Question

integral of square root of (4x^4 +4x^2 +2x)dx

Answer 1

\[\int\limits_{0}^{4}\sqrt{4x^4 + 4x^2 +2x}dx\]

Answer 2

I don't think this is integrable, so I guess you better just use cramer's rule or something

Answer 3

turing test, that shows poor algebra skills on your part.

Answer 4

I mean Simpson's rule
you can integrate this? I'm lost....

Answer 5

4x^4+4x^2+2x = u^2+2u+u if u = 2x^2...which means...you can pull out (u+1) from the radical and now you're integrating something simple

Answer 6

i would draw this out but the drawing tool is too slow

Answer 7

but that last term would not be u....

Answer 8

if u=2x^2 you would need a 2x^2 at the end

Answer 9

oh yea, my mistake. it seemed too good to be true i guess

Answer 10

i would normally ask, are you sure you wrote the problem down correctly? lol..

Answer 11

wolfram just says "no" to the indefinite integral
http://www.wolframalpha.com/input/?i=integral+of+sqrt%284x%5E4%2B4x%5E2%2B2x%29dx

Answer 12

it will approximate the definite one with some method I'm sure....

Answer 13

pretty sure i did it right up to there. it asks for the arc length of the curve r(t)=<t^2cos2t, t^2, t^2sin2t>

Answer 14

\[0\le t\le 1\]?

Answer 15

0 to 4

Answer 16

ok let me do it and see what I get...

Answer 17

\[\|\vec r'(t)\|=\sqrt{[2t\cos(2t)+2t^2\sin(2t)]^2+[(2t)^2+2t\sin(2t)+t^2\cos(2t)]^2}\]that is a very ugly expression, I hope we can simplify it

Answer 18

typo above, forgot minus sign in sine
look out for more please\[\|\vec r'(t)\|=\sqrt{[2t\cos(2t)-2t^2\sin(2t)]^2+[(2t)^2+2t\sin(2t)+t^2\cos(2t)]^2}\]

Answer 19

nvm...i messed up the addition

Answer 20

yeah this should simplify with a little diligence

Answer 21

it's supposed to be 4x^4 + 4x^2

Answer 22

i mean 8x^2

Answer 23

yes that's what I got :)

Answer 24

\[\|\vec r'(t)\|=\sqrt{[2t\cos(2t)+2t^2\sin(2t)]^2+[(2t)^2+2t\sin(2t)+t^2\cos(2t)]^2}\]
\[=\sqrt{4x^4+8x^2}\]which should not be hard to integrate

Answer 25

so it's 4x^4+8x^2+2x ??

Answer 26

no just what I wrote

Answer 27

ok so what is the actual problem here

Answer 28

i love it when people post typos lol

Answer 29

try it, the simplification's a pain but it turns out to\[\int_0^4\sqrt{4t^4+8t^2}dt=\int_0^42t\sqrt{t^2+2}dt\]or x or whatever
and now it's easy breezy :)

Answer 30

yes. thank you. sorry for the typo

Answer 31

no problem, happy to help !