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Where did I go wrong?

Okay, it's proving the integral of x.

x.dx = 1/2.x^2
2x.dx = x^2
x = x^2/2.dx
1 = x/2.dx
1/x = 1/2.dx
2.dx = x
2x = x
2 = 1

Oh shi...!

Someone help?

Originally posted by lord xyz
Where did I go wrong?

Okay, it's proving the integral of x.

x.dx = 1/2.x^2
2x.dx = x^2
x = x^2/2.dx
1 = x/2.dx
1/x = 1/2.dx
2.dx = x
2x = x
2 = 1

Oh shi...!

Someone help?

Take the derivative of 1/2x^2

Umm, what did you do with the dx?

Originally posted by Admiral Akbar
Take the derivative of 1/2x^2

Umm, what did you do with the dx? Integrated.

Ah **** it, 2 = 1.

Originally posted by lord xyz
Where did I go wrong?

Okay, it's proving the integral of x.

x.dx = 1/2.x^2
2x.dx = x^2
x = x^2/2.dx
1 = x/2.dx
1/x = 1/2.dx
2.dx = x
2x = x
2 = 1

Oh shi...!

Someone help?
Your notation is so horribly confusing, that I can't hope to help you here. Especially in this step, I don't have a clue what exactly you're trying to communicate (taking the derivative of both sides?)


2x.dx = x^2
x = x^2/2.dx

Grammar problems to boot.

Originally posted by King Kandy
Your notation is so horribly confusing, that I can't hope to help you here. Especially in this step, I don't have a clue what exactly you're trying to communicate (taking the derivative of both sides?)


2x.dx = x^2
x = x^2/2.dx

2x = d(x^2)/dx

I think that's my first error.

I solved for d and got x = x when I subed...

maybe you need to collect all the x's on one side then sub in d=1/2

also, I am assuming the periods in the first equation are multiplication?

even when I collected the x's, I just keep getting x^2 - x^2 = 0

I haven't done calculus in over 5 years....

EDIT- Removed

Originally posted by lord xyz
No, that isn't (x^2/2)dx, I divided by dx.

Which I believe should've been written as d(x^2)/dx integration, see.


Yeah, I did forget the constant. My bad.

Ah okay, I see, well you can't do that. So I guess you found your answer. I'm still not sure about the thing King Kandy posted a while back though.

if bardock eats 6 donuts in 60 seconds outta the dozen and x is what's left in the box, what does x=?

hehe..

Originally posted by Bardock42
Ah okay, I see, well you can't do that. So I guess you found your answer. I'm still not sure about the thing King Kandy posted a while back though.
I think the problem here is that he used the exact same notation for derivatives and integrals...

Originally posted by FistOfThe North
if bardock eats 6 donuts in 60 seconds outta the dozen and x is what's left in the box, what does x=?

hehe..

That's unsolvable. You gave a speed and an amount. We'll still need time to solve your equation.

Originally posted by Bardock42
That's unsolvable. You gave a speed and an amount. We'll still need time to solve your equation.

Well he said "in 60 seconds" not "per 60 seconds" so the answer should be 6.

Originally posted by Symmetric Chaos
Well he said "in 60 seconds" not "per 60 seconds" so the answer should be 6.


Well that still assumes that I stop after 60 seconds, or that it is counted after 60 seconds, which is not necessarily clear from the phrasing.

Ermm, bump?



I probably failed that question anyway.

You can't just "divide" by dx, that's not an operation that makes any kind of sense.

Is d a variable in calculus? I thought it was part of the notation.

ah, didn't think of that, just thought it was a second variable in the equation

dx denotes the integral.

All I wanted to know was how to prove |x dx =(x^2)/2

I know that you can't just times and divide by dx (now).

It was a stupid question that I probably failed, but I'm still curious as to how anyone can prove it. ermm

This page has a trivial proof using the fundamental theorem, and a sum-based one.

http://math2.org/math/integrals/more/x%5En.htm