Mathematics · May / June 2025 · Paper 21 · Question 3
YMA01
As printed in the paper
The question as text
3. y C P (1, 11) R Q (4, 5) O x Figure 1 In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable. Figure 1 shows a sketch of the curve C with equation y = 3x + 16 – 8 x > 0 2x The points P (1, 11) and Q (4, 5) lie on C and are shown in Figure 1. The region R, shown shaded in Figure 1, is bounded by C and line segment PQ. Use algebraic integration to find the area of R. (5)
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Show the answer and mark scheme
3
+
−
8 d
=
−
−
8
(
+
)
x
x
x
x
c
2
2
x
x
1
4
3
3
16
45
27
e.g. Area = (
)
2
5 11
+
−
−
−
8
=
24
−−
12
+
=
x
x
2
2
2
2
x
1
M1: Attempts to integrate the curve and achieves at least one term with the correct index i.e.
or
−→±
8
8x
(indices do not need to be processed for this mark e.g.
x
−
2
→
x
−
1
x
→
3
16
2
A1: Two correct terms from
−
−
8
, may be unsimplified but indices must be processed.
2 x
x
x
Allow equivalent forms.
3
16
2
A1:
−
−
8
, may be unsimplified. Allow the presence of the constant of integration.
2 x
x
x
Do not be concerned with spurious notation or any limits.
dM1: Full method including correct area (value or expression) of trapezium. The expression with correct
limits substituted in or partially evaluated as evidence is sufficient for this mark and condone subtracting
either way round. They cannot just proceed to the final answer without additional evidence of using the
limits. Note area of trapezium = 24, area under curve = 10.5.
Condone slips in evaluating some of the terms provided the intention is clear.
A1: 13.5 o.e following all previous marks scored. Note if they use the limits the wrong way round resulting
in a negative value for the integral then we must see that calculated correctly before they state the area is
13.5
Alternative: line – curve
4
2
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