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Numbers Mason 2
By: mudkipfucker
Source: Imgur
Ilt' s of
...
 
What do you think? Give us your opinion. Anonymous comments allowed.
#13

AnonymousDonor (11/06/2013) [+] (1 reply)
and then there's group theory
MFW i finally understand this language
MFW i finally understand this language
#46

anonymoose ONLINE (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)
Now:
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
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)
Now:
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
#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?
Not Master race, Degree race...get it? GET IT?
#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.
#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 !
I may be wrong, I'm not that good at maths, but I find x^6 when I calculate !
#42

nigeltheoutlaw (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.
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.
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.
#4

blasthardcheese (11/06/2013) [+] (8 replies)
Correct me if I'm wrong, (which I probably am) but 1/x * x=x?
#71

terrria **User deleted account** (11/06/2013) []
Goddamnit newton you and the maths you invented. Curse your brilliant mind, curses I say!
#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.