#1

is anyone here good at it becuase i need help...

#2

halp halp at kalkules i kan halp

#3

Yea i could probably be of some help, i passed it last semester with an A. What's the problem?

#4

I just got a 55 in AP Calculus, so I probably can't help you, lol.

#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!

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

note:

one question is

**S**(X^5-3X+(2/X)-2)dx=note:

**S**is the integral symbol
#7

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 isS(X^5-3X+(2/X)-2)dx=

note:Sis 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

f(x) = (x^6)/6 - (3/2)x^2 + 2*ln(x) - 2x + C

#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

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)

#11

isnt like 1/x the derivative of ln(x)? not sure..clarify?

#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

PENGUINEDIT: Meant to go to TS, not guy above me

#13

No.

#14

this next one seems trickier...

**S**Xdx / sqrt(3x^2 -1)
#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

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

#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.

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

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.

#18

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.

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.

#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.

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

I wrote this up, just because I can (and needed a break from homework).

Damn, you have good handwriting.

#23

Damn, you have good handwriting.

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

#24

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

You're a calculator?

#25

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.

#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

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

#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

^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

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

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

note:Sis the integral symbol

my head just exploded

#33

That's easy as fvck:

f(x) = (x^6)/6 - (3/2)x^2 + 2*ln(x) - 2x + C

well said

#35

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

#36

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

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)

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)

*Last edited by ludachris0606 at Apr 30, 2008,*

#38

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

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

*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.