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Simple calc question

G

gwlam12

Diamond Member
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?
 
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?
 
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?
 
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?
Click to expand...

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

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

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. 🙂
 
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.
Click to expand...

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.
Click to expand...

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?
 
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
 
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.
Click to expand...

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.
Click to expand...

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?
Click to expand...

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

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.
Click to expand...

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?
Click to expand...

it is a constant

ln(2x) = ln(2) + ln(x) => off by a constant ln(2)
 
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?
Click to expand...

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.
 
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.
Click to expand...

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.
Click to expand...

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?
Click to expand...

it is a constant

ln(2x) = ln(2) + ln(x) => off by a constant ln(2)
Click to expand...

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