Mathematics · Oct / Nov 2021 · Paper 41 · Question 6
YMA01
As printed in the paper
The question as text
6. In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable. y R O π π x 6 2 Figure 2 Figure 2 shows a sketch of the curve with equation 16 sin 2 x π 0 x 2 y = ( 3 + 4 sin ) x 2 The region R, shown shaded in Figure 2, is bounded by the curve, the x-axis and the line π with equation x = 6 Using the substitution u = 3 + 4 sin x , show that the area of R can be written in the form a + ln b , where a and b are rational constants to be found. (7)
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
u
=
3
+
4sin
x
⇒
=
4cos
x
d
x
(
)
∫
16sin 2
x
∫
32sin cos
x
x
∫
2
u
−
3
d
x
=
d
x
=
d
u
2
2
2
(3
+
4sin )
x
(3
+
4sin )
x
u
∫
2
6
6
=
−
d
u
=
2ln
u
+
u
u
u
2
6
6
12
49
⇒
2ln 7
+
−
2ln5
−
= −
+
ln
Uses limits of 5 and 7
7
5
35
25
B1: States or uses d
u
x =
4cos
x
o.e. This may be seen within the integrand.
d
M1: Attempts to write all in terms of u.
(
)
∫
16sin 2
x
∫
...sin cos
x
x
∫
...
u
±
3
Look for
d
x
=
d
x
= ±
d
u
2
2
2
(3
+
4sin )
x
(3
+
4sin )
x
u
It is common to lose the square on the
u which would be M0.
2
Only the penultimate mark would then be available.
(
)
∫
2
u
−
3 d
A1: Correct
u
o.e. Allow this to be unsimplified.
2
u
Condone a missing du this can be implied by further work.
(
)
∫
...
u
±
3 d
...
u
...lnu
±
dM1: Integrates
to
o.e Condone a missing du this can be implied by further work.
u
u
2
(
)
∫
...
u
±
3 d
...
2
...lnu
Note that
u
→
±
is acceptable
u
2
u
You may see attempts using parts.
∫
∫
1
1
1
1
(
)
(
)
(
)
×
2
u
−
6 d
u
= −
×
2
u
−
6
−
−
×
2 d
u
= −
×
2
u
−
6
+
2ln
u
u
u
u
u
2
6
6
1
2
2lnu
lnu
(
)
A1: Correct
+
. Note that
+
or
−
×
2
u
−
6
+
2ln
u
is also completely correct
u
u
u
M1: Uses limits 5 and 7 within their attempted integral and subtracts . Condone poor attempts at the integration
Alternatively converts their answer in u back to x's using the correct substitution and uses the given limits
12
49
12
−
+
ln
ln1.96
−
A1:
or equivalent such as
35
25
35
ISW after a correct answer.
4
−
4
λ
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