Page 51 - DMTH202_BASIC_MATHEMATICS_II

Basic Mathematics-II

Notes
2
1.  (2x  3)dx
1
2

f ( ) 2x  3,a  1,b  and nh  b a  2 1 1
x



a
f ( )  f (1) 5




( f a h  f (1 h ) 2(1 h ) 3  5 2h

)


h
( f a  2 )  f (1 2 ) 2(1 2 ) 3  5 2.2h
h




h

h
h
h

( f a  3 )  f (1  3 ) 2(1 3 ) 3  5 3.2h



:
:
h



h


( f a n  1 )  f (1 n  1 ) 2(1 n  1 ) 3  5 n  1.2h
h

b
h



h
x
a
)
Now , f ( )dx  lim h [ ( )  ( f a h  ( f a  2 )  ( f a  3 )   ( f a n  1 )]
f
h
a h 0
2
)
f
h
h
h



  (2x  3)dx lim h [ (1)  f (1 h  ( f a  2 ) f (1 3 )  f (1 n 1 )

h 0
1


h

h

 limh [5 5 2 .h  5 2.2h  5 3.2h   5 n  12 ]

h 0
h


 lim h [5n  2 (1 2 3   (n  1)]
h 0
 ( n n  1)
 lim h 5n  2h 
h 0   2 



 lim h [5n nh (n  1)] lim [5nh nh (nh h )]

h 0 h 0


 lim h [5 (1 h )] [ nh 1]

h 0


 5 1 6
1
2.  (x  3)dx
 1
1

x


f ( ) x  3,a  1,b  and nh  b a  1 1  2

a

f ( )  f ( 1) 2

3
( f a h  f ( 1 h   1 h   2 h


)

)

h
3
h
( f a  2 )  f ( 1 2 )   1 2h   2 2h




h
( f a  3 )  f ( 1  3 )   1 3h   2 3h


3
h

:
:

h
h
3



( f a n  1 )  f ( 1 n  1 )   1 n  1h   2 (n  1)h

b
h
Now , f ( )dx  lim h [ ( )  ( f a h  ( f a  2 )  ( f a  3 )   ( f a n  1 )]
x

a
h
f

)
h

a h 0
1






h

h

f
h


 (x  3)dx  limh [ ( 1)  f ( 1 h  f ( 1 2 ) ( 1 3 )   f ( 1 n  1 )]
)
 1 h 0
h

 limh [2 2 h  2 2h  2 3h   2 (n  1) ]




h 0
h

 lim h [2n h  2h  3h   (n  1) ]
h 0
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