Page 52 - DMTH202_BASIC_MATHEMATICS_II

Unit 4: Definite Integral

 lim h [2n h (1 2 3   (n  1)] Notes



h 0
)

 ( n n  1)  nh (nh h 
 lim h 2n h  lim h 2nh 

h 0   2   h 0   2  



 lim h [4 (2 h )] [ nh  2]
h 0
 4 2  6.

Example:
Evaluate the following definite integrals as limit of sums:
4 3 4
2
2
2
x
1.  (x  ) x dx 2.  (2x  5 )dx 3.  (2x  3x  1)dx
1 1 2
Solution:
4
2
1.  (x  ) x dx
1
2

x
f ( )  x  , x a  1,b  4andnh  b a  4 1  3

a
f ( )  f (1) 0

2
2
)
)



( f a h  f (1 h ) (1 h  (1 h ) 1 h 2 2h  1 h  h  h






2
2
2
2
2
h
h
h
h

( f a  2 )  f (1 2 ) (1 2 )  (1 2 ) 1 2 h  4h  1 2h  2 h  2h






2
2
2
2
2
h


h

h

h

( f a  3 )  f 91  3 ) (1 3 )  (1 3 ) 1 3 h  6h  1 3h  3 h  3h

:
:
2
h
h
h
h


( f a n  1 )  f (1 n  1 ) (1 n  1 )  (1 n  1 )



2
2
2
2
 1 (n  1) h  2(n  1)h  1 (n  1)h  (n  1) h  (n  1)h


b
x
a
h
f
h
)
h
Now, f ( )dx  lim h [ ( )  ( f a h  ( f a  2 )  ( f a  3 )   ( f a n  1 )]



h 0
a
2
h
)
4 (x  ) x dx  lim [ (1)  f (1 h  f (1 2 )  f (1 3 )

h
f

h

  h 0
h

1   f (1 n  1 )]
2
2
2
2
2

i.  lim [0 h   2 h  2 h  3h
h
h
h 0
2
2
  (n  1) h  (n  1) ]
h
2
2
2
2
2
h
h


 lim [ (1  2  3   (n  1)  h (1 2 3   (n  1))]
h 0
 2 (n n  1)(2n  1) ( n n  1)
 lim h h  h 

h 0  6 2 



 nh (nh h )2nh h nh (nh h 
 lim 
h 0   6 2  



)
 3(3 h )(6 h ) 3(3 h 

 lim  [ nh  3]
h 0   6 2  
3(3)(6) 3(3) 9 27
   9  
6 2 2 2
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