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• Mar 14th 2011, 02:55 PM
MaverickUK82
I can see that

$\displaystyle \int\frac{1}{u} du$

is easier to integrate, and can carry that integration out.

However, I can't see how your:

$\displaystyle \int\frac{\frac{1}{3}(u-2)}{u}\frac{1}{3} du$

corresponds to

$\displaystyle \int\frac{1}{u} du$

or how you got there.
• Mar 14th 2011, 04:40 PM
topsquark
Quote:

Originally Posted by MaverickUK82
I can see that

$\displaystyle \int\frac{1}{u} du$

is easier to integrate, and can carry that integration out.

However, I can't see how your:

$\displaystyle \int\frac{\frac{1}{3}(u-2)}{u}\frac{1}{3} du$

corresponds to

$\displaystyle \int\frac{1}{u} du$

or how you got there.

This is as step by step as I can make it.

The original integral is
$\displaystyle \displaystyle \int \frac{v}{2 + 3v}dv$

It's always a good idea to try to get the denominator to be as simple as possible. (Though there are exceptions to that.) So this gives the idea of the substitution
$\displaystyle u = 2 + 3v$

To sub this back into the integral we need to know what v is. So solving for v in terms of u we get:
$\displaystyle \displaystyle v = \frac{1}{3} \cdot (u - 2)$

But we are not done. We also need to find an expression for dv in terms of u also. So
$\displaystyle u = 2 + 3v \implies du = d(2 + 3v) = 3~dv$

Solving this for dv gives
$\displaystyle \diisplaystyle dv = \frac{1}{3} du$

Now we need to sub in our values for v and dv into the original integral:
$\displaystyle \displaystyle \int \frac{v}{2 + 3v}dv = \int \frac{ \left [ \frac{1}{3}(u - 2) \right ] }{[u]} \cdot \left [ \frac{1}{3}du} \right ]$

Factoring out the constants gives
$\displaystyle \displaystyle \frac{1}{9} \int \frac{u - 2}{u} du = \frac{1}{9} \int \left ( \frac{u}{u} - \frac{2}{u} \right ) du$

$\displaystyle \displaystyle = \frac{1}{9} \int du - \frac{1}{9} \int \frac{2}{u} du$

$\displaystyle \displaystyle = \frac{1}{9} \int du - \frac{2}{9} \int \frac{1}{u} du$

$\displaystyle \displaystyle = \frac{1}{9} \cdot u - \frac{2}{9} \cdot ln|u| + C$

And subbing back in $\displaystyle u = 2 + 3v$ gives
$\displaystyle \displaystyle = \frac{1}{9} \cdot (2 + 3v) - \frac{2}{9} \cdot ln|2 + 3v| + C$

-Dan
• Mar 15th 2011, 12:24 PM
MaverickUK82
THanks topsquark... Whilst I am able to follow that through, and can see what is being done. I would never have came up with that on my own. THanks for your patience and help!
• Mar 15th 2011, 06:24 PM
topsquark
Quote:

Originally Posted by MaverickUK82
THanks topsquark... Whilst I am able to follow that through, and can see what is being done. I would never have came up with that on my own. THanks for your patience and help!

For most students I've taught it's a case of you have to see how it's done a few times. You'll do better the next time you see one.

-Dan
• Mar 22nd 2011, 03:15 PM
MaverickUK82
Hi Topsquark, I've been following this, but what have you done here? Is this implicit differentiation?

Quote:

Originally Posted by topsquark
But we are not done. We also need to find an expression for dv in terms of u also. So
$\displaystyle u = 2 + 3v \implies du = d(2 + 3v) = 3~dv$

Solving this for dv gives
$\displaystyle \diisplaystyle dv = \frac{1}{3} du$-Dan

• Mar 22nd 2011, 03:46 PM
topsquark
Quote:

Originally Posted by MaverickUK82
Hi Topsquark, I've been following this, but what have you done here? Is this implicit differentiation?

It's called "Taking the differential" of both sides. The more pure Matheticians around here would probably rather have me use the following notation:
$\displaystyle \displaystyle u = 2 + 3v \implies \frac{du}{dv} = 3$

and you can put that into your integral. But I like the shorthand better. (But then I'm a Physicist...)

-Dan
• Mar 23rd 2011, 08:28 AM
MaverickUK82
Topsquark, thanks again, I have, hopefully, one final question with regards to the above.

We have

$\displaystyle \frac{du}{dv}=3$

yet, we have

$\displaystyle dv=\frac{1}{3}du$

I'm struggling with this, why does the derivative of v equal one third of the derivative of u (which of course would be 1)?

I think with a bit of clarification on this I will be done! I really appreciate your patience.
• Mar 23rd 2011, 04:50 PM
topsquark
Quote:

Originally Posted by MaverickUK82
Topsquark, thanks again, I have, hopefully, one final question with regards to the above.

We have

$\displaystyle \frac{du}{dv}=3$

yet, we have

$\displaystyle dv=\frac{1}{3}du$

I'm struggling with this, why does the derivative of v equal one third of the derivative of u (which of course would be 1)?

I think with a bit of clarification on this I will be done! I really appreciate your patience.

Now i'm really going annoy some of the members here....

Treat the derivative symbol as an actual fraction. (Mathematically this is not a correct treatment, but it works.) ie solve
$\displaystyle \frac{du}{dv}=3$

for dv. And why would du = 1?

-Dan
• Mar 24th 2011, 01:39 AM
MaverickUK82
Quote:

Originally Posted by topsquark
Now i'm really going p*ss off some of the members here....

Treat the derivative symbol as an actual fraction. (Mathematically this is not a correct treatment, but it works.) ie solve
$\displaystyle \frac{du}{dv}=3$

for dv. And why would du = 1?

-Dan

Because earlier, we calculated the derivative of du to be 3, so 1/3 of it would be 1. or have I gone completely mad?
• Mar 25th 2011, 03:52 PM
MaverickUK82
Topsquark.

Thanks for your help on this. I am still confused with the last bit (last post). I have found some good material on integration with by substitution and will continue to read up on it and hopefully it will become clearer.
• Mar 25th 2011, 08:56 PM
topsquark
Quote:

Originally Posted by MaverickUK82
Because earlier, we calculated the derivative of du to be 3, so 1/3 of it would be 1. or have I gone completely mad?

We calculated du/dv to be 3, not du.

-Dan
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