anti-derivative of SQRT(1+(9x- (x^3/3))^2)

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anti-derivative of SQRT(1+(9x- (x^3/3))^2)

Mathematics
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how you wanna do this lol the easy way or the hard way
lets go with easy because this is the part of problems i can never do
\[\sqrt{1+(9x-\frac{x^3}{3})^2}\] lets start by making sure we got the right problem is this it?

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yes
well; its gonna be a trig sub that turns to int(sec(x)) dx inthe end lol
omg i hate calculus lol
lol...trig sub just cleans it up thats all
yea but i integrated this on my calculator and i still got the wrong answer
\[\int_{} \sqrt{1+tan^2(u)}du\] ;\[u= \frac{27x -x^3}{3}\]
du = 27-3x^2/3 = 9-x^2 dx; maybe that aint a good idea lol
but it shows me another idea lol
yea im not so good with substitution either
u = 9x-x^3 ; du = 9-3x^2 dx {S} sqrt(1+(u/3)^2) du -------------- 9-x^2 maybe?
let me stare blindly at it for a few moments and let my thoughts congeal lol
ok im just trying to follow the process here
we might be better of expanding and combining everything under the radical into one shot
sure i couldnt go anymore wrong with this problem lol
\[\frac{\sqrt{x^6 + 729x^2 -54x^4 +9}}{3}\] is what i get with that ...maybe
yea i got something completely off but u r def prolly right lol
now we can factor under the radical to simply fy things
sooo...how should we factor it :) all polynomials higher then degrees 2 can factor to linears and or irreducibla quads
i honestly have no idea lol
ima try some synthetic stuff and hope i get lucky
lol ok im attempting to do other problems..im going to fail this quiz tomorrow lol
well, the good news is; wolfram couldnt even figure it out lol
whos that? some really smart kid im guessing :)
www.wolframalpa.com type in: int(sqrt(1+(9x-(x^3)/3)^2))dx
o gotcha lol..ok well maybe i went wrong somewhere..the question is find the length of the arch of the parabola y=9-x^2 that lies above the x-axis
wolframalpha.com
.... why you wanna keep secrets from me; i cant read your mind
you had the wrong stuff inputed lol
i did?? well the equation we use is SQRT(1+(dy/dx)^2
\[\int\limits_{} \sqrt{1 + [f'x]^2} dx\]
and is the derivative of 9-x^2 = to the integral of 9-x^2?
yea and the derivative of 9-x^2 is 9x-(x^3/3) right?
dy/dx = -2x
ooo i did the anti derivative didnt i ..i do that all the time because we keep having to switch from derivativeto anti derivative lol
anti derivative is a useless term; later chapters call them what they are inegrals lol
integrals
yea that what we do when we use integrals
{S} sqrt(1+4x^2) dx ; [a,b] whatever your interval s its up there some where
ok now i got it..wow i am retarded lol i do that a lot to
so now how to integrate SQRT(1+4x^2)
easier than you would that other monstrocity lol
i got 2/3(1+4x^2)^(3/2) and then do you also multiply that by the derivative of the inside??
sqrt(1+tan^2(u)) sec^2(u) du ------------------------- 4 sec(u). sec^2(u) du ---------------- 4 sec^3(u) du ---------- 4
got that outta whack a little
ok i g2g i got soccer..but thanks for the help :)
http://www.wolframalpha.com/input/?i=int%28+sqrt%28+1%2B+%28-2x%29^2%29%29+dx+from+-3+to+3

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