Page 69 - DMTH202_BASIC_MATHEMATICS_II

Basic Mathematics-II

Notes 4

(2x  1) 5/2 
 {60  6 5}  
5
.2 
2 2 
 9 5/2 5 5/2  243 57
 {60  6 5}      60  6 5   5 5   5
 5 5  5 5
e x  1 x logx 

7.  e   dx
1  x 
e 2 1 e 2 1
  .1dx   2 dx
x
2 log x 2 (log )
e 2 2 2
1  e 1 1 e 1
 .x    . .xdx   dx
2
x
x
log x e  2 (log ) x 2 (log ) 2
  e 2 e   e 2 1 e 2 1

  2    dx  e  2 dx
x
x
  log e log e   e (log ) (log )
e 2 e 2
  e   . e
2log e 2
b
3
Example: If  x dx  0
a
b 4 4
4
x  b a
   0    0
4  4 4
a
4
4
a
 b  a    b
b 2
2
Also  x dx 
a 3
b
3
x  2 b 3 a 3 2
3
3
       b  a  2
3  3 3 3 3
a
When a = b, then
3
3
a
a – a = 2 which is absurd    b
3
3
3
3
a
2
  a  a  2   2a   a   1    1
thus b = 1
Hence, a = – 1 and b = 1
k dx 
Example: If  2  , Find the value of k.

0 2 8x 16
1 k dx 
  
8 1 2 16
0  x
4
1 k dx 
  2 
8 0   2 2
1
   x
 2 
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