help me with this AP calc test problem!

[Jittles]

I'm preparing for the AP test on teusday and we have a bunch of take home practice tests but I don't get this problem.

if F(x) is the antiderivative of (ln x)^3 / x and F(0) = 1 then F(9) = ?


I have no clue. please help all you geniuses!

 

[StinkyMeat]

NOOOOO!!! I thought I got rid of you Calculus! Back! Back to the fires of hell!!!

 

[Rallispec]

if it wasnt 3am i'd help... right now my brain is drained and i'd probable do more harm than help. hopefully someone here can help though.


good luck on the AP test. it really isnt that bad. i think i got a 3/4 on it 4 years ago

 

[Jittles]

Damn, two responses in like 3 min, I was about to say DAMN you guys are quick

 

[gplanet]

i know how to do this give me 3 mins

 

[gplanet]

integral: (ln x)^3 / x

u = ln x
du = 1/x

integral: u^3 du

answer: (u^4)/4 + C
= (ln x)^4 / 4 + C

1 = F(0) = (ln 0)^4 / 4 + C
C = 1

Final Integral Answer: F(x) = [(ln x)^4 / 4] + 1

F(9) = [(ln 9)^4 / 4) + 1]

 

[slikmunks]

take the integral of lnx^3/x, plug in 0 to make sure you've got the rite answer, then plug in 9.... integral of lnx is xlnx+x or something, so i think it's like just chain rule, so just substitute U for lnx, and dU for the 1/x... then u just gotta integrate u^3, and substitute the lnx back in and plug 'n chug.... it's high school calc, that should be it....

 

[slikmunks]

yeah... like gplanet said... that's what i was talking about... except he did it for you... you're supposed to teach him how to do it, so he can do it on the test! hehehe, but then again, if i were asking this late at night, i'd be like screw it, gimme the answer and i'll try to figure it out tomorrow

 

[Jittles]

GAH!

I'm retarded, I totally didn't even think if u substitution! thanks!


Oh yeah and one more thing, the graph of |x|. Is that continuous @ x = 0, differentiable @ x = 0, ?

 

[gplanet]

it would be continuous but not differentiable i think

 

[SUBxWRX]

just wondering.......how would you get (ln 0)^ 4 ?

 

[Mday]

<< GAH!

I'm retarded, I totally didn't even think if u substitution! thanks!


Oh yeah and one more thing, the graph of |x|. Is that continuous @ x = 0, differentiable @ x = 0, ? >>



it's continuous, both limits (left and right) is 0.

it's not differentiable because the slopes are different to the left and right of 0. (epsilon away).

lim x-> 0 of ln(x) = -infinity....