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

- anonymous

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

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- schrodinger

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- amistre64

how you wanna do this lol the easy way or the hard way

- anonymous

lets go with easy because this is the part of problems i can never do

- amistre64

\[\sqrt{1+(9x-\frac{x^3}{3})^2}\] lets start by making sure we got the right problem is this it?

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## More answers

- anonymous

yes

- amistre64

well; its gonna be a trig sub that turns to int(sec(x)) dx inthe end lol

- anonymous

omg i hate calculus lol

- amistre64

lol...trig sub just cleans it up thats all

- anonymous

yea but i integrated this on my calculator and i still got the wrong answer

- amistre64

\[\int_{} \sqrt{1+tan^2(u)}du\] ;\[u= \frac{27x -x^3}{3}\]

- amistre64

du = 27-3x^2/3 = 9-x^2 dx; maybe that aint a good idea lol

- amistre64

but it shows me another idea lol

- anonymous

yea im not so good with substitution either

- amistre64

u = 9x-x^3 ; du = 9-3x^2 dx
{S} sqrt(1+(u/3)^2) du
--------------
9-x^2
maybe?

- amistre64

let me stare blindly at it for a few moments and let my thoughts congeal lol

- anonymous

ok im just trying to follow the process here

- amistre64

we might be better of expanding and combining everything under the radical into one shot

- anonymous

sure i couldnt go anymore wrong with this problem lol

- amistre64

\[\frac{\sqrt{x^6 + 729x^2 -54x^4 +9}}{3}\] is what i get with that ...maybe

- anonymous

yea i got something completely off but u r def prolly right lol

- amistre64

now we can factor under the radical to simply fy things

- amistre64

sooo...how should we factor it :) all polynomials higher then degrees 2 can factor to linears and or irreducibla quads

- anonymous

i honestly have no idea lol

- amistre64

ima try some synthetic stuff and hope i get lucky

- anonymous

lol ok im attempting to do other problems..im going to fail this quiz tomorrow lol

- amistre64

well, the good news is; wolfram couldnt even figure it out lol

- anonymous

whos that? some really smart kid im guessing :)

- amistre64

www.wolframalpa.com type in:
int(sqrt(1+(9x-(x^3)/3)^2))dx

- anonymous

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

- amistre64

wolframalpha.com

- amistre64

.... why you wanna keep secrets from me; i cant read your mind

- amistre64

you had the wrong stuff inputed lol

- anonymous

i did?? well the equation we use is SQRT(1+(dy/dx)^2

- amistre64

\[\int\limits_{} \sqrt{1 + [f'x]^2} dx\]

- amistre64

and is the derivative of 9-x^2 = to the integral of 9-x^2?

- anonymous

yea and the derivative of 9-x^2 is 9x-(x^3/3) right?

- amistre64

dy/dx = -2x

- anonymous

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

- amistre64

anti derivative is a useless term; later chapters call them what they are inegrals lol

- amistre64

integrals

- anonymous

yea that what we do when we use integrals

- amistre64

{S} sqrt(1+4x^2) dx ; [a,b] whatever your interval s its up there some where

- anonymous

ok now i got it..wow i am retarded lol i do that a lot to

- anonymous

so now how to integrate SQRT(1+4x^2)

- amistre64

easier than you would that other monstrocity lol

- anonymous

i got 2/3(1+4x^2)^(3/2) and then do you also multiply that by the derivative of the inside??

- amistre64

sqrt(1+tan^2(u)) sec^2(u) du
-------------------------
4
sec(u). sec^2(u) du
----------------
4
sec^3(u) du
----------
4

- amistre64

got that outta whack a little

- anonymous

ok i g2g i got soccer..but thanks for the help :)

- amistre64

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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