12:21 AM
Philip
: relativity
That shit makes my mind go WTF?!?!?!
12:22 AM
me
: haha mine too
that's when i stopped taking physics
12:23 AM
Philip
: I managed to get through it barely
but it never really made sense
12:24 AM
me
: it's like that saying about math
"you never fully understand it. You just get used to it."
Philip
: hahahaha
12:25 AM
me
: like do you fully understand why a riemann sum is equal to an integral?
12:26 AM
Philip
: yup
:)
me
: please explain
Philip
: It's just binpacking
you have a space
right?
and you want to shove things into it until it's full.
12:27 AM
So you start with a rectangular block
and you then keep shoving in smaller rectangles
until it works.
me
: ok, i understand the geometrical concept
keep taking smaller rectangles
sum up their areas
take the limit
yeah, geometrically that makes sense
but why is that the sum as taking an integral?
12:28 AM
how come that sum of delta x's as delta x approaches 0 is equal to an integral?
*the same
Philip
: Because
12:29 AM
when you keep shoving in rectangles
you know the total area of rectangles you've shoved in
equals the size of the container
when the size of the rectangles you're shoving in approaches 0
12:30 AM
me
: ok, but how come this integration operator finds you the area under the curve?
i understand derivatives
it gives you the slope at any point
Philip
: Did you just understand what I said?
lol
Ok
me
: and i understand integration is the reverse of that
Philip
: so you have a container
say a jar
right?
the depth is simple
so it's a prism
that's open on one end so we can shove shit into it
12:31 AM
we start shoving in rectangles of size, say 1cm^2
me
: yo
Philip
: what's the cross-sectional area of the container if we can fit in 5 rectangels?
it's > 5
right?
me
: i understand why a riemann sum equals the area under a curve
Philip
: now let's say our rectangles are .01cm^2
an integral
12:32 AM
is just the riemann sum when the size of the rectangles goes to 0
me
: ok, sure, i agree you can define it that way
but the integral operator has all these properties
how come taking this riemann sum
an infinite sum
taking the limit as delta x --> 0
12:33 AM
how come that's the same as applying an integral
and these antiderivative rules
Philip
: Ok....
me
: like, you know, the integral of x^2 = x^3/3
Philip
: so the question you're really asking is
me
: etc.
Philip
: how is an integral an anti-derivative?
me
: yeah
an integral in sense of a riemann sum
Philip
: well yeah.
Ok
me
: do you understand why?
cuz I don't
12:34 AM
Philip
: Yeah.
So you're going from left to right
on the area right?
me
: yeah
Philip
: and you take a segment of size 1
x-wise
how much area do you add to your current total?
me
: the area under the curve
12:35 AM
for that segment of size 1
Philip
: which is approximated by?
me
: the riemann sum
Philip
: the value of the function at that point.
me
: sure
Philip
: lights flash
make sense?
me
: but how does integrating at that point give you the area under the curve?
12:36 AM
Philip
: look
me
: how does integrating, finding the function whose slope is the integrand
Philip
: draw this with me
Draw some random curve
me
: how is this the same as the area under the curve?
Philip
: in the first quadrant
simple x, y
starting from origin
shade some random area from 0 to x*
where x* is some random x* you pick on the x axis
we call the curve you drew f'(x)
12:37 AM
we call the area under the curve up to x* f(x)
me
: uh
how is the area = x * f(x) ?
Philip
: so f(x) = Integral from 0 to x* of f'(x)
12:38 AM
no the area is f(x)
me
: yeah
A = integral from 0 to x* of f'(x)
Philip
: We call the area under the curve up to x*, f(x)
yup
me
: i think you should call it A
Philip
: ok sure
lol
me
: it's a definite integral
Philip
: call it whatever you want
hahahaha
so now extend
the shading
12:39 AM
so that you add a vertical segment from x* to x*+1
how much area did you just add?
12:40 AM
me
: that's the integral from x* to x* + 1 of f'(x)
12:41 AM
ok, let me give an example of what i'm talking about
let's say f(x) = 2x
f'(x) = 2
2 easy graphs to draw
the indefinite integral of f'(x) is 2x
12:42 AM
f'(x) represents the value of the slope of f(x) at every point
all right, fine.
now, let's say we want to find the area under the curve of f'(x)
between say, 0 and 2
Philip
: yup
12:43 AM
me
: well, we know from geometric arguments that it's 4
Philip
: yup
me
: but
by our rules of integration
we also know
it's the same as finding the indefinite integral of f'(x)
which is f(x), which is 2x
Philip
: yup
me
: and then plugging in f(2) - f(0)
how is it that finding the area under the curve is the same as evaluating the antiderivative at two points and taking their difference?
12:44 AM
Philip
: What I'm trying to convince you of
is that
geometrically
you're accumulating the area by "undoing" the derivative.
Because each segment you add
is = to the value of f'(x) at the point.
me
: right
12:45 AM
Philip
: and since what we're adding is the "rate of change" of the area.
then the value of the curve at that point is the derivative of the area function we are trying to find.
Thus integral = anti-derivative.
and since we sorta assumed integral = riemann sum
riemann sum = anti-derivative.
12:46 AM
me
: hold on, let me think about this
12:51 AM
oh, i think that makes sense to me
this area function, let's call it A
if you just did A(x) for arbitrary x
that would tell you area under the curve for -infinity up to point x
under the curve f'(x)
the area function is like a cumulative sum
12:52 AM
and so if you're only interested in finding the area under the curve for one part of the curve f'(x), say between x1 and x2, that's why you do A(x2) - A(x1)
12:53 AM
Philip
: sure
:)
me
: it was very helpful for me when you said above the infinitesimal segment we're adding is the "rate of change" of the area
that made sense
and I understand how rate of change is the slope
hence a derivative
12:54 AM
Philip
: :D
me
: hence "rate of chagne of the area" means "rate of change of the area function"
= "derviative of the area function"
thus, I understand that integral is antiderivative
so we are taking the integral of the "derivative of the area function" in order to recover the area function
12:55 AM
this is pretty cool
thanks for helping me understand
Philip
: :D