Integration Solutions
1.
Let the volume of the solid of revolution be V.
 (ax  2)
a
V=
=

2
0
a

 ( x 2  2) 2 dx
(M1)
(a2x2 + 4ax + 4 – x4 – 4x2 –4)dx
0
(M1)
a
1
4 
1
=   a 2 x 3  2ax 2  x 5  x 3 
5
3 0
3
2 
2
=   a 5  a 3  units3
3 
 15
(M1)
(A1)
2a 3 π 2
(a + 5)
15
=
(C4)
Note: The last line is not required
[4]
3.
(a)
y
2=k
–1
0
1
2
k 2
1
3 x
–1
–2
3
Notes: Award (A1) for the correct intercepts
(A1) for graphing over the correct interval
(A1) for the correct x-coordinate of the maximum point.
1
(b)
Required area =

k

( x  x 2 )dx – ( x  x 2 )dx
0
1
1
k
x
x
x 
x 
=      
3 0  2
3 1
2
2
3
(M1)
2
3
1 k2 k3 1 1 k3 k2


  

6 2
3 6 3 3
2
1
= (2 + 2k3 – 3k2)
6
=
(A1)
(M1)
(A1)
OR
k
Required Area =
 ( x  x )dx
2
(M1)
1
k
 x2 x3 
=   
3 1
 2
(M1)(A1)
k3 k2 1


3
2 6
(A1)
=
4
[7]
1
a2
1
dx = 0.22
a 1  x2
2
[arctan x ] aa
Then
= 0.22
4.
If

(M1)
2
arctan a – arctan a – 0.22 = 0
(A1)
a = 2.04 or a = 2.62
(A1) (C3)
Notes: Award final (A1) only if both correct answers are shown.
If no working is shown and only one answer is correct, award
(C1).
GDC example: finding solutions from a graph.
[3]
5.
(a)
y = ln x5 – 3x2
(i)
2.5
–0.5
0.5
1
1.5
2
–2.5
–5
–7.5
–10
–12.5
asymptote
asymptote
(G2)
Note: Award (G1) for correct shape, including three zeros, and (G1)
for both asymptotes
(ii)
(b)
f (x) = 0 for x = 0.599, 1.35, 1.51
f (x) is undefined for
(x5 – 3x2) = 0
x2(x3 – 3) = 0
Therefore, x = 0 or x = 31/3
f (x) =
(c)
(G1)(G1)(G1)
5
(M1)
(A2)
5x 4  6 x  5x 3  6 
 or

x 5  3x 2  x 4  3x 
3
(M1)(A1)
f (x) is undefined at x = 0 and x = 31/3
(d) For the x-coordinate of the local maximum of f (x), where
0 < x < 1.5 put f (x) = 0
5x3 – 6 = 0
(A1)
3
(R1)
(M1)
1
 6 3
x=  
5
(e)
(A1)
3
(A2)
2
The required area is
1.35
A=
 f ( x)dx
0.599
Note: Award (A1) for each correct limit.
[16]
6.
Let
dv
 ln xdx   u dx dx where u = lnx and
Then
du 1
and v = x.

dx x
dv
=1
dx
(M1)
Using integration by parts,
2
1
 ln xdx  x ln x   x x dx
(A1)
= x ln x – x + C
(A1)
[3]
7.
x-intercepts are = π, 2π, 3π.
Area required =

2π
π
(A1)
sin x
dx 
x

3π sin
2π
x
x
dx
(M1)
= 0.4338 + 0.2566
= 0.690 units2
(G1) (C3)
[3]
The curves meet when x = –1.5247 and x = 0.74757.
x

0.74757
2
3

 dx

e
The required area =

1.5247  1  x 2


= 1.22.
(G1)
9.
(G1)

(M1)
(C3)
[3]
10.
(a)
=
(b)
 x3

ln x  –
3

 x2 ln x dx = 
x3 1
 3 x dx
(M1)(A1)(A1)
x3
x3
ln x – (Constant of integration not required.)
3
9

2
1
x 2 ln x dx = 1.07
(A1) (C4)
7
 8
 or ln 2 – 
9
 3
(A2) (C2)
[6]
11.
Using integration by parts u = 
du = d
=>  cos d =  sin – sind
=>  cos d =  sin + cos
v = sin θ
dv = cos d
Therefore, => ( cos – )d =  sin + cos –
Note:
(M1)
(M1)(A1)
(A1)
2
+c
2
Award (C5) for  sin + cos –
(A2) (C6)
2
, ie
2
penalize omission of + c by [1 mark].
[6]
3
12.
(a)
(A1)(A1)
2
6
4
A
2
g ( x)
–4
–3
–2
–1
1
2
3
4
5
6
–2
–4
f(x)
–6
Note: Award (A1) for showing the basic shape of f (x).
Award (A1) for showing both the vertical asymptote and
the basic shape of g (x).
(b)
(c)
(d)
(e)
(i)
x = –3 is the vertical asymptote.
(A1)
(ii)
x-intercept: x = 4.39 ( = e2 – 3)
y-intercept: y = –0.901 ( = ln 3 – 2)
(G1)
(G1)
3
(G1)(G1)
2
f (x) = g (x)
x = –1.34 or x = 3.05
(i)
See graph
(ii)
Area of A =
(iii)
Area of A = 10.6
 4 – 1 – x  
3.05
2
0
– (ln (x + 3) – 2)dx
(M1)(A1)
(G1)
4
y = f (x) – g (x)
y = 5 + 2x – x2 – ln(x + 3)
dy
1
 2 – 2x –
dx
x3
(M1)
Maximum occurs when
2 – 2x =
dy
=0
dx
1
x3
5 – 4x – 2x2 = 0
x = 0.871
y = 4.63
OR
Vertical distance is the difference f (x) – g (x).
Maximum of f (x) – g (x) occurs at x = 0.871.
The maximum value is 4.63.
(A1)
(A1)
(M1)
(G1)
(G1) 3
[14]
14.
(a)
 R cos = 1, R sin =
π
 R = 2,  =
3
cosx +
3 sinx = R cos cosx + R sin sinx (M1)
3
(A1)(A1)
Note:
3
Award (M1)(A1)(A0) if degrees used instead of radians.
4
(b)
π

,
3

π

when x  
fmax = 2 
3  ; fmin = 1 (when x = 0)
Since f (x) = 2 cos  x –
(i)
Range is [1, 2]
(ii)
(c)
(A1)(A1)
(A1)
Inverse does not exist because f is not 1:1
(R2)
Notes: Award (R2) for a correct answer with a valid reason.
Award (R1) for a correct answer with an attempt at a valid
reason, eg horizontal line test.
Award (R0) for just saying inverse does not exist, without any
reason.
2
π

 =
3
2

π
π
x– = 
3
4
π
x=
12
2  cos  x –
f (x) =
OR
f (x) =
5
(M1)
(A1)
(A1)
2
(M1)
x = 0.262
(G1)
π
12
 x=
(A1)
3
π
(d)
I=
1 2 
π
sec x – dx

0
2
3

(M1)
π
1  
π
π   2

= ln  sec x –   tan  x –  
2  
3
3   0

1 
 2



1  3
3
ln
=
2  2– 3 




1 
3 2 3 
 = 1 ln (3 + 2 3 ).
ln 
=
2  2 – 3 2  3 
2




Note:

(A1)
(A1)(A1)
(M1)(AG)
5
Award zero marks for any work using GDC.
[16]
5
15.
y = ex – e
y = ln x
Curves intersect at x = 0.233
and x = 1
Area =

1
0.233
(G1)
(G1)
(ln x  e x  e)dx
(M1)(A1)
= 0.201
(G2) (C6)
[6]
17.
METHOD 1
Region required is given by
y
(2, 5)
(4, 7)
x
(–3, 0)
from gdc outer intersections are at x = –3 and x = 4
Area =

4
(A1)(A1)
y1 – y 2 dx
–3
(M1)
= 101.75
(A3) (C6)
METHOD 2
From gdc intersections are at x = –3, x = 2, x = 4
Area =
 x
2
3
3

– 3 x 2 – 9 x  27 – ( x  3) dx 
= 101.75
 x  3 – ( x
4
2
(A1)(A1)(A1)
3

– 3x 2 – 9 x  27 ) dx (M1)(M1)
(A1) (C6)
[6]
6
18.
Using the chain rule f (x) =  2 cos  5 x    5
2 


(a)
= 10 cos  5 x   
2

(b)
f (x) =
= 
(M1)
A1
2
 f ( x) dx
2
π

cos  5 x   + c
5
2

A1
 π π
Substituting to find c, f  π  = – 2 cos  5    + c = 1
5
2
 2 2
c = 1 + 2 cos 2 = 1 + 2 = 7
5
5
5
M1
(A1)
f (x) = – 2 cos  5 x    + 7
2
5
5

A1
N2
4
[6]
19.
Substituting u = x + 2  u – 2 = x, du = dx

(M1)
(u  2) 3
x3
d
x

du
( x  2) 2
u2

A1
3
2
 u  6u 2 12u  8 du
u




A1

 u du  (6) du  12 du  8u 2 du
u
A1
2
 u  6u  12 ln u  8u 1  c
2

( x  2) 2
8
 6( x  2)  12 ln x  2 
c
2
x2
A1
A1
N0
[6]
20.
e
x
cos x dx  e x cos x   e x sin x dx
= e x cos x  e x sin x   e x cos x dx
2  e x cos x dx  e x cos x  sin x   c
x
 e cos x dx 
ex
cos x  sin x   k
2
Note:
(M1)(A1)
(M1)(A1)
(M1)
(A1) (C6)
Do not penalize for missing integration
constants.
[6]
21.
Attempting to find point of intersection
Intersection at x = 2
Note: Award M1A1 if x = 2 is seen as upper
limit of an integral.
Using appropriate definite integrals
Area = 1.66
(M1)
(A1)
M2
A2
N2
7
[6]
+c=1
M1
c = 1 + 2 2 7 2  5 x    7
2  5 22.
5 5 5 5 
e
2x

METHOD 1
sin x dx   e 2 x cos x  2 e 2 x cos x dx

=  e 2 x cos x  2 e 2 x sin x  2  e 2 x sin x dx
(M1)A1

(M1)A1

=  e 2 x cos x  2 e 2 x sin x  4 e 2 x sin x dx
5 e 2 x sin x dx  e 2 x 2 sin x  cos x 


e 2 x sin x dx 
(M1)
e2x
2 sin x  cos x   C
5
A1
N0
METHOD 2
e
2x
1
1 2x
sin x dx  e 2 x sin x 
e cos x dx
2
2

=

1 2x
1
1
e sin x   e 2 x cos x   e 2 x sin x dx
2
4
4
5 2x
1
1
e sin x dx  e 2 x sin x  e 2 x cos x

4
2
4
e2x
 e sin x dx  5 2 sin x  cos x   C
2x
Note:
23.
(a)
(M1)A1
(M1)
A1
Do not penalize the absence of constants
of integration.
[6]
1
f x   ln x  x  1
 x
(M1)
= ln x
(b)
(M1)A1
A1
N2
Using integration by parts
METHOD 1
 ln x
2
dx  ln x   x  x 

2
= x ln x   2
2
2
ln x dx
x
 ln xdx
= x (ln x)2  2(x ln x  x) + C
A1A1
(A1)
A1
(= x (ln x)2  2x ln x + 2x + C)
METHOD 2
 ln x
2
dx  x ln x   x ln x 
2
 ln x 1dx
= x (ln x)2 x ln x  (x ln x  x  x) + C
A1A1A1
A1
(= x (ln x)2  2 x ln x + 2x + C)
Note: Do not penalize the absence of + C.
[6]
8
9