434 INTEGRAL CALCULUS

(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 bet-
  ween the upper and lower limits, i.e. 3− 1 =2.
  Using the trapezoidal rule with 4 intervals gives
  an interval widthd=
  3 − 1
  4
  = 0 .5 and ordinates
  situated at 1.0, 1.5, 2.0, 2.5 and 3.0. Corre-
  sponding 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 problem demonstrates that even with just 4
     intervals 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 shown in the table below.
     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 num-
     ber of intervals chosen (i.e. the smaller the interval
     width) the more accurate will be the value of the
     definite integral. The exact value is found when the
     number of intervals is infinite, which is, of course,
     what the process of integration is based upon.
     Problem 3. Use the trapezoidal rule to evalu-
     ate
     ∫ π
     2
     0
     1
     1 +sinx
     dxusing 6 intervals. Give the
     answer correct to 4 significant figures.
     With 6 intervals, each will have a width of
     π
     2
     − 0
     6
     i.e.
     π
     12
     rad (or 15◦) and the ordinates occur at
     0,
     π
     12
     ,
     π
     6
     ,
     π
     4
     ,
     π
     3
     ,
     5 π
     12
     and
     π
     2