Correct Du and Dx Notation in Regards to Integration

[Jason76]

$\displaystyle u = 9x^{2}$

$\displaystyle du = 18x^{2} dx$

$\displaystyle 18xdu = 2dx$

$\displaystyle du = \dfrac{2x}{18x}dx$

$\displaystyle du = \dfrac{1}{9}dx$

Is this logic right as far as where I put the symbols?

 

[Deleted member 4993]

$\displaystyle u = 9x^{2}$

$\displaystyle du = 18x^{2} dx$ ..................Incorrect

du = 18 * x dx

$\displaystyle 18xdu = 2dx$

$\displaystyle du = \dfrac{2x}{18x}dx$

$\displaystyle du = \dfrac{1}{9}dx$

Is this logic right as far as where I put the symbols?

.

 

[HallsofIvy]

$\displaystyle u = 9x^{2}$

$\displaystyle du = 18x^{2} dx$

du= 18 x dx

I have no clue how you suddenly got "18xdu". Algebraically, could divide both sides by 18 x to
get $\displaystyle \frac{1}{18x}du= dx$

$\displaystyle 18xdu = 2dx$

$\displaystyle du = \dfrac{2x}{18x}dx$

$\displaystyle du = \dfrac{1}{9}dx$

Is this logic right as far as where I put the symbols?

 

[Jason76]

du= 18 x dx

I have no clue how you suddenly got "18xdu". Algebraically, could divide both sides by 18 x to
get $\displaystyle \frac{1}{18x}du= dx$

I don't know the logic behind it, but it does give me my "constant of integration". But what is the correct way to get the "constant of integration"?

 

[srmichael]

I don't know the logic behind it, but it does give me my "constant of integration". But what is the correct way to get the "constant of integration"?

Do you have the actual problem that this question is coming from? It seems like you are doing an integral using u-substitution and you are letting u = 9x². If we had the problem, I think we can clear up your confusion on this pretty quickly.

 

[JeffM]

$\displaystyle u = 9x^{2}$

$\displaystyle u = 9x^2 \implies \dfrac{du}{dx} = 9 * (2 x^{(2 - 1)}) = 18x.$ This is basic differentiation.

$\displaystyle u = 9x^2 \implies \dfrac{du}{dx} = 18x \implies du = 18xdx.$ Straightforward.

$\displaystyle du = 18x^{2} dx$

This is simply an inexplicable error because you know differential calculus.

$\displaystyle 18xdu = 2dx$

This does not follow from your previous statement. Where did the 2 come from? Where did the x^2 go? How did the 18 jump from one side of the equation to the other?

$\displaystyle du = \dfrac{2x}{18x}dx$

This does not follow from your previous statement. Where did the x in the numerator come from?

$\displaystyle du = \dfrac{1}{9}dx$

Is this logic right as far as where I put the symbols?

No. The calculus and algebra are both wrong. It looks as though you are just doing things to arrive at a predetermined result. What are you trying to do? In other words, what does the problem say exactly?

 

[DrPhil]

When you submit the same problem more than once,
and when you completely disregard the complete solution already given,
you wind up irritating us.

Kahn told you yesterday afternoon that the proper substitution is $\displaystyle u=3x$, and I might add, $\displaystyle a=4$. You will then have an integral of the form

$\displaystyle \displaystyle \int \dfrac{du}{\sqrt{u^2 + a^2}}$

for which you were given a formula.

There is a multiplier of (1/3) resulting from the substitution. That coefficient has NOTHING TO DO with the constant of integration, which is always added to the result of integration because the derivative of the constant is zero so it is completely invisible and unknown in the integral.

 

[Jason76]

When you submit the same problem more than once,
and when you completely disregard the complete solution already given,
you wind up irritating us.

Kahn told you yesterday afternoon that the proper substitution is $\displaystyle u=3x$, and I might add, $\displaystyle a=4$. You will then have an integral of the form

$\displaystyle \displaystyle \int \dfrac{du}{\sqrt{u^2 + a^2}}$

for which you were given a formula.

There is a multiplier of (1/3) resulting from the substitution. That coefficient has NOTHING TO DO with the constant of integration, which is always added to the result of integration because the derivative of the constant is zero so it is completely invisible and unknown in the integral.

Sorry, I will just stick with the answer you guys gave me. Also when I said "constant of integration" I accidentally meant to say multiplier.