NCEA Level 3 Calculus (91579) 2016 — page 1 of 6 Assessment Schedule – 2016 Calculus: Apply integration methods in solving problems (91579) Evidence Statement Q1
Expected Coverage
Achievement
(u)
(a)
x 2 − ln x + c
Correct
solution.
Absolute value
signs not
required.
c not required.
(b)
1
sec ( 3x ) + c
3
Correct
solution.
Accept 0.3, 0.33
c not required.
(c)
∫ 3y dy = ∫ cos x dx
Correct integral.
c not required.
3y 2
= sin x + c
2
π
3
⎛ π⎞
x = , y = 1 ⇒ = sin ⎜ ⎟ + c
⎝ 6⎠
6
2
3 1
= +c
2 2
c =1
3y 2
= sin x + 1
2
7π
3y 2
⎛ 7π ⎞
x=
⇒
= sin ⎜ ⎟ + 1
⎝ 6 ⎠
6
2
= −0.5 + 1
Merit
(r)
= 0.5
3y = 1
Correct solution
with correct
integral.
Accept positive
solution only.
2
y2 =
1
3
y=±
(d)
1.2
Area =
∫ (e
2x
1
±1
or
or ± 0.577
3
3
)
− e −3x dx
0
Correct integral
[inside square
brackets].
1.2
⎡ e 2 x e −3x ⎤
=⎢
+
3 ⎥⎦ 0
⎣ 2
⎛ e 2.4 e −3.6 ⎞ ⎛ 1 1 ⎞
=⎜
+
−⎜ + ⎟
3 ⎟⎠ ⎝ 2 3 ⎠
⎝ 2
= 4.69
c not required.
Correct solution
with correct
integral.
Accept any
correct
numerical
substitution
(2nd last line)
Excellence
(t)
NCEA Level 3 Calculus (91579) 2016 — page 2 of 6 (e)
Correct integral.
dv
= −kvt
dt
1
∫ vdv = − ∫ kt dy
−kt 2
ln v =
+c
2
t = 0 ⇒ c = ln 3000
t = 20
−k × 20 2
+ ln 3000
2
ln 3000 − ln 2400 = 200k
ln1.25
k=
= 0.00112
200
t = 96
ln 2400 =
Correct integral
and correct
value of k.
−0.00112 × 96 2
+ ln 3000
2
= 2.8454
v = e 2.8454 = 17.2 mL
ln v =
Correct integral
and correct
solution.
Accept any
correct
numerical
substitution
NØ
N1
N2
A3
A4
M5
M6
E7
E8
No response;
no relevant
evidence.
ONE answer
demonstrating
limited
knowledge of
integration
techniques.
1u
2u
3u
1r
2r
1t with minor
error(s).
1t
NCEA Level 3 Calculus (91579) 2016 — page 3 of 6 Q2
(a)
(b)
(c)
Expected Coverage
5x 5 −
Achievement
(u)
Merit
(r)
Correct
solution.
10x 3
+x+c
3
7
Correct
solution.
1
a(t) = 0.2t + 0.3t 2
3
v(t) = 0.1t 2 + 0.2t 2 + c
t=4
v(4) = 0.1× 16 + 0.2 × 8 + c = 5
3.2 + c = 5
c = 1.8
Correct integral.
3
v(t) = 0.1t 2 + 0.2t 2 + 1.8
Correct solution
with correct
integral.
Accept 59.9,
60.
2nd last line
acceptable.
5
2
3
0.1t
+ 0.08t + 1.8t + k
3
Distance travelled in first 9 seconds
5
0.1× 9 3
s(9) =
+ 0.08 × 9 2 + 1.8 × 9
3
= 59.94 m
s(t) =
(d)
2m
2
3
⎡
⎤
∫ ( 2x − m ) = ⎢ ( 2x − m ) + c ⎥
m
1
⎣6
2m
⎦m
1⎡
( 4m − m )3 − ( 2m − m )3 ⎤⎦
6⎣
1
3
= ⎡⎣( 3m ) − m 3 ⎤⎦
6
26m 3
=
6
13m 3
=
3
=
13m 3
= 117
3
117 × 3
m3 =
= 27
13
m=3
∴
Correct
integration
[inside square
brackets].
Correct solution
with correct
integration.
Excellence
(t)
NCEA Level 3 Calculus (91579) 2016 — page 4 of 6 (e)
( k − 1) x 2 = 9 − x 2
kx 2 = 9
9
x2 =
k
±3
x=
k
Area = 2 ×
3
k
∫ (( 9 − x ) − ( k − 1) x
2
0
= 2×
2
)dx
3
k
∫ ( 9 − kx )dx
2
0
3
k
⎡
kx 3 ⎤
= 2 × ⎢ 9x −
3 ⎥⎦ 0
⎣
9 ⎞
⎛ 27
= 2×⎜
−
⎟
⎝ k
k⎠
36
=
k
∴
kx 3
− 9x
3
Correct integral.
Accept
kx 3
− 9x
3
Correct solution.
For E7:
3
k
∫ = 24
0
leads to
18
= 24
k
3
k=
4
9
k=
16
(E7)
36
= 24
k
k = 1.5
k = 2.25
NØ
N1
N2
A3
A4
M5
M6
E7
E8
No response;
no relevant
evidence.
ONE answer
demonstrating
limited
knowledge of
integration
techniques.
1u
2u
3u
1r
2r
1t with minor
error(s).
1t
NCEA Level 3 Calculus (91579) 2016 — page 5 of 6 Q3
(a)
Expected Coverage
Achievement
(u)
Merit
(r)
4
4
k⎞
k⎤
⎛
⎡
∫1 ⎜⎝ 4 + x 2 ⎟⎠ dx = ⎢⎣ 4x − x ⎥⎦1
k⎞
⎛
= ⎜ 16 − ⎟ − ( 4 − k )
⎝
4⎠
= 12 +
3k
4
12 +
3k
=0
4
k = −16
Correct solution
with correct
integration.
(b)
7.8
Correct
solution.
(c)
2 − x 2 = −x
x2 − x − 2 = 0
( x − 2 )( x + 1) = 0
x = 2, − 1
2
Area =
∫ [( 2 − x ) − ( −x )]dx
2
−1
2
=
∫ (2 − x
−1
2
)
+ x dx
2
⎡
x3 x2 ⎤
= ⎢ 2x − + ⎥
3 2 ⎦ −1
⎣
8
1 1⎞
⎛
⎞ ⎛
= ⎜ 4 − + 2 ⎟ − ⎜ −2 + + ⎟
⎝
3 ⎠ ⎝
3 2⎠
= 4.5
(d)
⎛ e 3x − x 2 ⎞
1 ⎛ 3e 3x − 3x 2 ⎞
dx
=
dx
∫ ⎜⎝ e 3x − x 3 ⎟⎠
3 ∫ ⎜⎝ e 3x − x 3 ⎟⎠
1
= ln e 3x − x 3 + c
3
(
)
Correct
integration
[inside square
brackets].
Both integrals
completed
separately is ok
for u.
Correct solution
with correct
integration
Correct
solution.
Absolute value
signs not
required.
Brackets not
required.
Excellence
(t)
NCEA Level 3 Calculus (91579) 2016 — page 6 of 6 (e)
dy
= e y + sin x
dx
1 dy
⋅ = e y ⋅ esin x
cos x dx
sec x ⋅
1
∫e
∫e
y
dy = ∫ cos x ⋅ esin x dx
−y
dy = ∫ cos x ⋅ esin x dx
Correct
integration of 1
side.
−e − y = esin x + c
x= 0, y = –1
−e1 = e 0 + c
Correct solution
with correct
integration.
Answer may not
be exactly zero
if the value of c
is evaluated
numerically.
E.g. c = –3.7183
leads to
y = –0.0611.
Accept
continuity.
Correct
integration of
both sides.
c = −e1 − 1
∴−e − y = esin x − e − 1
e − y = e + 1− esin x
π
x=
2
−y
e = e + 1− e
e− y = 1
y=0
NØ
N1
N2
A3
A4
M5
M6
E7
E8
No response;
no relevant
evidence.
ONE answer
demonstrating
limited
knowledge of
integration
techniques.
1u
2u
3u
1r
2r
1t with minor
error(s).
1t
Cut Scores Not Achieved Achievement Achievement with Merit Achievement with Excellence 0–6 7–13 14–19 20–24