Simple calc question

[gwlam12]

If I integrate 1/2x dx...

Just straight integration, I get (ln x)/2

But if I use u-sub where u = 2x, then I get (ln 2x)/2

So what's the problem?

 

[cchen]

you're doing the subsitution incorrectly

u = 2x
du = 2

 

[gwlam12]

u = 2x
du = 2dx

so dx = du/2

So we're integrating 1/2u du which is (ln u )/ 2 which equals (ln 2x) / 2

No?

 

[Scrooge2]

I think you are doing it wrong i get (1/2) lnx

 

[gwlam12]

Can you provide the steps please? For u-sub. Or at least point out my error. Thanks.

 

[Scrooge2]

don't use u substitution here just think of it as (1/2) x^-1 and integrate. Also if you need further help, use The Integrator

You can thank me later.

 

[gwlam12]

Well...thanks, I guess.

But why is the answer different when using u-sub?

That's my original question. Is there some fundamental law of calculus being broken or something?

 

[TuxDave]

Originally posted by: gwlam12
If I integrate 1/2x dx...

Just straight integration, I get (ln x)/2

But if I use u-sub where u = 2x, then I get (ln 2x)/2

So what's the problem?

You're almost right.

a) 1st part should get ln(x)/2+C1
b) 2nd part should get ln(2x)/2+C2 = ln(2)/2+ln(x)/2 + C2 = ln(x)/2 + C3

So they turn out the same if you just lump together the constants.

 

[Scrooge2]

TuxDave that one flew right over my head, what did you do? I think he's justt taking calc 2.

 

[TuxDave]

Originally posted by: Scrooge2
TuxDave that one flew right over my head, what did you do? I think he's justt taking calc 2.

He forgot the PLUS C part at the end of any indefinite integral. The second integral will turn out to be the same as the first if he just lumps together all the constants.

 

[gwlam12]

The constant...the constant. Hah. So you're saying with u-sub and without u-sub, the constants are different? Cool. I just graphed it, and you're right, there's a difference by .35 all the way through.

Thanks.

And Scrooge2, no, I am not taking Calc 2. I missed the constant, just like you did.

 

[Scrooge2]

Heh, some help i turned out to be . What are you taking then?

 

[gwlam12]

But wait...

If you consider the definite integral..shouldn't they be the same?

<edit>nevermind on this</edit>


BUT

so if you were taking a test and it asked you to find the integral of 1/2X...what would YOU put as your answer?

 

[gwlam12]

Originally posted by: Scrooge2
Heh, some help i turned out to be . What are you taking then?

I'm not taking any math right now. My friend is studying for her GRE and popped that question to me. I tried it out, had the same problem, thought I'd go to geek site, as I like to refer to this place as.

 

[gwlam12]

Originally posted by: TuxDave

Originally posted by: Scrooge2
TuxDave that one flew right over my head, what did you do? I think he's justt taking calc 2.

He forgot the PLUS C part at the end of any indefinite integral. The second integral will turn out to be the same as the first if he just lumps together all the constants.

Hey TuxDave, I'm lost again. I thought the difference between lnx/2 +C1 and ln2x/2 + C2 was constant, but it's not. The difference gets bigger as the numbers get larger. Am I missing something?

 

[electricJ]

Well, try this:

Let 1/(2*x) = (1/2)*(1/x)

1/2 is pulled outside of the integral, thus you only need to integrate (1/x). This integration leaves you with ln(x), which must be multiplied by the factor you pulled out giving you (1/2)*ln(x) (assuming that it is a definate integral, if not, then it must be (1/2)*ln(x) + C

 

[gwlam12]

Thanks...but...

My question is, why is it that when I do u-sub, I don't get 1/2 * lnx

 

[TuxDave]

Originally posted by: gwlam12

Originally posted by: TuxDave

Originally posted by: Scrooge2
TuxDave that one flew right over my head, what did you do? I think he's justt taking calc 2.

He forgot the PLUS C part at the end of any indefinite integral. The second integral will turn out to be the same as the first if he just lumps together all the constants.

Hey TuxDave, I'm lost again. I thought the difference between lnx/2 +C1 and ln2x/2 + C2 was constant, but it's not. The difference gets bigger as the numbers get larger. Am I missing something?

uhh... the difference between lx(x)/2 and ln(2x)/2 should be a constant. It should be off by exactly ln2/2

 

[dighn]

Originally posted by: gwlam12

Originally posted by: TuxDave

Originally posted by: Scrooge2
TuxDave that one flew right over my head, what did you do? I think he's justt taking calc 2.

He forgot the PLUS C part at the end of any indefinite integral. The second integral will turn out to be the same as the first if he just lumps together all the constants.

Hey TuxDave, I'm lost again. I thought the difference between lnx/2 +C1 and ln2x/2 + C2 was constant, but it's not. The difference gets bigger as the numbers get larger. Am I missing something?

it is a constant

ln(2x) = ln(2) + ln(x) => off by a constant ln(2)

 

[TuxDave]

Originally posted by: gwlam12
But wait...

If you consider the definite integral..shouldn't they be the same?

<edit>nevermind on this</edit>


BUT

so if you were taking a test and it asked you to find the integral of 1/2X...what would YOU put as your answer?

I would put ln(x)/2 + C because that's the simpliest solution.
ln(x)/2+A+B+C+D+Es also correct, but not as simple.

 

[electricJ]

Originally posted by: dighn

Originally posted by: gwlam12

Originally posted by: TuxDave

Originally posted by: Scrooge2
TuxDave that one flew right over my head, what did you do? I think he's justt taking calc 2.

He forgot the PLUS C part at the end of any indefinite integral. The second integral will turn out to be the same as the first if he just lumps together all the constants.

Hey TuxDave, I'm lost again. I thought the difference between lnx/2 +C1 and ln2x/2 + C2 was constant, but it's not. The difference gets bigger as the numbers get larger. Am I missing something?

it is a constant

ln(2x) = ln(2) + ln(x) => off by a constant ln(2)

Ding, ding, ding... we have a winner! :beer:

 

[gwlam12]

Thanks all, it is constant. I feel stupid now, but thanks.

 

[TuxDave]

Originally posted by: gwlam12
Thanks all, it is constant. I feel stupid now, but thanks.

:beer: It's ok. Even I was confused for a second when you first posted the problem.