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ducduyaktt · 08-31-2008, 05:08 PM

let f be a function that the graph of f is semicircle with endpoints (a,0) and (b,0) where a<b ( a, b are real numbers)

18. The improper integral :[ integral(from a to b) f(x)f'(x)dx ] is necessarily ZERO.

Could anyone please explain it for me? Thanks

And ONE more question :

which letter is not homomorphic to letter C? J, N, S, O, U

Thanks so much

D13E12 · 09-23-2008, 07:16 PM

Hi answer for Q no. 18 is by integrating the integrand f(x)f'(x)dx we get (f^2)(x)/2| with limits to substitute as a to b. so, the integral equal to ((f(b))^2-(f(a))^2)/2=0, because f(a)=0, f(b)=0.
note:-It's not an improper integral. integral is called improper only when one or both of the limits are infinity.
I do not the know answer to the second question. But if you come to know please do post it (with explanation).
Thanks.
cheers,
Siva Nagi

oldmathguy · 12-17-2008, 12:59 PM

The question must have been about "homeomorphic" not "homomorphic":
It's a topology question.
C is homeomorphic to N, S and U since all four of these letters can't
be straightened out to (are homeomorphic to) a line segment.

C is not homeomorphic to J because if we take away the top middle point
of the J we are left with three connected components, while this cannot
happen with any point of C. (Using: If X is homeomorphic to Y and x is
element of X and y is element of Y then X-{x} is homeomorphic to Y-{y}.)

C is not homeomorphic O because the latter has a loop in it that can't
be shrunk to a point in it. (Technically, they have different fundamental
groups -- not a concept that would be on the GRE.)

08-31-2008, 05:08 PM	#1 (permalink)
ducduyaktt Trying to make mom and pop proud Join Date: Mar 2006 Posts: 10	#18 Practice book let f be a function that the graph of f is semicircle with endpoints (a,0) and (b,0) where a<b ( a, b are real numbers) 18. The improper integral :[ integral(from a to b) f(x)f'(x)dx ] is necessarily ZERO. Could anyone please explain it for me? Thanks And ONE more question : which letter is not homomorphic to letter C? J, N, S, O, U Thanks so much Last edited by ducduyaktt : 08-31-2008 at 05:51 PM.

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09-23-2008, 07:16 PM	#2 (permalink)
D13E12 Trying to make mom and pop proud Join Date: Sep 2008 Posts: 1	Hi answer for Q no. 18 is by integrating the integrand f(x)f'(x)dx we get (f^2)(x)/2\| with limits to substitute as a to b. so, the integral equal to ((f(b))^2-(f(a))^2)/2=0, because f(a)=0, f(b)=0. note:-It's not an improper integral. integral is called improper only when one or both of the limits are infinity. I do not the know answer to the second question. But if you come to know please do post it (with explanation). Thanks. cheers, Siva Nagi

12-17-2008, 12:59 PM	#3 (permalink)
oldmathguy Trying to make mom and pop proud Join Date: Dec 2008 Posts: 5	The question must have been about "homeomorphic" not "homomorphic": It's a topology question. C is homeomorphic to N, S and U since all four of these letters can't be straightened out to (are homeomorphic to) a line segment. C is not homeomorphic to J because if we take away the top middle point of the J we are left with three connected components, while this cannot happen with any point of C. (Using: If X is homeomorphic to Y and x is element of X and y is element of Y then X-{x} is homeomorphic to Y-{y}.) C is not homeomorphic O because the latter has a loop in it that can't be shrunk to a point in it. (Technically, they have different fundamental groups -- not a concept that would be on the GRE.)

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