ESM1144-SOLUTION
SEMESTER 2, 2009/2010
QUESTION 1 [8 marks]question 1a.
Solutions

3
1
2 x  x 2 dx
   2 x  x 2 dx   2 x  x 2 dx    2 x  x 2 dx
0
2
1
3
0
0
2
2
3
 x3



x3   x3
   x2   x2      x2 
3 0  3
3
 1 
2
8
8

 0  1   4    [( 9  9)  (  4)]
3
3

16 5
7

3 3

5
0
f ( x )dx
 1(3)  2(3) 
 12 


2
(12 ) 
1
( 2)( 3)
2
2
DEPARTMENT OF ENGINEERING
2
ESM1144-SOLUTION
SEMESTER 2, 2009/2010
Question 1b
Solutions
(a)
F ( 2)
2
  f (t ) dt
2

1
1
( 4)( 3)   (12 )
2
2
6

2
(b)
DEPARTMENT OF ENGINEERING
3
ESM1144-SOLUTION
SEMESTER 2, 2009/2010
F ' ( x)
dF ( x )
dx
d  x

f (t ) dt 

dx  2
 f (t )

 F ' ( 3)
 f ( 3)
 1
Given the following graph of g(x).
Evaluate

7
0
g ( x)dx
DEPARTMENT OF ENGINEERING
4
ESM1144-SOLUTION
SEMESTER 2, 2009/2010
CHAPTER 4
1.
A spring has a natural length of 22 cm. If a force of 15 N is required to keep it stretched to a length
of 32 cm, how much work is required to stretch it from 22 cm to 40 cm?
F  kx
15  k (0.1)
k  150
Work  
0.18
0
2.

150 xdx  75 x 2

0.18
0
 75(0.0324)  2.43J
A tank is full of water. Find the work required to pump the water out of the outlet.
Round the answer to the nearest thousand.
h=1m,r=1m,d=5m
Select the correct answer.
a. W = 30,000 J
b. W = 308,000 J
c. W = 757,000 J
DEPARTMENT OF ENGINEERING
d. W = 431,000 J
e. W = 305,000 J
5
ESM1144-SOLUTION
SEMESTER 2, 2009/2010
The region bounded by the given curves is rotated about the specified axis. Find the volume of the
resulting solid by any method.
3.
y  x 2  2x  3 ; about the x-axis
4. Find the area enclosed by the line y = x - 1 and the parabola y2 = 2x + 6
Solutions
4
A
2
xR  xL  dy
   y  1  12 y 2  3 dy
2
4

4
2

1
2
y 2  y  4  dy
4

1  y3  y 2
      4 y
2 3  2
 2
  16 (64)  8  16   43  2  8   18
DEPARTMENT OF ENGINEERING
6
ESM1144-SOLUTION
SEMESTER 2, 2009/2010
Find the volume of solid generated when area bounded by
y
3 2
3 2
x  3, y 
x  3 and
16
16
y  axis
is revolved about the line y = 2
DEPARTMENT OF ENGINEERING
7
ESM1144-SOLUTION
 1 2 x
FINAL CHAP
2
SEMESTER 2, 2009/2010
x5 dx   1  x 2 x 4  x dx   u (u  1) 2

1
2

1
2
du
2
 u (u  2u  1) du
 (u  2u  u ) du
2
5/ 2
3/ 2
1/ 2
 12 ( 72 u 7 / 2  2  52 u 5/ 2  23 u 3/ 2 )  C
 17 (1  x 2 )7 / 2  52 (1  x 2 )5/ 2
 13 (1  x 2 )3/ 2  C
FINAL CHP 3
M1
M1
Question 3(a) [6 marks]

ln 2
0
4e x sinh( 2 x) dx
ln 2
 e2 x  e2 x 
dx
4e 
2


=

=
 2e
0
ln 2
0
x
3x

 2e x dx
S1(t.id)
A1(simp)
DEPARTMENT OF ENGINEERING
8
ESM1144-SOLUTION
SEMESTER 2, 2009/2010
ln 2
 e3 x e  x 
= 2


 1 0
 3
1
 8 1

 2 (  )  (  1) 
3
 3 2

M1(u subs)A1(int)
M1(ky in lim)A1(cor ans only, CAO)
Question 3(b) [3 marks]

cos(sec 1 x)
x x2  1
dx
let u = sec-1 x
1
dx
du =
x x2  1
M1(subs rule)
=  cos u du
S1 (in terms u)
-1
= sin u + c = sin (sec x) + c A1 (CAO)
Question 4 [6 marks]
Evaluate
x

x  2x  3
2
x
dx
  ( x  1) 2  2
dx . Express your answer in terms of ln.
M1(Csq)A1
let u = x+1
du = dx

u 1
u 2
2
du  
u
u 2
2
du  
du
u 2
1 1
 u 
dw  sinh 1 


2
w
 2
 x  1
= w  sinh 1 

 2 
=
=
S1
2
A1 (doesn’t matter in what terms)
A1 (for
w)
( x  1) 2  2  ln ( x  1)  ( x  1) 2  2  c A1 (in terms of ln)
DEPARTMENT OF ENGINEERING
9
ESM1144-SOLUTION
SEMESTER 2, 2009/2010
Question 5(a) [8 marks]
ln x
dx
e x2
let u = ln x
du = 1/x dx

e
dv = 1/x2
v = -1/x
ln x
1 1
   . dx
x
x x
e
ln x
1
=
  2 dx
x e
x
=
1 1
 1
= 
ln e    
e
e
 x
M1 (correct u & dv)
A1A1 (diff & int)
M1(IBP)
S1(simp)
e
A1(int)
e
1
1
1 1
3 e 4
= 
 

e 2 e e
2e
e
M1(key in) A1(COA)
Question 5(b) [4 marks]
 x sin
3
( x 2 ) dx =
1
sin 3 u du
2
let u = x2
du = 2x dx
1 1 2
2

 sin u cos u   sin udu 

2 3
3

1 1
2

=  sin 2 u cos u  ( cos u )  c
2 3
3

1
1
=  sin 2 ( x 2 ) cos( x 2 )  cos( x 2 )  c
6
3
=
DEPARTMENT OF ENGINEERING
M1(subs rule)
M1(RF)
A1(u terms)
A1(x terms)
10
ESM1144-SOLUTION
question 2
Evaluate

3
0
x 
SEMESTER 2, 2009/2010
( x3  6 x) dx
ba 3

n
n

3
0
( x3  6 x)dx
n
 lim  f ( xi )x
n 
i 1
n
 3i  3
 lim  f  
n 
 n n
i 1
3
3 n  3i 
 3i  
 lim     6   
n  n
 n  
i 1 
 n 
(Eqn. 9 with c  3 / n)
3 n  27 3 18 

 3 i  n i 
n  n
i 1  n
 lim
54 n 
 81 n
 lim  4  i 3  2  i 
n  n
n i 1 
 i 1
(Eqns. 11 & 9)
 81  n(n  1)  2 54 n(n  1) 
 lim  4 

  n2
n  n
2 
  2 
(Eqns. 7 & 5)
 81  1  2
 1 
 lim  1    27 1   
n 
 n  
 4  n 
81
27
  27  
 6.75
4
4
DEPARTMENT OF ENGINEERING
11