Calculus 2 Help Needed

[dleccord 5,000+ posts]

How the heck can the K factor be 4 when sinx/cosx are bounded between -1 to 1?

 

[benchambers80]

Do this with steps:

Integral of dx/x(ln x)^2 a=2 b=4

 

[revrider1 5,000+ posts]

Id go crosseyed tryin that

 

[cj232 10+ year member]

Id go crosseyed tryin that

What he said.

 

[benchambers80]

Its actually really simple just missing a key factor.

Integral of dx/x is lnx

 

[benchambers80]

I did u substitution and was left with integral of dx/2u u=lnx^2

 

[▼)Ñ╩fZΣµ& 10+ year member]

join a math forum, and ask them this shit

whats pre algebra??? //content.invisioncic.com/y282845/emoticons/confused.gif.e820e0216602db4765798ac39d28caa9.gif

 

[GabeC 10+ year member]

i got 4 as well

 

[benchambers80]

some body gots to be edjumacated

dleccord come one mane

 

[cobra65427289 10+ year member]

to solve for K you need to take the molarity of the products multiplied by the quantity (RT) raised to the change in "n" divided by molarity of the reactants... er that's chemistry, never mind.

 

[dleccord 5,000+ posts]

To solve for the K factor, you would need to take the second derivative and absolute value that, which is sinx.

Since x is greater or equal to 0, and it is lesser or equal to pi, we can say the curve is bounded under 1.

Therefore, when x has to be pi/2. Plug that into the absolute value of the second derivative and you get 1.

Hence, K factor = 1.