/sci/ - Science & Math

[Anonymous No.6590001]

So guys I'm trying to understand why integration gives me the surface area under a function. A /sci/entist gave me a very good and simple explanation some days ago but I forgot it(I know I'm dumb...). It had to do with a function f(x) and then another function F(x) which (doesn't matter in what way) always gives me the area under f(x).

[Anonymous No.6590003]

>>6590001
>A /sci/entist gave me a very good and simple explanation some days ago but I forgot it

Check the archive, should be pretty easy to find if you know the (approximate) date and some phrases used in the discussion.
>>/sci/

[Anonymous No.6590004]

the rate of the increase of the area is given by the height
so if you take the summation of the of all the heights from a to b then you get the area

[Anonymous No.6590005]

>>6590003
Ok, thanks. Will post in the thread when I find it.

[Anonymous No.6590009]

>found it

Imagine a function f(x) is an ordinary bog-standard curve. Now imagine another function F(x) that tells you how much area f(x) has traced out between 0 and x (it doesn't matter how, it just does).

If f(x) is some positive constant, then the area it traces out between 0 and x increases linearly as x does. Thus, F(x) will be a positive-sloped straight line. What is the derivative of such a line?

If f(x) is a linear function of x, then the area it traces out will increase in proportion to the square of x (the area is triangle shaped, and a triangle is half a rectangle). Therefore, F(x) will be a quadratic. What is the derivative of a quadratic?

In each case, the derivative of the "area function" gives you the original function. Therefore, to get the area function, you take the anti derivative.

>inb4 eaten alive because "antiderivative ≠ integral" and other autism

[Anonymous No.6590010]

>>6590001
You're basically summing up the areas of an infinite amount and infinitely thin vertical rectangles to find the total area under the curve.

Infinite sum = integral

[Anonymous No.6590040]

>>6590004
This is the best intuitive reason. The area function A(x) has a rate of change dA/dx.

By looking at the picture you can see dA/dx is f(x). So to get A you do the reverse process as differentiation to f(x)

[Anonymous No.6590160]

>So guys I'm trying to understand why integration gives me the surface area under a function.
Because it was defined that way. It is the notion of primitives that was linked to it later with the fundamental theorem of calculus. Here’s a proof if you like (I find it much more satisfying than what I saw on Wikipedia and Khan Academy)

Let <span class="math">f[/spoiler] be continuous on <span class="math">[a;b][/spoiler] and <span class="math">F:x \mapsto \int_a^x f(t) \mathrm{d}t[/spoiler]. Thus for all <span class="math">(x;y) \in [a;b]^2[/spoiler] and <span class="math">\varepsilon > 0[/spoiler] there exists <span class="math">\eta > 0[/spoiler] such that
<div class="math">|y-x| < \eta \Rightarrow |f(y) - f(x)| < \varepsilon</div>
Let <span class="math">h \in [a;b][/spoiler] such that <span class="math">|h| < \eta[/spoiler]. Thus <span class="math">|f(x+h) - f(x)| < \varepsilon[/spoiler]
<div class="math">\left| \fraction{F(x+h) - F(x)}{h} - f(x)\right| = \left| \fraction{1}{h} \left[ \int_a^{x+h} f(t) \mathrm{d}t - \int_a^x f(t) \mathrm{d}t \right] - \fraction{1}{h} \int_x^{x+h} f(x) \mathrm{d}t \right| = \fraction{1}{|h|} \left| \int_x^{x+h} [f(t) - f(x)] \mathrm{d}t \right| </div>

Given that <span class="math">|t-x| \leq |h| < \eta[/spoiler], <span class="math">|f(t) - f(x)| < \varepsilon[/spoiler], thus
<div class="math">\fraction{1}{|h|} \left| \int_x^{x+h} [f(t) - f(x)] \mathrm{d}t \right| < \fraction{1}{|h|} \left| \int_x^{x+h} \varepsilon \mathrm{d}t \right| = \varepsilon</div>
By definition, <span class="math">F'(x) = \lim_{h \rightarrow 0} \fraction{F(x+h) - F(x)}{h} = f(x)[/spoiler].

In case I failed, http://mathsprepa.free.fr/Mathsprepa0607/Cours/Chap3/C16.pdf (Theorem 16.1.2)

[Anonymous No.6590177]

>>6590010

it's that "simple". finding the total S area is just summing the infite amount of infinitely smalls rectangle of f(x)*dx (the infinitely small amount dx large and f(x) high) for x going from a point a to a point b.

you can understand it with Riemann sums:
http://fr.wikipedia.org/wiki/Somme_de_Riemann
(watch calculus line)

when n tend to infinite, the amount (b-a)k/n represents the x variation ("integral from a to b...") and the amount (b-a)/n represents the dx (very small width of the rectangle). it's why the limit of the sum is just equal to the integral of f.

[Anonymous No.6590251]

>>6590001
Here is some shitty stuff I made for explanations. It's not really a proof but you can use the idea to make one and I hope it will help you.
It's not detailed nor rigorous but I'm lazy.

[Anonymous No.6590285]

>>6590251
What is C and why did it suddenly become the lower bound?

[Anonymous No.6590321]

>>6590009
hey im the anon who wrote that, glad you found it useful c:

[Anonymous No.6590447]

>>6590160
This is not as strong a statement as the other proof though which uses the mean value theorem

You don't need f continuous you really only need f riemann integrable.

It is a more intuitive proof though.
Intuitively what that proof is saying, if F(x) is the area under the curve then F(x+h)-F(x) is approximately a rectangle of height f(x) and width h

So,
F(x+h)-F(x)~f(x)*h for small h.
Therefore, [F(x+h)-F(x)]/h~f(x) for small h

[Anonymous No.6590454]

>>6590001
>surface area
no.

[Anonymous No.6590485]

I like to think of it in terms of geometry. Integration is analagous to finding the area of a shape defined by a function. Differentiation is analogous to tracing a perimeter of that shape.

Another way to think of it is in terms of linear spaces. When you integrate, you are adding a dimension to a space spanned by a function. When you differentiate, you are projecting your function onto a lower dimensional space.

[mothergoose !!DUwV6rmVW4k No.6590496]

understand the formal definition of a limit, then the formal definition of a derivative, then the formal definition of an integral and it all should come together.