Question

Hi, how could we solve this integration?
∫(dy/dx)^2 dx

Answer 1

\[ \Large \int \frac{d^2y}{dx^2} \cdot dx \] ?

Answer 2

http://www.wolframalpha.com/input/?i=integrate+%28dy%2Fdx%29^2+with+respect+to+x
i would try if i know there is solution.

Answer 3

seems like an awkward notation to me in general.

\[(f'(x))^2dx = f'(x)\cdot f'(x)dx \]
Not that it doesn't exist, just haven't encountered it yet.

Answer 4

I don't think you can simplify this as it is without more info about the function

Answer 5

isn't that equal to \[\frac{dy}{dx}+C\]

Answer 6

according to wolf ...
http://www3.wolframalpha.com/Calculate/MSP/MSP49031a2ebcc22c7ig5if0000371e29592cf2a32e?MSPStoreType=image/gif&s=43&w=1773&h=35

Answer 7

how would the exponent get lower @mukushla ?

Answer 8

that's exactly what I would have derived next @experimentX q-:

Answer 9

but I am really curious about the statement of the integral in general.

Answer 10

lol....misread

Answer 11

just expand the integrand wrt. Taylor series and integrate ... perhaps.

Answer 12

i dont think they are equal @experimentX:
\[\int\limits \left ( \frac{dy}{dx} \right )^2dx \neq \int\limits \left ( \frac{\partial y}{\partial x} \right )^2dx\] are they?

Answer 13

it depends, like I said we need to know more about the function

Answer 14

There is no closed solution for your integral.

Answer 15

they could be equal I think, if the function is as such...

Answer 16

That is what I think as well @eliassaab
though I am having fun trying to integrate it by parts
funny things are happening :)

Answer 17

@TuringTest me too...parts

Answer 18

yep ... so i scrupulously asked wolf.

Answer 19

great minds think alike @mukushla ;)

Answer 20

Do not waste your time. If Mathematica cannot do it, then it very probable that it is not feasible.

Answer 21

its actually this one:
\[\int\limits\limits v^2dt\]

Answer 22

It may not be a waste of time trying something futile if you can learn exactly *why* it is futile I would say

Answer 23

what is v?

Answer 24

well what is \(v\) ?

Answer 25

velocity

Answer 26

\[v=\frac{dx}{dt}\]
in physics

Answer 27

then we are back to where we started

Answer 28

x is some function of time right?

Answer 29

yes

Answer 30

should be if it's moving

Answer 31

do you know that function?

Answer 32

or even if it's not lol

Answer 33

BTW why would you want to evaluate v^2?

Answer 34

its the orginal problem:

\[F=C\int\limits v^2dt\]
where F is the force as a function of V and c is a constant

Answer 35

can you state the original problem ... full, maybe ,,, we are missing something,

Answer 36

oh! it would be so hard to explain!

Answer 37

just try ... maybe we 'could' find the solution.

Answer 38

so we could do something:
\[dF=Cv^2dt \rightarrow \frac{dF}{dt}=m \frac{da}{dt}=m\frac{d^3x}{dt^3}=C \left( \frac{dx}{dt} \right)^2\]I think its not an integration anymore! it looks like a differential equation! can anyone solve this?
\[\frac{d^3x}{dt^3}-\frac{C}{m} \left( \frac{dx}{dt} \right)^2=0\]or maybe i should ask it in an other question?

Answer 39

non linear DE ...

Answer 40

\[x'''-\frac Cm(x')^2=0\]let\[x'=y\]\[y''-\frac Cmy^2=0\implies y''=\frac Cmy^2\]I think we can solve that with undetermined coefficients