20th January 2009

integration mein ek point se koi farak nahin padta if that point is an end point

and do remember this also

IF A FUNCTION CONTAIN POINT DISCONTINUITY THEN ITS AREA REMAINS THE SAME

could some one help me out in this question[17][17]

hmm yes i told that in post#4 itself

no one saw [17]

SIR PLEASE CHECK #14,17,19 AND PLEASE DO EXPLAIN WITH SOME CLARITY WHAT KAYMANT SIR MEANT[7]

IF U R CONTRADICTING SKYGIRL THEN SHE REPLACED X WITH 3-X
SO THE SHE IS TAKING IT AS END POINT

4 MORE INFORMATION REFER TO #14 1ST POINT[1]

HERE THE DISCONTINUITY IS NOT A POINT DISCONTINUITY
THERE THE DISCONTINUITY IS NON REMOVABLE DISCONTINUITY AS AT X=∩/2 TAN X MOVES UP TO ∞ AND FROM LATER IT STARTS FROM INFINITY
U CAN CHECK FROM ITS GRAPH

wat abt integral tanx from 0 to pi

TAKE THE EXAMPLE OF THIS GIVEN CASE
ITS AREA FROM 0 TO P( WHERE THE CURVE CUTS +VE X AXIS) WILL REMAIN SAME WHETHER THE DISCONTINUITY AT X₀ IS PRESENT OR NOT

@ DEEPANSHU

PLEASE DONT TELL US WHAT U THINK

GIVE US THE FULL FLEDGE SOLUTION TO THE WHOLE PROBLEM

so finally,

2I = 1/4 ₀∫³ dx/ [(x-3/2)² - (1/√2)²]

bus [1]

neglecting fallacy ans cums to be

ln ( 3 -√2/3 +√2) /4√2

this one is exactly same as my doubt a few days ago....

jus replace x by (3-x) .

then add.

we will get ...

2I = ₀∫³ dx/(4x² - 12x + 7)