Page 54 - DMTH202_BASIC_MATHEMATICS_II

Unit 4: Definite Integral

2
h
h
h
h
( f a n  1 )  f (2 n  1 )  2(2 n 1 )  3(2 n 1 ) 1 Notes





2


 2(4 2(n  1)h  (n  1) h 2 ) 6 3(n 1)h 1

2

 15 (n  1).11h  (n  1) .2h 2
b
)
a
h
x
h
h

f
Now,  f ( )dx  lim [ ( )  ( f a h  ( f a  2 )  ( f a  3 )
a h 0
  ( f a n  1 )]

h
4
2
  (2x  3x  1)dx
2
h
 lim [ (2)  f (2 h  f (2 2 )  f (2 3 )  f (2 n  1 )]
h
f



)
h
h

h 0
2
2
2
2



h
 lim [15 15 11h  2h  15 2.11h  2 .2h  15 3.11h  3 .2h 2

h 0
  15 (n  1)11h  (n  1) .2h 2 ]
2

2
2
2
2
2


h
h

h
 lim [15n  11 (1 2 3   (n  1)) 2 (1  2  3   (n  1) )]
h 0
 ( n n  1) 2 (n n  1)(2n  1)
 lim h 15n  11h  2h
h 0   2 6  
)


 11 nh (nh h )(2nh h 
 lim h 15nh  nh (nh h 

)
h 0   2 3  


)
 2(2 h )(4 h  16 172
 lim 30 11(2 h   30 22  
)



h 0   3   3 3
Example:
2
3
Evaluate as limit of sums  x  1dx
1
Solution:
2
3
 x  1dx
1
3


f ( )  x  1,a  1,b  2andnh  b a  2 1 1

x
f ( )  f (1)  2
a
2
3
3
2




)
)



( f a h  f (1 h ) (1 h  1 1 3h  3h  h  1  2 3h  3h  h 3
3
2
3
3
2
h


h

( f a  2 )  f (1 2 ) (1 2 )  1 1 2.3h  2 .3h  2 h  1


h
2
2
3

 2 2.3h  2 .3h  2 h 3
2
2
2
3
3
h
h
h





( f a  3 )  f (1 3 ) (1 3 )  1 1 3.3h  3 .3h  3 .h  1
2
3
2

 2 3.3h  3 .3h  3 h 3
:
:
3
3
2
2
3
h
h
h



( f a n  1 )  f (1 n  1 ) (1 n 1 )  1 1 (n 1)3h (n 1) .3h  (n 1) h  1



3
2
2
 2 (n  1).3h  (n  1) .3h  (n 1) .h 3

b
x
f
h
h
a
Now,  f ( )dx  lim [ ( )  ( f a h  ( f a  2 )  ( f a  3 )  ( f a n  1 )]

h

)
h
a h 0
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