Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

yes

Answer 5

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

Answer 6

omg i hate calculus lol

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

but it shows me another idea lol

Answer 12

yea im not so good with substitution either

Answer 13

u = 9x-x^3 ; du = 9-3x^2 dx


{S} sqrt(1+(u/3)^2) du
--------------
9-x^2

maybe?

Answer 14

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

Answer 15

ok im just trying to follow the process here

Answer 16

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

Answer 17

sure i couldnt go anymore wrong with this problem lol

Answer 18

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

Answer 19

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

Answer 20

now we can factor under the radical to simply fy things

Answer 21

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

Answer 22

i honestly have no idea lol

Answer 23

ima try some synthetic stuff and hope i get lucky

Answer 24

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

Answer 25

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

Answer 26

whos that? some really smart kid im guessing :)

Answer 27

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

Answer 28

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

Answer 29

wolframalpha.com

Answer 30

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

Answer 31

you had the wrong stuff inputed lol

Answer 32

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

Answer 33

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

Answer 34

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

Answer 35

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

Answer 36

dy/dx = -2x

Answer 37

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

Answer 38

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

Answer 39

integrals

Answer 40

yea that what we do when we use integrals

Answer 41

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

Answer 42

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

Answer 43

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

Answer 44

easier than you would that other monstrocity lol

Answer 45

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

Answer 46

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

sec(u). sec^2(u) du
----------------
4


sec^3(u) du
----------
4

Answer 47

got that outta whack a little

Answer 48

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

Answer 49

http://www.wolframalpha.com/input/?i=int%28+sqrt%28+1%2B+%28-2x%29^2%29%29+dx+from+-3+to+3