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General Forums [Trash Bin] Come on peeps get with the name. :P It's basically a place to move all the dead topics over here... However, To keep people's posts around. The topics here are not deleted. o.o
View Poll Results: see below
ln(3) + ln(5) 4 33.33%
2*ln(3) + ln(5) 5 41.67%
ln(3) + 2*ln(5) 2 16.67%
ln(5) - ln(3) 1 8.33%
2*ln(5) - ln(3) 0 0%
Voters: 12. You may not vote on this poll

 
 
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  (#11 (permalink)) Old
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Default 04-04-2008

YEAH!! only one day... tomorrow


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Default 04-05-2008

The answer should be B


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Default 04-05-2008

We have only a few more hours to see who got it right sweet


I HAVE SEEN DEATH AND IT IS ME


^sig by blackmamba thank you ^_^

   
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Default 04-05-2008

why do I always see your quizez too late...now I dont have any time to solve it...


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I will be checking to see later tonight if I am right


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Default 04-05-2008

yay today it ends the waiting


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  (#17 (permalink)) Old
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Default 04-05-2008

just minutes left
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Cool The Answer - 04-05-2008

Quote:
Use partial fractions to evaluate the following.
5
⌠ ( 4x - 9 ) * dx
⌡ 2(x^2) - 9x + 10
3
The first step for evaluating the integral is to break up the fraction so that the denominator is simpler.

The denominator can be factored into (2x-5)(x-2) so …

5
⌠ ( 4x - 9 ) * dx
⌡ (2x-5)(x-2)
3

This fraction can now be broken in half…

A/(2x-5) + B/(x-2)

A and B have taken the place of the numerator, because we do not know what their values are. To find their values we set the original fraction equal to A/(2x-5) + B/(x-2).

( 4x - 9 )/((2x-5)(x-2)) = A/(2x-5) + B/(x-2)

4x-9 = A(x-2) + B(2x-5)

Then you substitute in numbers for x, because this equation should be true for all numbers. So…

If x =2
-1 = -1*B
B=1

If x = 5/2
1 = ˝ *A
A = 2

Now we return to our split-up fraction and substitute them in
5
∫ (A/(2x-5) + B/(x-2))*dx =
3
5
∫ (2/(2x-5) + 3/(x-2))*dx
3

Now we can proceed with taking the integral…

5
∫ (2/(2x-5) + 3/(x-2))*dx =
3

We will solve this in two parts
5
∫ (2/(2x-5))*dx
3

u = 2x-5
du = 2dx
dx = du/2
so we substitute in u for 2x-5 and du/2 for dx …

∫ (2/u)*du/2 =

∫ (1/u)*du = ln(u)

We then substitute the 2x-5 back in…

ln (2x-5) from 3 to 5 =
ln (10-5) – ln (6-5) = ln(5) – ln(1)

now we solve the other half…

∫dx/(x-2)

u = x-2
du = dx

∫du/u = ln(u)

ln(u) = ln(x-2)

ln(x-2) from 3 to 5 = ln(5-2) – ln(3-2) = ln(3) – ln(1)

now we add the two halves together

ln(5) – ln(1) + ln(3) – ln(1)

at some point you need to remember that ln(1) = 0

so you get ………. Answer choice A) ln(5)+ln(3)
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Default 04-05-2008

congratulations to grandLuffy, Kobushi, Rukia_Kuchiki18, Thrax for getting the correct answer.
we had a total of 12 participants and 33.3% of participants answered correctly.

new quiz will be posted tomorrow
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Default 04-05-2008

DANG, NO WONDER I DIDN'T MAJOR IN MATH


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