Page 58 - DMTH202_BASIC_MATHEMATICS_II

Unit 4: Definite Integral

4 Notes
x
  2 dx  lim [ (2)  f (2 h  f (2 2 )  f (2 3 )   f (2 n  1 )]


)
h
h

h
h
f

2 h 0
3h
h
h

 lim [4 4.2  4.2 2h  4.2   4.2 (n 1)h ]
h 0
nh
4(2  1) h
nh
 lim h  lim 4(2  1)
h
h
h 0 2  1 h 0 2  1
h
2
 lim h 4 (2  1)
h
h 0 2  1
x
1  a  1 
 4(3)   lim  log a 
log 2  h 0 x 
12

log 2
2
x
2.  5 x
 1
x
x
f ( ) 5 ,a   1,b  2 and nh b a  2 ( 1) 3






a
f ( )  f ( 1)  5  1


1

)
( f a h  f ( 1 h  e  1 h  5 .5 h

)


1


( f a  2 )  f ( 1 2 )  5  1 2h  5 .5 2h

h
h


1 3h

h
h

( f a  3 )  f ( 1 3 )  5  1 3h  5 5

:
:

1



h
( f a n  1 ) f ( 1 n  1 ) 5  1 (n 1)h  5 5 (n 1)h
h



b
)
a
f
x

h
h

h
Now,  f ( )dx  lim [ ( )  ( f a h  ( f a  2 )  ( f a  3 )  ( f a n  1 )]
h
a h 0
2
x


h



h
h



)
  5 dx  lim [ ( 1)  f ( 1 h  f ( 1 2 )  f ( 1 3 )   f ( 1 n  1 )]

h
f
h 0
 1
1
2h
3h
1
1
1 h
1




h
 lim [5  5 5  5 .5  5 .5   5 5 (n 1)h ]
h 0
nh
1

5 (5  1) h

3
1
 lim h  lim 5 (5  1)
h
h
h 0 5  1 h 0 5  1
h
2
 lim h 4 (2  1)
h
h 0 2  1
1  1  124
2
  5   
log 5  5  5log 5
Example:
Evaluate the following definite integrals as limit of sums:
b  /4  2 2
1.  sinxdx 2.  cosxdx 3.  sin xdx
a 0  6
Solution:
b
1.  sinxdx
a
f(x) = sin x and nh = b – a
f(a) = sina
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