Page 57 - DMTH202_BASIC_MATHEMATICS_II

Basic Mathematics-II

Notes
1

4.  e 2 3x dx
0

f ( )  e 2 3x ,a  0,b  1 and nh b a  1 0  1
x



a
f ( )  f (0)  e 2


( f a h  f ( )  e 2 3h  e 2 .e  3h
)
h

h
h
( f a  2 )  f (2 )  e 2 3h  e 2 .e  3h

h
h
( f a  3 )  f (3 )  e 2 9h  e 2 .e  9h
:
:

h


h


( f a n  1 ) f (1 n  1 ) e 2 3(n 1)h  e 2 .e  ( 3 n 1)h
b
h
f
Now,  f ( )dx  lim [ ( ) f (a h ) f (a  2 ) f (a  3 )  f (a n  h
h
h
1 )]
x
a





h 0
a
1

f
  e 2 3x dx  lim [ (0)  f ( )  f (2 )  f (3 )   ( f a n  1 )]
h
h
h

h
h
0 h 0
2
2
 lim [e  e 2 .e  3h  e 2 .e  6h  e 2 .e  9h   e  e  3(n 1)h ]
h
h 0
e 2 (e  3nh  1) h
 lim h  lim e  2 (e  3nh  1)
h 0 e  3nh  1 h 0 e  3h  1
 3h  1 
3


 lim h    e 2(e  1) [ nh  1]
h 0 e  3h  1   3 
1 1 1  1 
3
1 2
2


 e 2 (e  1) (e e )   e  
3 3 3  e 
Example:
Evaluate the following definite integrals as limit of sums:
4 2
x
x
1.  2 x 2.  5 x
2  1
Solution:
4
1.  2 x
2
x
f ( ) 2 ,a  2,b  4 and nh b a  4 2  2

x



2
a
f ( )  f (2)  e  4
h
2



( f a h  f (2 h  e 2 3h  2 .2  4.2 h
)
)
2

h

h

( f a  2 )  f (2 2 ) 2 2 2h  2 .2 2h  4.2 2h

h
h

( f a  3 )  f (2 3 ) 2 2 3h  22.2 3h  4.2 3h

:
:
2




h
( f a n  1 ) f (2 n  1 ) e 2 (n 1)h  2 .2 (n 1)h  4.2 (n 1)h
h

b
f
Now,  f ( )dx  lim [ ( ) f (a h ) f (a  2 ) f (a  3 )  f (a n  1 )]
h
h
x
h
a
h





h 0
a
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