Numbers Mason 2. Source: Imgur. Ilt' s of. MFW I read the comments.



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#17 - iRoflcopterU (11/06/2013) [-]
stickied by mudkipfucker
MFW I read the comments.
MFW I read the comments.
#13 - AnonymousDonor (11/06/2013) [+] (1 reply)
and then there's group theory   
MFW i finally understand this language
and then there's group theory

MFW i finally understand this language
User avatar #46 - anonymoose (11/06/2013) [-]
Gonna try and explain this for y'all.

Guide for what the terms mean:
ln(x) - "ln" is a logarithm of base e.
logarithm (from the definition of ln(x) ) - logarithm is a function, basically we have multiplication and division and we also have logarithms and powers. Lets say Log(x) is logarithm to the base 10 of a value x. 10^(log(x)) is equal to x. log (1000) is 3 because 10^3 is 1000.
e (from the definition of ln(x) ) - e is an important number in maths, like pi. It has infinite decimal places, like pi. It is roughly 2.718
dx - Differentiation is another function. When you differentiate x^2 you take the number in the power (2 here) and multiply by that, then you take 1 away from the power, then add "dx" to the end. If you differentiate u^2 it's 2u du. So differentiating x^2 = 2x dx. In basic terms, dx is just something you add to the end of the equation after differentiating.
Integration - Integration is the opposite of differentiation. Integrating a number, we add 1 to the power (the power of the number beside the d) and then divide by the number in the power. E.g. integrating x^2, we get (x^3)/3 and then we remove the dx or d(whatever letter we're integrating with respect to)


u = ln(x)
On the left, we're differentiating u, so we multiply by the power (which is just 1, here) then take away 1 from the power, leaving us with u^0. Any number to the power of 0 is just 1. We're left with 1 * du, or just du.

On the right, we have ln(x). Some values, when differentiated have different rules than just taking one from the power. For a logarithm, the differentiation is 1/x giving us 1/x dx

So, differentiating u = ln(x) we get du = 1/x dx

The bottom equation, we're integrating.

dv is really 1*dv, so integrating 1 (with respect to v because of the dv) we just get v and remove the dv.
x^5 dx, we add 1 to the x and divide by the number in the power, to get (x^6)/6
User avatar #27 - johncaveson **User deleted account** (11/06/2013) [+] (3 replies)
Mathematics student Degree Race reporting in.

Not Master race, Degree race...get it? GET IT?
#31 - agentmoleman (11/06/2013) [+] (8 replies)
Math thread..
User avatar #1 - stafforadam (11/05/2013) [+] (6 replies)
shouldn't it be x^6/6 + c?
#48 - anonymous (11/06/2013) [+] (3 replies)
This is really stupid, dx is a notation for differential calculus, it is by no means a variable that can be replaced with a value.
User avatar #52 to #48 - anonymoose (11/06/2013) [-]
But dx is technically 1*dx. Integrating 1*dx is x (+c).
#44 - Her (11/06/2013) [-]
Everyone else here is speaking math.   
But then there's me, speaking elvish and 						****					.
Everyone else here is speaking math.
But then there's me, speaking elvish and **** .
User avatar #73 - strelokkk (11/06/2013) [+] (2 replies)
Isn't the result of the 2nd equation v = x^6 instead of v = (x^6)/6 ?
I may be wrong, I'm not that good at maths, but I find x^6 when I calculate !
User avatar #74 to #73 - anonymoose (11/06/2013) [-]
For integration, you add one to the power and divide by whatever the power is. So x^5, you add 1 and it becomes x^6 and then you divide by 6 to give you (x^6)/6
#59 - cobrien ONLINE (11/06/2013) [+] (2 replies)
Lol @ people who are confused by calc 1
#49 - howaitoenjeru (11/06/2013) [-]
Made me think of this
User avatar #42 - nigeltheoutlaw ONLINE (11/06/2013) [+] (4 replies)
Aw, I miss that level of integration. We just finished triple integrals in Calculus 3.
#10 - icekingg (11/06/2013) [+] (2 replies)
Mathfag here to translate this **** .
My best guess would be that the original integrand was x^5*ln(x)dx. Integrating it by parts as suggested in the image spits out the answer of 1/6*x^6[ln(x)-1/6]+c
The top line in the image is the original problem and the second line is the formula used when integrating by parts.
In the third line, I plugged in all of my v's, u's, and du's then factored out 1/6.
In the fourth line, I simplified, then evaluated the integrand and shat out a constant.
The fifth line is the answer, with x^6 factored out of both terms in brackets.

#25 to #10 - quesocnkane (11/06/2013) [-]
the original equation must have been :

Int(u*dv = int((ln(x))*x^5*dx
You then break it into u, v, du, and dv.
u = ln(x)
du = 1/x
v = (x^6)/6
dv = x^5
Integration by parts states that the int(u*dv = uv - int(v*du)
so multiply u times v, then integrate v*du with respect to your original variable. (dx)
so you'd have ln(x)*((x^6)/6) - int(((x^6)/6)*(1/x)dx
we know that x^6 / 6x can be simplified to 1/6 * x^6/x which is just x^5
so bring out your constant of 1/6 and integrate x^5 with respect to x to get 1/6x^6
bring back your constant....
you have 1/36*x^6
So your entire final product should be ln(x)*([x^6]/6)-[(x^6)/36]
If you want common denominators to simplify past this, go for it. But from here you can incorporate any limits to solve for the area under the curve!

TLDR I think we did the same thing.
User avatar #4 - blasthardcheese (11/06/2013) [+] (8 replies)
Correct me if I'm wrong, (which I probably am) but 1/x * x=x?
User avatar #39 to #20 - goldenglimmer (11/06/2013) [-]
The less you know*
User avatar #71 - terrria **User deleted account** (11/06/2013) [-]
Goddamnit newton you and the maths you invented. Curse your brilliant mind, curses I say!
#70 - fourtitwo (11/06/2013) [-]
User avatar #16 - davidokuro (11/06/2013) [-]
That's the basics.
#12 - anonymous (11/06/2013) [-]
yeah, this is the process for integration by parts ****** . when you need the antiderivative of two functions of a variable multiplied together.
#11 - muffinssnuffims (11/06/2013) [+] (2 replies)
Just wait till you get to Multivariable calculus and Differential Equations. That **** looks like 2+2 comparatively.
#9 - yubb (11/06/2013) [-]
MFW I understand because my Calculus teacher taught us antiderivatives today
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