Can anyone explain integration?

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I am OK with differentiation, but integration beyond x squared gets a little tricky.

I mean all the numbers a re nice like x = x squared/2
and x2 = x cubed/3 etc

but the proof of this is a bit horrible as you continue to x cubed and beyond, although
the results are 'nice' is 4 to the 4th/4.

I would have thought there would be a nice easy mathematical explaination, that did
not require any 'fancy' maths, as the results are 'elegant', however the maths ain't
or at least it aint to me anyway.

Just thought I'd ask :smile:
 

Answers and Replies

352
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Do you know that the derivative of [tex]x^n[/tex] is [tex]nx^{n-1}[/tex]? If so, those integration formulas follow directly from this.
 
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Do you know that the derivative of [tex]x^n[/tex] is [tex]nx^{n-1}[/tex]? If so, those integration formulas follow directly from this.
Yes, what I mean is, it is easy to see that the area under y=x is half a square (x squared/2)
But not so easy to see the area under x squared is x cubed/3.

I can visualise what that would look like, like a cube sliced at two 45 degree angles, and it
would look like about 1/3 the size of a 'full cube', but it is hard to prove that visually
in the way you can do with a square, because it is basically 2 equal triangles.

Actually I was thinking about it just now and I think I might be able to do it for a cube,
but I am not sure, I might be able to visualise a cube split into 6 parts, 2 of which
would be what remains giving the 1/3 but I am not sure of this, I expect I am wrong.

I would like to see it expressed in a visual sense rather that a mathematical formula
sense, if you see what I mean. I mean it is easy to see that the iintegral of y=x is x2/2.
 
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It's going to be hard to think of it visually since we are talking about area of a plane region. So when you have the graph of a function you are talking about the area under the graph from one x to another x. Say, [tex] f(x) = 2x [/tex] from [tex]x = 0[/tex] to [tex]x = 2[/tex], then you can see that this is a straight line with slope of two. From [tex]x = 0[/tex] to [tex]x = 2[/tex] it is easy to see that the area under the line is 4 since this is a triangle and we take the area for this triangle. But using integration we can see: [tex]\int^2_0 2x dx | = x^2|^2_0 = ((2)^2 - (0)^2) = 4 [/tex].
 
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It can't be easily visualized, at least not that I can tell.

If it were easy to visually determine the area under a curve, no one would have needed to invent integration.
 
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It can't be easily visualized, at least not that I can tell.

If it were easy to visually determine the area under a curve, no one would have needed to invent integration.
Well it can be be visualised for y=1, and y=x, so why not y=x squared?
Actually I can visualise it for that but not easilly.
Y= x to the fourth is where things get tricky.
 
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Well it can be be visualised for y=1, and y=x, so why not y=x squared?
Actually I can visualise it for that but not easilly.
Y= x to the fourth is where things get tricky.
Have you gone through Riemann sums yet, in your discussion of integration? I certainly hope you were lead through the ideas behind them before being introduced to the fundamental theorems of calculus. Certainly, 'visualizing' the effects of integration in the normal sense becomes difficult after a power or two (or going into roots and other functions), but you should easily be able to think about summing all the infinitesimal areas of Riemann sums as the number of divisions goes to infinite (or as the limit of the widths, dx, goes to zero)
 
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Have you gone through Riemann sums yet, in your discussion of integration? I certainly hope you were lead through the ideas behind them before being introduced to the fundamental theorems of calculus. Certainly, 'visualizing' the effects of integration in the normal sense becomes difficult after a power or two (or going into roots and other functions), but you should easily be able to think about summing all the infinitesimal areas of Riemann sums as the number of divisions goes to infinite (or as the limit of the widths, dx, goes to zero)
Yes I have came across Rieman sums, but I am not really interested in that method.

I mean I can see it for:-
y=1
y=x
y=x.x ( not as obvious but it can be done)

You can do all these without any calculus whatsoever so it's not that much
of a stretch to say you can do the other higher powers without calculus, is it?
Especially as the result follows a very simple rule.
 
HallsofIvy
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Yes I have came across Rieman sums, but I am not really interested in that method.
Then you do not understand it. Knowing how to express an integral in terms of Riemann sums is the basic way you set up an integral to solve an application.

I mean I can see it for:-
y=1
y=x
y=x.x ( not as obvious but it can be done)

You can do all these without any calculus whatsoever so it's not that much
of a stretch to say you can do the other higher powers without calculus, is it?
Especially as the result follows a very simple rule.
 
84
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Then you do not understand it. Knowing how to express an integral in terms of Riemann sums is the basic way you set up an integral to solve an application.

OF course I know how to underatand it, you do that kind of stuff in primary school.
I am just not interested in doing it using integrals.
 
352
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You're trolling, right?

If there were a country where children learned Riemann integration in primary school I would move there.

[tex]\int dx[/tex] and [tex]\int x\,dx[/tex] are easy to visualize because they denote the area of a rectangle and a triangle, respectively. The graph of [tex]y=x^2[/tex] is not a simple geometric shape and so the area under the curve cannot be easily "visualized" and must be understood by an analytic proof using the definition of integration.
 
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You're trolling, right?

If there were a country where children learned Riemann integration in primary school I would move there.

[tex]\int dx[/tex] and [tex]\int x\,dx[/tex] are easy to visualize because they denote the area of a rectangle and a triangle, respectively. The graph of [tex]y=x^2[/tex] is not a simple geometric shape and so the area under the curve cannot be easily "visualized" and must be understood by an analytic proof using the definition of integration.
In primary school you calculate the area by count the squares under the line
on graph paper, that is the basis or Rieman.

You can visualise [tex]y=x^2[/tex] because it is a pyramid.
 
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Ok.........
You can say that the 'area' for for a power of x, is ((power of x) times x) but that if you
'plot' this from the origin, zero, you lose area in evey 'dimesnion' of x in equal value.
Thus the integral is x times that power of x (x to the n+1) divided by the number of powers +1. ie x(n+1)/(n+1).

So you can say that when you increase the power of x the total volume' would increase
by times x, but you would have an extra dimension by with to divide by because it would
also lose 'area' in that dimension too.

You see what I am thinking is that because the answer is simple and elegant the
explaination/proof should also be simple and elegant, not horrible and nasty.....like calculus :smile:


Do you get me? :rofl::approve:
 
352
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I learned about natural numbers in primary school too; does that mean I should pooh-pooh any math that uses them?

The graph of [tex]x^2[/tex] is certainly not a pyramid.

I don't understand your second post. Notably we're talking about area and I don't know why you're bringing volume into this.

The proof that [tex]a^n+b^n=c^n[/tex] has no solutions for [tex]n>2[/tex] is extremely long and complex. Simple result doesn't imply simple proof.

If you don't think going through the fundamental theorem of calculus to figure out the integral of [tex]x^n[/tex], there is a proof of it from the definition of an integral on page 66 of Apostol's "Calculus" (first edition) and probably in many other books as well.
 
Gib Z
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esbo, I am a bit confused about what you actually want? If its a way to find a nice friendly geometric interpretation in your head for the integral of powers of x, then all I can tell you is that there isn't one. Sure, y= c is some rectangle, and y= x gives some triangle, but I have no idea what you mean about what you think for a quadratic, and there isn't any more nice easy shapes any more afterwards. The best way to visualise an integral is summing rectangles, with the number of rectangles used being inversely proportional to the width of each rectangle. The integral is then the limit as the number of these rectangles goes to infinity.

If you think using Calculus in a proof for integrals is horrible and nasty, you are seriously misguided.
 
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HallsofIvy
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In primary school you calculate the area by count the squares under the line
on graph paper, that is the basis or Rieman.
If you believe that, then, as I said, you do not understand Riemann sums!

You can visualise [tex]y=x^2[/tex] because it is a pyramid.
Now you've got me! In what sense is x2 a pyramid? Archimedes could find the area long before Newton and Leibniz but he had to use a complicated "exhaustion" method.
 
341
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If you cannot visualize it, probably you are trying in the wrong context. Take for example, just for the fun of it...

[tex]\int^\pi_0{\tan\sqrt{\theta}d\theta}[/tex]

Why should you visualize this integral? Make something good today and think about visualizing complex integrals and the real solutions of them instead for example.

Also, if you want crazy visualizations I strongly recommend (even throw it on your head if don't read it) Tristan Needham's Visual Complex Analysis. It is a true gem for geometric thinkers
 
445
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So you are saying y(x)=4x^3 looks like a reasonable slope graph of y(x)=x^4?

You will not be able to see most of the results from calc up.
 
1,222
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Most of the posts in this thread so far have been very negative towards the op, but think that this is an excellent question worthy of attention. I also have no trouble understanding the question asked in the first post, and I am also not interested in judging someone who prefers to reason with geometry rather than analysis.

If its a way to find a nice friendly geometric interpretation in your head for the integral of powers of x, then all I can tell you is that there isn't one.
You will not be able to see most of the results from calc up.
The mathematical standard for claims of impossibility is no less than a rigorous proof. Even if it is the case that no purely geometric proof has been found (noting that the Archimedes method of exhaustion is essentially based on the limit concept), that doesn't mean it isn't worthwhile to search for one.

If you think using Calculus in a proof for integrals is horrible and nasty, you are seriously misguided.
Anyone can see that using analysis and the theory of integration to solve a problem of finding areas is like using a helicopter to travel across town. The act itself is simple, but the machinery required to support it is excessive for this application.
 
1,341
3
Yes I have came across Rieman sums, but I am not really interested in that method.

I mean I can see it for:-
y=1
y=x
y=x.x ( not as obvious but it can be done)

You can do all these without any calculus whatsoever so it's not that much
of a stretch to say you can do the other higher powers without calculus, is it?
Especially as the result follows a very simple rule.
As it's all ready been stated methods to find the area under all of theses curves was discovered. Archimedes's would as able to show that the area under the graph of x^2 was x^3/3. But I mean after that there really isn't a known way to geometrically show the areas of polynomial curves.
 
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445
3
You said you visualized them as a series of cubes. That is alluding to pixels. The thing with real numbers is that you can't pin them down to the last element, because they are everywhere dense.

There are ways to help us visualize higher dimensions, via projections and such. Even those are limited though. What I'm saying is sooner or later you will have to give up on a geometric interpretation for everything. It simply won't exist. You will need to trust logic and theorems.
 

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