| Exam Board | Edexcel |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2022 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Numerical integration |
| Type | Complete table then apply trapezium rule |
| Difficulty | Moderate -0.3 This is a straightforward numerical integration question requiring routine application of the trapezium rule. Part (a) involves simple calculator substitution, part (b) is standard trapezium rule application with given ordinates, and part (c) requires recognizing that the integral of (3 + log₁₀(sin x)) relates to the original through linearity of integration. All techniques are standard P2 content with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.06d Natural logarithm: ln(x) function and properties1.09f Trapezium rule: numerical integration |
| \(x\) | 0.5 | 1 | 1.5 | 2 | 2.5 | 3 |
| \(y\) | 1.319 | 1.001 | 1.223 | 1.850 |
| Answer | Marks | Guidance |
|---|---|---|
| 2(a) | x | 0.5 |
| Answer | Marks | Guidance |
|---|---|---|
| y | 1.319 | 1.075 |
| Answer | Marks | Guidance |
|---|---|---|
| (b) | h=0.5 seen or implied | B1 |
| Answer | Marks |
|---|---|
| 2 | M1 |
| = a w r t 2 . 9 6 | A1 |
| Answer | Marks |
|---|---|
| (c) | 3+log (sin(x))=4−( 1−log (sin(x))) |
| 10 10 | B1 |
| Answer | Marks |
|---|---|
| 0 .5 | M1 |
| = 7.04 follow through 1 0 − t h e i r 2 . 9 6 | A1ft |
| Answer | Marks |
|---|---|
| Alt (c) | 3 3 |
| Answer | Marks |
|---|---|
| 0 .5 0 .5 | B1 |
| Answer | Marks |
|---|---|
| 0 .5 | M1 |
| = 7.04 follow through 1 0 − t h e i r 2 . 9 6 | A1ft |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Scheme | Marks |
Question 2:
--- 2(a) ---
2(a) | x | 0.5 | 1 | 1.5 | 2 | 2.5 | 3 | B1
B1
y | 1.319 | 1.075 | 1.001 | 1.041 | 1.223 | 1.850
(2)
(b) | h=0.5 seen or implied | B1
0 .5
Area ( 1 .3 1 9 + 2 ( " 1 .0 7 5 " + 1 .0 0 1 + " 1 .0 4 1 " + 1 .2 2 3 ) + 1 .8 5 0 )
2 | M1
= a w r t 2 . 9 6 | A1
(3)
(c) | 3+log (sin(x))=4−( 1−log (sin(x)))
10 10 | B1
3
3 + l o g ( s i n ( x ) ) d x = 4 x 3 − " 2 .9 6 " = 1 2 − 2 − " 2 .9 6 "
1 0 0 .5
0 .5 | M1
= 7.04 follow through 1 0 − t h e i r 2 . 9 6 | A1ft
(3)
Alt (c) | 3 3
l o g ( s i n ( x ) ) d x = 1 d x − " 2 . 9 6 " = 2 . 5 − " 2 . 9 6 "
1 0
0 .5 0 .5 | B1
3
3 + l o g ( s i n ( x ) ) d x = 3 x 3 + ( 2 . 5 − " 2 . 9 6 " ) = 9 − 1 . 5 + 2 . 5 − " 2 . 9 6 "
1 0 0 .5
0 .5 | M1
= 7.04 follow through 1 0 − t h e i r 2 . 9 6 | A1ft
(3)
(8 marks)
Notes:
(a)
B1: Either of awrt 1.075 or awrt.1.041 correct
B1: Both of awrt 1.075 and awrt.1.041 correct
Remember to check the body of the script for the values if the table has not been completed.
Note that the incorrect values of 2.758 and 2.457 come from the use of degrees.
(Which gives 4.51 in (b))
(b)
B1: Correct interval width seen or implied by working.
M1: Correct application of the trapezium rule. Must all be y values, and should be first + last +
2(sum of middle values), but if one middle value is omitted allow as a slip. A repeated value is
M0. Follow through their values from (a). Allow obvious copying slips.
"0.5"
Note that 1.319+2("1.075"+1.001+"1.041"+1.223)+1.850 scores M0 unless the
2
missing brackets are implied by subsequent work.
Allow this mark if they add the areas of individual trapezia e.g.
" 0 .5 " " 0 .5 "
( 1 .3 1 9 + " 1 .0 7 5 " ) + ( " 1 .0 7 5 " + 1 .0 0 1 ) + ...
2 2
A1: For awrt 2.96. Allow to come from answers in (a) incorrectly rounded or wrong accuracy.
FYI: Use of rounded figure gives 2.96225, use of calculated figures gives 2.9625726… Actual value is 2.89
to 2 decimal places.
( )
(c) Condone all work in (c) written without the “10” in l o g s i n ( x )
1 0
B1: Writes 3 + l o g ( s i n ( x ) ) correctly in terms of 1 − l o g ( s i n ( x ) ) . May be implied.
1 0 1 0
M1: Integrates the K in K ( 1 − l o g ( s i n ( x ) ) ) t o K x , K 3 , substitutes in the limits 3 and 0.5 and
1 0
subtracts oe e.g. 4(3 – 0.5) and uses the answer to (b).
A1ft: Correct answer awrt 7.04, follow through their 2.96 if given to 3sf or better.
Alt:
3
B1: Rearranges integral to find l o g ( s i n ( x ) ) d x = 2 . 5 − " 2 . 9 6 " . May be implied.
1 0
0 .5
M1: Integrates the 3 and substitutes in given limits and subtracts oe e.g. 3(3 – 0.5) and adds their attempt at
3
l o g ( s i n ( x ) ) d x found using the answer to (b).
1 0
0 .5
A1ft: Correct answer awrt 7.04, follow through their 2.96 if given to 3sf or better.
A common incorrect method in (c):
3
3 + l o g ( s i n ( x ) ) d x = 3 x 3 + " 2 . 9 6 " = " 1 0 . 4 6 "
1 0 0 .5
0 .5
Scores no marks as they have not used the integral in part (b)
Attempts to use the trapezium rule again in (c) score no marks
Question | Scheme | Marks\includegraphics{figure_1}
Figure 1 shows the graph of
$$y = 1 - \log_{10}(\sin x) \quad 0 < x < \pi$$
where $x$ is in radians.
The table below shows some values of $x$ and $y$ for this graph, with values of $y$ given to 3 decimal places.
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline
$x$ & 0.5 & 1 & 1.5 & 2 & 2.5 & 3 \\
\hline
$y$ & 1.319 & & 1.001 & & 1.223 & 1.850 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Complete the table above, giving values of $y$ to 3 decimal places. [2]
\item Use the trapezium rule with all the $y$ values in the completed table to find, to 2 decimal places, an estimate for
$$\int_{0.5}^{3} \left(1 - \log_{10}(\sin x)\right) dx$$
[3]
\item Use your answer to part (b) to find an estimate for
$$\int_{0.5}^{3} \left(3 + \log_{10}(\sin x)\right) dx$$
[3]
\end{enumerate}
\hfill \mbox{\textit{Edexcel P2 2022 Q2 [8]}}