# Integration! (1 Viewer)

#### hscishard

##### Active Member
$\bg_white \frac{dy}{dx}= 2x$
$\bg_white \frac{dx}{dy}= 1/2x$
Therefore x = [log(2x)]/2 ??

The question that involved this
$\bg_white \frac{dr}{dt}= \frac{1}{2\pi r^2}$
$\bg_white \frac{dt}{dr}= 2\pi r^2$
$\bg_white t=\frac{2\pi r^3}{3} + C$
WTF is wrong with my example?

#### hscishard

##### Active Member
Is it because dy is on the bottom?

#### Tofuu

##### Member
$\bg_white \frac{dy}{dx}= 2x$
$\bg_white \frac{dx}{dy}= 1/2x$
Therefore x = [log(2x)]/2 ??

i don't think you can do that. You're integrating 1/2x in respect to y not x

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#### hscishard

##### Active Member
i don't think you can do that. You're integrating 1/2x in respect to y not x
But then how can you get a derivative of something in terms of itself? [2nd example]

#### kcqn93

##### Member
$\bg_white \frac{dy}{dx}= 2x$

therefore y = x^2 + C

LOL

EDIT: ohh man LOL, thanks to the guy below!

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No, y = x^2 + C

#### random-1006

##### Banned
$\bg_white \frac{dy}{dx}= 2x$
$\bg_white \frac{dx}{dy}= 1/2x$
Therefore x = [log(2x)]/2 ??

The question that involved this
$\bg_white \frac{dr}{dt}= \frac{1}{2\pi r^2}$
$\bg_white \frac{dt}{dr}= 2\pi r^2$
$\bg_white t=\frac{2\pi r^3}{3} + C$
WTF is wrong with my example?
for your example to be correct , you are integrating (1/2x) with respect to y, NOT x !

TO GO FROM dx/dy= 1 / (2x) , you must integrate both sides with respect to y, NOT x

so its x= integral 1/ (2x) dy ( very important its dy), and i dont know you do that, think its some uni crap

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#### hscishard

##### Active Member
$\bg_white \frac{dy}{dx}= 2x$

therefore y = x + C

LOL
x^2, doesn't really answer my question

#### Tofuu

##### Member
But then how can you get a derivative of something in terms of itself? [2nd example]
not sure what you mean, but in the first example this is what you're doing

$\bg_white x=\int \frac{1}{2x}dy$

you cant integrate 1/2x if its dy

the second example is pretty much correct

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#### kcqn93

##### Member
x^2, doesn't really answer my question
for the second example

$\bg_white \frac{dr}{dt}= \frac{1}{2\pi r^2}$
$\bg_white \frac{dt}{dr}= 2\pi r^2$

dt = 2 pi r^2 dr

integrate both sides

t = 2/3 pi r^3 + c

#### MetroMattums

##### Member
x^2, doesn't really answer my question
Notice how the dy is on the bottom? You can move it to the RHS of the equation and move the 2x on the otherside to the LHS and chuck integration operators in front to get an integration.

#### hscishard

##### Active Member
Okay. I get it.
So..
$\bg_white \frac{dy}{dx}=y$

$\bg_white \frac{dx}{dy}=1/y$
x = logy + C

And
$\bg_white x = \int \frac{1}{2x}dy$

#### random-1006

##### Banned
Okay. I get it.
So..
$\bg_white \frac{dy}{dx}=y$

$\bg_white \frac{dx}{dy}=1/y$
x = logy + C

And
$\bg_white x = \int \frac{1}{2x}dy$

what is the actual question, i dont think anyone actually knows

#### Tofuu

##### Member
Okay. I get it.
So..
$\bg_white \frac{dy}{dx}=y$

$\bg_white \frac{dx}{dy}=1/y$
x = logy + C

And
$\bg_white x = \int \frac{1}{2x}dy$
the first part is correct

the second part is not the answer
you can only integrate if its 1/2y, not 1/2x

#### hscishard

##### Active Member
the second part is not the answer
you can only integrate if its 1/2y, not 1/2x
Shit. Now I'm confused.

$\bg_white \frac{dr}{dt}= \frac{1}{2\pi r^2}$
$\bg_white \frac{dt}{dr}= 2\pi r^2$
$\bg_white t=\frac{2\pi r^3}{3} + C$
WTF is wrong with my example?
In this one, you can't find r= (blah blah blah), can you?

#### Tofuu

##### Member
You can do this
$\bg_white x=\int \frac{1}{2y}dy$

you cannot do this because its integrating in respect to y and yet you have x
$\bg_white x=\int \frac{1}{2x}dy$

#### hscishard

##### Active Member
what is the actual question, i dont think anyone actually knows
The dy/dx = y I just amde up myself.
The real example was in the first post and was from a rates of change question.

#### random-1006

##### Banned
Shit. Now I'm confused.

In this one, you can't find r= (blah blah blah), can you?

why not, they would give an initial condition to find the C, then rearrange, take cubed root

#### hscishard

##### Active Member
why not, they would give an initial condition to find the C, then rearrange, take cubed root
But it's respect to t

#### Tofuu

##### Member
In this one, you can't find r= (blah blah blah), can you?
you can, you just have to rearrange the equation you have for t