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![[Post New]](/templates/default/images/icon_minipost_new.gif) 13 Mar 2008 22:09:27 IST
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I1 = 1 + cos2 t  x .f { x (2-x) } dx sin2 t I2 = 1 + cos2 t f { x (2-x) } dx sin2 t find I1/I2
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<TABLE CELLSPACING="1" CELLPADDING="1" BORDER="0">
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<DIV ALIGN="right">Glitter Graphics</DIV></TD></TR></TABLE>
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13 Mar 2008 22:17:26 IST
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is the answer 2???????????
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13 Mar 2008 22:27:15 IST
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i think its 1
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<TABLE CELLSPACING="1" CELLPADDING="1" BORDER="0">
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<DIV ALIGN="right">Glitter Graphics</DIV></TD></TR></TABLE>
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13 Mar 2008 22:29:13 IST
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ya got it.its 1.
see apply the law :integral from a to b f(x) = integral from a to b f(a+b-x) in I1.it will cum equal to I2
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13 Mar 2008 22:35:06 IST
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yeah.. you'll get
I1 = 2I2 - I1
after applying f(x) = f(a+b-x)
 I1 = I2
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13 Mar 2008 22:42:05 IST
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Will nip in at times to solve problems :)
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13 Mar 2008 22:43:29 IST
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edit - the term inside the I2 integral must be f(2-x) and not x-2, its a typo.
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Will nip in at times to solve problems :)
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