436 Higher Engineering Mathematics

Problem 1. (a) Use integration to evaluate,

correct to 3 decimal places,

∫ 3

1

2

√

x

dx(b) Use the

trapezoidal rule with 4 intervals to evaluate the

integral in part (a), correct to 3 decimal places.

(a)

∫ 3

1

2

√

x

dx=

∫ 3

1

2 x−

1

(^2) dx

⎡
⎢
⎣
2 x
(− 1
2
)

• 1
  −
  1
  2


• 
  

  1
  ⎤
  ⎥
  ⎦
  3
  1

  
  

  [
  4 x
  1
  2
  ] 3
  1
  = 4
  [√
  x
  ] 3
  1 =^4
  [√
  3 −
  √
  1
  ]
  =2.928,correct to 3 decimal places
  (b) The range of integration is the difference between
  the upper and lower limits, i.e. 3− 1 =2. Using
  the trapezoidal rule with 4 intervals gives an inter-
  val widthd=
  3 − 1
  4
  = 0 .5 and ordinates situated
  at 1.0, 1.5, 2.0, 2.5 and 3.0. Corresponding values
  of
  2
  √
  x
  are shown in the table below, each correct
  to 4 decimal places (which is one more decimal
  place than required in the problem).
  x
  2
  √
  x
  1.0 2.0000
  1.5 1.6330
  2.0 1.4142
  2.5 1.2649
  3.0 1.1547
  From equation (1):
  ∫ 3
  1
  2
  √
  x
  dx≈( 0. 5 )
  {
  1
  2
  ( 2. 0000 + 1. 1547 )

  
  


• 
  
  
  
  1. 6330 + 1. 4142 + 1. 2649
     }
     =2.945,correct to 3 decimal places
     This problemdemonstrates that even withjust 4 inter-
     vals a close approximation to the true value of 2.928
     (correct to 3 decimal places) is obtained using the
     trapezoidal rule.
     Problem 2. Use the trapezoidal rule with 8
     intervals to evaluate,
     ∫ 3
     1
     2
     √
     x
     dxcorrect to 3
     decimal places.
     With 8 intervals, the width of each is
     3 − 1
     8
     i.e. 0.25
     giving ordinates at 1.00, 1.25, 1.50, 1.75, 2.00, 2.25,
     2.50, 2.75 and 3.00. Corresponding values of
     2
     √
     x
     are
     showninthetablebelow.
     x
     2
     √
     x
     1.00 2.0000
     1.25 1.7889
     1.50 1.6330
     1.75 1.5119
     2.00 1.4142
     2.25 1.3333
     2.50 1.2649
     2.75 1.2060
     3.00 1.1547
     From equation (1):
     ∫ 3
     1
     2
     √
     x
     dx≈( 0. 25 )
     {
     1
     2
     ( 2. 000 + 1. 1547 )+ 1. 7889
  
  
  
  


• 
  
  
  
  1. 6330 + 1. 5119 + 1. 4142
  
  
  
  


• 
  
  
  
  1. 3333 + 1. 2649 + 1. 2060
     }
     =2.932,correct to 3 decimal places.
     This problem demonstrates that the greater the number
     of intervals chosen (i.e. the smaller the interval width)
     the more accurate will be the value of the definite inte-
     gral. The exact value is found when the number of
     intervalsis infinite, whichis, ofcourse, what the process
     of integration is based upon.
     Problem 3. Use the trapezoidal rule to evaluate
     ∫ π
     2
     0
     1
     1 +sinx
     dxusing 6 intervals. Give the answer
     correct to 4 significant figures.