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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.
#4

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

xxxsonic fanxxx (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 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.
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.
#29

Bluemistake (11/06/2013) []
I love how whenever there's a math post on FJ everyone has to do their best to solve the question, like doesn't even matter that this is simple, basic high school level calculus. Everyone here needs to feel like a genius because of it just to show off 'da smartz'
#19

scrumdillyicious (11/06/2013) []
It looks like they're in the middle of integration by parts. Calc 1 **** . Watch Khan academy and you'll be fine.