| Exam Board | AQA |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2019 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Numerical integration |
| Type | Trapezium rule then exact integration comparison |
| Difficulty | Standard +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. |
| Spec | 1.08d Evaluate definite integrals: between limits1.08e Area between curve and x-axis: using definite integrals1.09f Trapezium rule: numerical integration |
| Answer | Marks | Guidance |
|---|---|---|
| (a)(i) | States correct number of | |
| ordinates | 1.2 | B1 |
| Answer | Marks |
|---|---|
| 14(b) | Obtains at least 4 correct y |
| Answer | Marks | Guidance |
|---|---|---|
| and correct h | 1.1b | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| (Exact value 1136/85) | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| marks CAO | 1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| x2 +1 | 3.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Ignore limits | 1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| obtaining an ln term correctly. | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Condone missing limits. | 1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| notation with AG | 2.1 | R1 |
| Answer | Marks |
|---|---|
| 14(c) | Explains that as n increases the |
| Answer | Marks | Guidance |
|---|---|---|
| OE | 2.4 | E1 |
| Total | 10 | |
| Q | Marking Instructions | AO |
| Answer | Marks |
|---|---|
| (a)(ii) | Uses model with t=0 to find correct |
| Answer | Marks | Guidance |
|---|---|---|
| AWRT 5.9 | AO3.4 | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| equation for t | AO3.4 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| PI correct answer | AO1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| NMS can score 3 | AO3.2a | 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]}}