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#1
is anyone here good at it becuase i need help...
Later in the evenin, as you lie awake in bed, hear the echoes of the amplifiers ringin' in your head, smoke the days last cigarette rememberin' what she said...
#2
halp halp at kalkules i kan halp
Quote by denizenz
I'll logic you right in the thyroid.

Art & Lutherie
#3
Yea i could probably be of some help, i passed it last semester with an A. What's the problem?
#5
Im in calc III right now, thats the one with multivariable vector valued functions, greens/stokes stuff; I should be able to help...

Shoot!
#6
i missed class like all last week because i was sick and my teacher handed out a takehome quiz... its on integrals and pretty much the only thing i learned about it was that you take the anti-derivative, but ik theres way more to it.

one question is
S(X^5-3X+(2/X)-2)dx=

note:S is the integral symbol
Later in the evenin, as you lie awake in bed, hear the echoes of the amplifiers ringin' in your head, smoke the days last cigarette rememberin' what she said...
#7
Quote by ludachris0606
i missed class like all last week because i was sick and my teacher handed out a takehome quiz... its on integrals and pretty much the only thing i learned about it was that you take the anti-derivative, but ik theres way more to it.

one question is
S(X^5-3X+(2/X)-2)dx=

note:S is the integral symbol


Find the anti-derivative? And if you have bounds, solve. Thats what I would do.

I guess I know what to do, but the reason I'm failing is cause I would get the wrong anti-derivative.
#8
That's easy as fvck:

f(x) = (x^6)/6 - (3/2)x^2 + 2*ln(x) - 2x + C
Quote by denizenz
I'll logic you right in the thyroid.

Art & Lutherie
#9
All you have to do is integrate. If you can do that, you're good.

The answer is (x^6)/6 - (3x^2)/2 + 2ln(x) - 2x+c

EDIT: If you have limits of integration there's no c, you evaluate
#10
hmm i understand the most part except 2ln(x)... is there another way to write that because i dont think my teacher went over natural logs(i think thats what it means)
Later in the evenin, as you lie awake in bed, hear the echoes of the amplifiers ringin' in your head, smoke the days last cigarette rememberin' what she said...
#12
^No, the derivative of lnx is 1/x, therefore, the antiderivative of 1/x is lnx.

PENGUINEDIT: Meant to go to TS, not guy above me
#14
this next one seems trickier...
S Xdx / sqrt(3x^2 -1)
Later in the evenin, as you lie awake in bed, hear the echoes of the amplifiers ringin' in your head, smoke the days last cigarette rememberin' what she said...
#15
Use substitution.

u = 3x^2 - 1
du = 6x dx ===> dx = du / 6x

Replacing dx with du and sqrt(3x^2 - 1) with u makes the integral

integral( x * (du / 6x) * 1/u)

Xs cancel out, leaving: 1/6 * integral(du / u)

Integrate to get (1/6)*ln(3x^2 - 1) + C
Quote by denizenz
I'll logic you right in the thyroid.

Art & Lutherie
#16
1/6 sqrt(3x^2 -1)

derivative of 1/6 sqrt(3x^2 -1) =
[1/6]*[1/sqrt(3x^2 -1) ]*[6x] = x/sqrt(3x^2 -1)

I'm pretty sure.
#17
Quote by gibsonpenguin
1/6 sqrt(3x^2 -1)

derivative of 1/6 sqrt(3x^2 -1) =
[1/6]*[1/sqrt(3x^2 -1) ]*[6x] = x/sqrt(3x^2 -1)

I'm pretty sure.


LOLOLOL No.
Quote by denizenz
I'll logic you right in the thyroid.

Art & Lutherie
#18
Quote by darkstar2466
Use substitution.

u = 3x^2 - 1
du = 6x dx ===> dx = du / 6x

Replacing dx with du and sqrt(3x^2 - 1) with u makes the integral

integral( x * (du / 6x) * 1/u)

Xs cancel out, leaving: 1/6 * integral(du / u)

Integrate to get (1/6)*ln(3x^2 - 1) + C

No, the derivative of that would be
x/(3x^2 -1), you're missing the sqrt.

The problem is that you replaced sqrt(3x^2 -1) with 3x^2 -1 (u instead of sqrt(u)).
#19
Oh lulz, I messed it up.

integral( x * (du / 6x) * 1/sqrt(u))

Xs cancel out, leaving: 1/6 * integral(du / sqrt(u))

Integrate to get (1/6)*2sqrt(3x^2 - 1) + C

which gives (1/3)sqrt(3x^2 - 1) + C

Thanks gibsonpengiun for pointing out my stupidity and carelessness in reading posts.
Quote by denizenz
I'll logic you right in the thyroid.

Art & Lutherie
#20
I wrote this up, just because I can (and needed a break from homework).

Last edited by Logic Smogic at Apr 30, 2008,
#21
Alright, so integral of:
x((3x^2 -1)^(-1/2))
= ((3x^2 -1)^(1/2))*1/6
When you take the derivative of that, you get:
((3x^2 -1)^((1/2)-1))*1/6*(derivative of (3x^2 -1))
derivative of (3x^2 -1) = 6x
=((3x^2 -1)^(-1/2))*1/6*6x
6s cancel, replace ((3x^2 -1)^(-1/2)) with 1/((3x^2 -1)^(1/2))
= x/((3x^2 -1)^(1/2))
= x/sqrt(3x^2 -1)

PENGUINEDIT: Darkstar wrote it out better than I did, I do stuff like that pretty much mentally because I've been doing it for a while now. And the user above me has nice handwriting.
#22
Quote by Logic Smogic
I wrote this up, just because I can (and needed a break from homework).


Damn, you have good handwriting.

...modes and scales are still useless.


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#23
Quote by Xiaoxi
Damn, you have good handwriting.


I do this for a living, so I have to.

#24
Quote by Logic Smogic
I do this for a living, so I have to.


You're a calculator?

...modes and scales are still useless.


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Thanks
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#25
Quote by Xiaoxi
You're a calculator?


Heh, not exactly. Physics graduate student.

Here's an integral I did last night:

#26
^^^I do that in my sleep.

...modes and scales are still useless.


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#27
That looks like some spherical coordinate stuff I have to do for my calc III final tomorrow morning.... There's so much stuff on the exam and I'm so screwed.
#28
alright cool, thanks
i have one last question and ill try to figure out the rest,

Evaluate dy/dt for the function at the point
x^3+y^3=9; dx/dt=-3, x=1, y=2
Later in the evenin, as you lie awake in bed, hear the echoes of the amplifiers ringin' in your head, smoke the days last cigarette rememberin' what she said...
#29
That looks like some spherical coordinate stuff I have to do for my calc III final tomorrow morning.... There's so much stuff on the exam and I'm so screwed.


^Yeah. Had to integrate by parts 4 times - but if you're good at bookkeeping, it's not so bad. Don't forget you Jacobian, either (r^2 * Sin(theta) in spherical, usually).
#30
Quote by Logic Smogic
^Yeah. Had to integrate by parts 4 times - but if you're good at bookkeeping, it's not so bad. Don't forget you Jacobian, either (r^2 * Sin(theta) in spherical, usually).

Yeah, that's the easy part.

I've been having a much harder time with these line integrals, integrals of vector fields, integrals with respect to arc length, integrals over a surface, etc. etc.
#31
Quote by ludachris0606
alright cool, thanks
i have one last question and ill try to figure out the rest,

Evaluate dy/dt for the function at the point
x^3+y^3=9; dx/dt=-3, x=1, y=2


sorry it's blurry



Please make sure you understand what I did, and if you have questions - ask!
#32
Quote by ludachris0606


one question is
S(X^5-3X+(2/X)-2)dx=

note:S is the integral symbol


my head just exploded
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The Truth: True. God is the CEO and Jesus does QC at Mesa...yup.


#34
Quote by Logic Smogic
I wrote this up, just because I can (and needed a break from homework).






I think you messed up on the leading term there buddy. Exponent on the x for x/6 should 6. LaTeX ftw
#35
Logic Smogic, in the line after you say chain rule, how did you get the constants of 3 on both sides?
Later in the evenin, as you lie awake in bed, hear the echoes of the amplifiers ringin' in your head, smoke the days last cigarette rememberin' what she said...
#36
Quote by Negative Burn



I think you messed up on the leading term there buddy. Exponent on the x for x/6 should 6. LaTeX ftw


right on. it should be a 6. Fixed the original in paint.

Quote by ludachris0606
Logic Smogic, in the line after you say chain rule, how did you get the constants of 3 on both sides?


When you differentiate y(t)^3 with respect to t, you must differentiate the function (in this case y^3), and then multiply by the derivative of the embedded function (in this case y(t)).

So the derivative of y^3 is 3*y^2.
The derivative of y(t) is dy/dt.

So the result is (3*y^2)*(dy/dt).
Same goes for the right side.
Last edited by Logic Smogic at Apr 30, 2008,
#37
ohh okay...
Yeah the power looked like a 3 to me after that still so I was confused...

and the final answer should be 3/4 right?
because 3((1^2)/(2^2)) = 3(1/4) = (3/4)
Later in the evenin, as you lie awake in bed, hear the echoes of the amplifiers ringin' in your head, smoke the days last cigarette rememberin' what she said...
Last edited by ludachris0606 at Apr 30, 2008,
#38
Quote by ludachris0606
ohh okay...
Yeah the power looked like a 3 to me after that still so I was confused...


Heh, sorry it's so blurry. Shaky hands, I guess - and the flash washes it all out.

But yeah, after doing the chain rule, they should both be 2's.
#39
and the final answer should be 3/4 right?
because 3((1^2)/(2^2)) = 3(1/4) = (3/4)

nah dont worry about it you helped me out a lot...
thank you so much
atleast now i have a basic understanding
Later in the evenin, as you lie awake in bed, hear the echoes of the amplifiers ringin' in your head, smoke the days last cigarette rememberin' what she said...
Last edited by ludachris0606 at Apr 30, 2008,
#40
hopefully when i take calc in the fall, you guys will be here to help me too.
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