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
Yes, but In this case, it looks like it's just some side work for an equation where integration by parts is needed. If that's the case, adding the constant in there is unnecessary because there will be a further integration. Since the constant is arbitrary, you can keep it until you've integrated fully.
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
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.
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!
Yup. It looks like we did the same thing, with the exception of you being less lazy and showing your work. If you factor out a 1/6*x^6 from your answer, then we end up with the same thing.
I'm going to make a wild guess and say that you were going to use limits, and that is utter ******** . If you put the entire equation inside the context of a limit where x->0, then you could in theory say that 1/x=0. Even so, the nominator is x, because (1/x)*x = x/x, and in that case the limit WOULD BE 1. Dividing by zero is never allowed.
I study Econometrics & Operational Research and studied Technical physics last year, and this doesn't even come close to the **** I have to put up with x.x
This is integration by parts... honestly pretty simple stuff but not something I would expect people to remember or care about unless they're late in high school or a technical major.
Just started Uni at Heriot Watt doing Physics and it's all just advanced higher maths again. God am I bad at maths....hence the choice in Physics...hmmm....
Well good for you! That's because it's not hard! Don't let everyone on FJ trying to solve it make you feel smart, because people on here are mostly early-mid teens!