {$x$};
    \draw[line width=0.8pt, ->] (0,0) -- (0,9.6) node[left] {$y$};

    % 5. Ticks and Labels
    \node[below left] at (0,0) {0};

    \foreach \x in {2,4} {
        \draw[line width=0.8pt] (\x, 0) -- (\x, -0.1) node[below] {\x};
    }

    \foreach \y in {2,4,6,8} {
        \draw[line width=0.8pt] (0, \y) -- (-0.1, \y) node[left] {\y};
    }

    % 6. Label 'A'
    \node at (2.5, 1.2) {$A$};

\end{tikzpicture}

Caroline is attempting to approximate the shaded area, A, under the curve using the trapezium rule by splitting the area into $n$ trapezia.

\begin{enumerate}[label=(\alph*)]
\item When $n = 4$
\begin{enumerate}[label=(\roman*)]
\item State the number of ordinates that Caroline uses. [1 mark]
\item Calculate the area that Caroline should obtain using this method.

Give your answer correct to two decimal places. [3 marks]
\end{enumerate}

\item Show that the exact area of $A$ is

$$16 - \ln 17$$

Fully justify your answer. [5 marks]

\item Explain what would happen to Caroline's answer to part (a)(ii) as $n \to \infty$ [1 mark]
\end{enumerate}

\hfill \mbox{\textit{AQA Paper 1 2019 Q14 [10]}}