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/sci/ - Science & Math

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7129825 No.7129825 [Reply] [Original]

How do I get full intuition to why derivation and integration rules work the way they do?

>> No.7129827

Look up the proofs, and understand each step.

>> No.7129829

re-read definitions and study the applications

>> No.7129834

>>7129825
>rules

I fucking hate you, people.

>> No.7129842

>>7129825
What is mathematics by richard courant. The chapter on calculus.

>> No.7129848

You don't. You Plug and chug then get your engineering degree where you'll be making 100k like everyone else.

>> No.7129882

Integration cannot be proven without the concept of differentiation.

>> No.7129916

>>7129825
It's actually very simply. I usually struggle with proofs, but integration for simple curves is a piece of piss.
Try this for starters: http://en.wikipedia.org/wiki/Trapezoidal_rule

>>7129882
Wrong, integration is just area under the curve, it has nothing to do with derivatives at it's core.

>> No.7129949

>>7129916

>what is arc length

>> No.7129964

>>7129949
And how does that relate to simple curve examples, the kind used for proofs?

I'm talking like y=2x between 1 and 5, or something equally easy

>> No.7129967

>>7129825
>Understand how a limit works
>Use ground definition of a limit to derive derivation rules
>Use Riemann sums to derive integration rules

>> No.7129984

>>7129825

Learn infinitesimals

>> No.7130009

>>7129964

i'm not referring to simple curve examples.

it was more of a reply to your second comment.

>> No.7130013

>>7129825
Study the proofs for derivatives and integration.

>> No.7130866

>>7129825
https://www.youtube.com/watch?v=8pTEmbeENF4

>> No.7131379

>>7129916
it's impossible to determine the area under a curve without derivatives.

>> No.7131392

>>7131379
>it's impossible to determine the area under a curve without derivatives

But that's wrong. The function IS the derivative of the function of the area under it. Assuming that the integral constant is 0, of course.

>> No.7131398

>>7129848
>You don't. You Plug and chug then get your engineering degree where you'll be making 100k like everyone else.

nice meme mathfag, but, at least on my uni, they don't let you pass if you can't explain it. 50% of the grade is on problems and the other 50% you get if you answer the questions the professor gives you, in front of him. You need to understand it to pass, if you just start writing equations on the board he will drop you in a second

>> No.7131434

I can remember the moment in highschool when I fully understood how integration relates to differentiation. It was pretty awesome.

>> No.7131478
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7131478

Here's something that can help.
Notice that the function on top is the derivative of the function on the bottom. Also, notice that any value at a point on the function on the bottom correlates to the area under the top function. As soon as the area starts decreasing, so does the slope on the bottom function.
It's not very rigorous, but does provide some insight.

>> No.7131612

>>7131478
The function is not even derivable.

>> No.7131722

>>7131398
kek what's the point in that? it's work for the sake of work, an engineer has literally no reason to "understand" math.

>> No.7131747

>>7131612
differentiable

>> No.7131769

>>7131612
do you even Fourier series?

>> No.7131944

>>7129825

For me, the derivative becomes intuitive when I look at the definition this way:

dy/dx = lim (Δx -> 0) (Δy/Δx)

(Note that the definition of Δy is f(x+Δx)-f(x).)

The "dy/dx" looks very much like the "Δy/Δx", so it's easy for me to understand the derivative as a slope or a rate. Informally, just think of dx and dy as being "infinitesimally small" values of Δx and Δy respectively (due to the "lim").

-------

To show that the integral represents the area, it takes a little more work. It's hard to describe without a diagram, but I'll give it a shot:

Let's say you have a function f(x). Define another function F(x) that represents the area bounded by the x axis on the bottom and f(x) on the top, between two vertical lines x=a and x=b.

Now imagine a tiny sliver of that area that's bounded on the left and right by x and x+Δx respectively. Think of Δx as being very small.

You can calculate the area of that tiny sliver in two completely different ways:

(1) The tiny sliver is approximately shaped like a rectangle, with a height of approximately f(x), and a width of Δx. That means its area is approximately f(x) * Δx.

(2) The tiny sliver also has an area of F(x+Δx) - F(x).

Now, set (1) and (2) equal to each other, equating the two different ways you can represent the area of that tiny sliver:

F(x+Δx) - F(x) = f(x) * Δx

Now, divide both sides of the equation by Δx:

[ F(x+Δx) - F(x) ] / Δx = f(x)

Now, take the limit as Δx approaches 0:

lim (Δx->0) [ F(x+Δx) - F(x) ] / Δx = f(x)

Notice that the left side of the equation is simply the definition of the derivative of F(x), so substitute it:

dF/dx = f(x)

And finally, take the anti-derivative of both sides of the equation, with respect to x:

F(x) = ∫ f(x) dx + C

Now you can see that the area F(x) is the integral of f(x) -- plus or minus some arbitrary constant, because you can choose the left boundary of the area (x=a) to be anything.

>> No.7132583
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7132583

>>7131944

didn't read lmao

>> No.7133090

>>7131392
again, it is impossible to determine the area under a curve without the concept of derivatives.

>> No.7133110

>>7133090
still wrong.
Archimedes found areas under curves 1000 years before anything resembling a derivative existed

>> No.7133112

>>7133110
>Archimedes found areas under curves 1000 years before anything resembling a derivative existed
approximately
:^)

>> No.7133119

>>7133112
exactly actually.

Apostol teaches integrals before derivatives.
Most high schools teach rectangle methods which don't rely on derivatives.

>> No.7133230

>>7133119
Only for simple curves, though.

His methods more or less worked with using limits of geometric functions approaching infinity, but without defining the concept of limits. This didn't show any relations between the area beneath a function and how it relates to the function itself, other than that it was the area.