Page 55 - DMTH202_BASIC_MATHEMATICS_II

Basic Mathematics-II

Notes 2
3

h
h
f


h
h
)

  (x  1)dx  lim [ (1)  f (1 h  f (1 2 )  f (1 3 )  f (1 n  1 )]
1 h 0
3
2
2
2
3
2
2
3
3
h




 lim [2 2 3h  3h  h  2 2.3h  2 .3h  2 .h  2 3.3h  3 .3h  3 h 3
h 0
  2 (n  1)3h  (n  1)  3h  (n  1) h 3 ]
2
3
2

2
2
2
2
h
h

h



 lim [2n  3 (1 2 3   (n  1)) 3 (1  2  3  (n  1) 2
h 0
3
3
3
3
 h 3 (1  2  3 ).  (n  1) ]
 ( n n  1) 2 (n n  1)(2n  1) 3 (n n  1)
 lim h  2n  3h  3h  h 
h 0  2 6 2 

 lim h  2nh  3 nh (nh h )(2nh h  1 [nh (nh h )] 2 

)

h 0 2 4
 3 1 1 2 
)
)

 lim 2  (1 h  (1 h )(2 h  (1 h )



h 0   2 2 4  
2 1 2 1 47
 2   1   3   
3 4 3 4 12
Example: Evaluate the following definite integrals as limit of sums:
5 1
2x
x
1.  e dx 2.  e dx
2  1
3 1
x


3.  e dx 4.  e 2 3x
1 0
Solution:
5
x
1.  e dx
2
x

f ( )  ex ,a  2,b  5andnh b a  2  3

f ( )  f (2)  e 2
a

)
)

( f a h  f (2 h  e 2 h  e 2 .e 2h


h
h
( f a  2 )  f (2 2 )  e 2 2h  e 2 .e 2h


h
( f a  3 )  f (2 3 )  e 2 3h  e 2 .e 3h
h

:
:

h
h
n



( f a n  1 ) f (2   1 ) e 2 (n 1) h  e 2 .e (n 1)h
b
h
h
h
f
h
Now,  f ( )dx  lim [ ( )  ( f a h  ( f a  2 )  ( f a  3 )  ( f a n  1 )]

a

)
x
h 0
a
3
2x
3h
2h
2
h
h
 e dx  lim [e  e 2 .e  e 2 .e  e 2 .e   e 2 .e (n 1)h ]
2 h 0
nh
n
e 2 (e  1)  ( a r  1)
2

 lim h   a ar  ar   ar n 1  
h
h 0 e  1  r  1 
h 2 3

e
 lim . (e  1) [ nh  3]
h
h 0 e  1
x
 e  1 
3
5
2
 e (e  1)  e  e 2  Using lim  1 
 x 0 x 
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