Mathematics · May / June 2022 · Paper 11 · Question 1
YMA01
As printed in the paper
The question as text
1. Find ∫ 3 10 x 5 + 6 x 3 − d x x 2 giving your answer in simplest form. (4)
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
10
x
5
+
6
x
3
−
d
x
=
10
×
+ ×
6
−×
3
(
+
c
)
.
x
2
6
4
1
−
5
x
6
3
x
4
3
=
+
+
+
c
3
2
x
Notes:
x
5
→
..
x
6
x
3
→
..
x
4
x
−
2
→
..
x
−
1
M1: For raising any power by 1, e.g.
,
or
.
x + etc
5 1
Accept unprocessed indices such as
A1: For two of the three terms in x correctly integrated (but may be left unsimplified). See below
x
6
x−
1
Accept for this mark terms like
10
×
and
−×−
3
6
1
This may be implied by a correct simplified answer.
x +
5 1
Do NOT accept for this mark ''unprocessed terms'' such as
10
×
5
+
1
A1: For two correct and simplified terms in x all on one line.
3x− for 3
3
x
4
1
1.5x for
4
Accept for this mark equivalent terms such as
and
x
2
1
x−
Do NOT accept for this mark terms like
−×−
3
and
1.67x
6
1
A1: Fully correct and simplified with +c all on one line.
3x− for 3
Accept simplified equivalents, e.g.
1
.
x
5
6
3
4
3
5
6
3
4
3
x
x
x
x
Do NOT accept with spurious symbols like
+
+
+
c
d
x
or
+
+
+
c
3
2
x
3
2
x
Do NOT allow if they then go on and multiply to get rid of fractions
21
cm,
13 c
m,
25 ,
A
B
=
B
C
=
∠
BAC
=
°
∠
A
C
B
= °
x
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