Edexcel PURE 2024 October — Question 1

Exam BoardEdexcel
ModulePURE
Year2024
SessionOctober
PaperDownload PDF ↗
TopicNumerical integration
TypeApply trapezium rule to given table
DifficultyModerate -0.8 This is a straightforward application of the trapezium rule with given values, followed by a simple manipulation using linearity of integration. Part (a) requires only substitution into the standard formula, and part (b) uses the fact that the integral of a sum equals the sum of integrals, where the second integral can be computed exactly. Both parts are routine procedural questions with no problem-solving or conceptual challenges beyond basic recall.
Spec1.08d Evaluate definite integrals: between limits1.09f Trapezium rule: numerical integration
  1. A continuous curve has equation \(y = \mathrm { f } ( x )\).
A table of values of \(x\) and \(y\) for \(y = \mathrm { f } ( x )\) is shown below.
\(x\)0.51.7534.255.5
\(y\)3.4796.1017.4486.8235.182
Using the trapezium rule with all the values of \(y\) in the given table,
  1. find an estimate for $$\int _ { 0.5 } ^ { 5.5 } \mathrm { f } ( x ) \mathrm { d } x$$ giving your answer to one decimal place.
  2. Using your answer to part (a) and making your method clear, estimate $$\int _ { 0.5 } ^ { 5.5 } ( \mathrm { f } ( x ) + 4 x ) \mathrm { d } x$$
Question 1:
AnswerMarks Guidance
1(a)h=1.25 B1
A 1 1.25  3.479+5.182+2 ( 6.101+7.448+6.823 )
AnswerMarks
2M1
= 30.9A1
(3)
AnswerMarks
(b)5 .5
( f ( x ) + 4 x ) d x = 3 0 .9 +  2 x 2  5 .5 = 3 0 .9 + 2  5 .5 2 − 2  0 .5 2
0 .5
0 .5
or
5 .5 5 .5
  1
( f ( x ) + 4 x ) d x = f ( x ) d x + ( 4  0 .5 + 4  5 .5 )  5 = ...
2
AnswerMarks
0 .5 0 .5M1
= 30.9 + 60 = 90.9A1ft
(2)

Total 5

(b)
M1: Attempts their answer to (a) +  ... x 2  5
0
.5
.5
= t h e i r a n s w e r t o ( a ) + ...5 .5 2 − .. .0 . 5 2
May be implied by e.g. their answer to
(
a
)+25.52−20.52
if the integration is not
seen explicitly.
Or
their answer to (a) +
1
2
( 4  0 . 5 + 4  5 . 5 )  5 (trapezium)
1
oe e.g. their answer to (a) +( 40.55 )+ ( 5( 55.5−50.5 )) (rectangle + triangle)
2
May be implied by their answer to (a) + 60
Condone clear misreads of the 4 in the 4x but do not condone a misread of e.g. 4x for 4.
Condone clear mis-copy/mis-read of limits as long as they are non-zero E.g. 5 for 5.5.
5.5 5.5
 
Condone poor notation e.g. ( f (x)+4x ) dx=30.9+ 2x2dx=30.9+25.52−20.52
0.5 0.5
A1ft: Correct answer of awrt 90.9 or correct ft e.g. 60 + their answer to part (a)
Allow exact or exact ft answers e.g.
2 9
3
0
2
8
0
1
for 90.9
Correct answer only scores no marks as the questions says “making your method
clear”
Attempts to use the trapezium rule again by adding 4x to the y values score M0 in (b)
Note that 3 0 .9 + 4 ( 0 .5 + 1 .7 5 + 3 + 4 .2 5 + 5 .5 ) = 9 0 .9 is fortuitous and scores M0
But 3 0 . 9 +
1
2
5
4
( 2 + 2 2 + 2 ( 7 + 1 2 + 1 7 ) ) = 9 0 . 9 scores M1A1
Condone an error with the strip width as in part (a) for this method but the ft is not available
in this case.
Examples of minimal acceptable working in (b) for both marks:
30.9 + 60 = 90.9
3 0 .9 +  2 x 2 
5
0
.5
.5
= 9 0 .9
PMT
5.5 5.5
 
f (x) dx+ 4x dx=30.9+60=90.9
0.5 0.5
30.9+25.52 −20.52 =90.9
All score M1A1
Question 1:
--- 1(a) ---
1(a) | h=1.25 | B1
A 1 1.25  3.479+5.182+2 ( 6.101+7.448+6.823 )
2 | M1
= 30.9 | A1
(3)
(b) | 5 .5

( f ( x ) + 4 x ) d x = 3 0 .9 +  2 x 2  5 .5 = 3 0 .9 + 2  5 .5 2 − 2  0 .5 2
0 .5
0 .5
or
5 .5 5 .5
  1
( f ( x ) + 4 x ) d x = f ( x ) d x + ( 4  0 .5 + 4  5 .5 )  5 = ...
2
0 .5 0 .5 | M1
= 30.9 + 60 = 90.9 | A1ft
(2)
Total 5
(b)
M1: Attempts their answer to (a) +  ... x 2  5
0
.5
.5
= t h e i r a n s w e r t o ( a ) + ...5 .5 2 − .. .0 . 5 2
May be implied by e.g. their answer to
(
a
)+25.52−20.52
if the integration is not
seen explicitly.
Or
their answer to (a) +
1
2
( 4  0 . 5 + 4  5 . 5 )  5 (trapezium)
1
oe e.g. their answer to (a) +( 40.55 )+ ( 5( 55.5−50.5 )) (rectangle + triangle)
2
May be implied by their answer to (a) + 60
Condone clear misreads of the 4 in the 4x but do not condone a misread of e.g. 4x for 4.
Condone clear mis-copy/mis-read of limits as long as they are non-zero E.g. 5 for 5.5.
5.5 5.5
 
Condone poor notation e.g. ( f (x)+4x ) dx=30.9+ 2x2dx=30.9+25.52−20.52
0.5 0.5
A1ft: Correct answer of awrt 90.9 or correct ft e.g. 60 + their answer to part (a)
Allow exact or exact ft answers e.g.
2 9
3
0
2
8
0
1
for 90.9
Correct answer only scores no marks as the questions says “making your method
clear”
Attempts to use the trapezium rule again by adding 4x to the y values score M0 in (b)
Note that 3 0 .9 + 4 ( 0 .5 + 1 .7 5 + 3 + 4 .2 5 + 5 .5 ) = 9 0 .9 is fortuitous and scores M0
But 3 0 . 9 +
1
2

5
4
( 2 + 2 2 + 2 ( 7 + 1 2 + 1 7 ) ) = 9 0 . 9 scores M1A1
Condone an error with the strip width as in part (a) for this method but the ft is not available
in this case.
Examples of minimal acceptable working in (b) for both marks:
30.9 + 60 = 90.9
3 0 .9 +  2 x 2 
5
0
.5
.5
= 9 0 .9
PMT
5.5 5.5
 
f (x) dx+ 4x dx=30.9+60=90.9
0.5 0.5
30.9+25.52 −20.52 =90.9
All score M1A1
\begin{enumerate}
  \item A continuous curve has equation $y = \mathrm { f } ( x )$.
\end{enumerate}

A table of values of $x$ and $y$ for $y = \mathrm { f } ( x )$ is shown below.

\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | }
\hline
$x$ & 0.5 & 1.75 & 3 & 4.25 & 5.5 \\
\hline
$y$ & 3.479 & 6.101 & 7.448 & 6.823 & 5.182 \\
\hline
\end{tabular}
\end{center}

Using the trapezium rule with all the values of $y$ in the given table,\\
(a) find an estimate for

$$\int _ { 0.5 } ^ { 5.5 } \mathrm { f } ( x ) \mathrm { d } x$$

giving your answer to one decimal place.\\
(b) Using your answer to part (a) and making your method clear, estimate

$$\int _ { 0.5 } ^ { 5.5 } ( \mathrm { f } ( x ) + 4 x ) \mathrm { d } x$$

\hfill \mbox{\textit{Edexcel PURE 2024 Q1}}
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