1ans is not none dear :)
if
I = 0∫1 dx/ [(5+2x-2x2)(1+ e2-4x)]
and,
I1 = 0∫1 dx/ (5+2x-2x2) ,
then...
a) I = I1
b) I = 2 I1
c) 3I = 5 I1
d) not.
-
UP 0 DOWN 0 0 11
11 Answers
1this sum can be done by using the properties of integration
let i be the given integration
the value of I does'nt change when x is replaced by( 0+1-x)
0∫1dx/[(5+2x-2x2)(1+e(2-4x))]
so this is equal to
0∫1 dx/[(5+2x-2x2)(1+e-(2-4x))]
so
I=0∫1e(2-4x)dx/[(5+2x-2x2)(1+e(2-4x))]
by adding both
2I=I1
so the answer is none
11REplace x by 1-x in I......
Then add original I and this I..........
This will eliminate enonsense from the denominator.....
I think so [12]
Didn't do it as yet......... or in the words of our Minister.......Tukka lagaya [3]
1chazi u did a chhhhhhhhota sa calculation mistake ...
or else wonderful..
but u cud have got a negative marking..
its [a].