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## #26 2012-04-25 17:13:25

bob bundy
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### Re: Calculus - Area of a curve

I saw it. I still do not get why the two small parts next to the rectangle don't mess it up.

You are not integrating between the two crossing points on the x axis.  Integration theory works out areas between two ordinates.  Drive those little areas from your mind completely.

you will get the area below the curve down to the x axis.  That's what area type integration does.
(Why is a much longer post for another day I think!)

If you ask,  I'll go through the 'why' for you.

Bob

You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

## #27 2012-04-25 17:24:47

anonimnystefy
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### Re: Calculus - Area of a curve

Oh,yes.I forgot that! Ok,thanks for the clarification,K5.

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

## #28 2012-04-25 19:00:58

amberzak
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### Re: Calculus - Area of a curve

I'm going to ask why Bob. Because I always want to know why.

Don't think outside the box. Think there is no box

## #29 2012-04-25 19:46:04

anonimnystefy
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### Re: Calculus - Area of a curve

Hi Bob

Look at his picture on page 1.He integrates the function from -1 to 3,which gives him the area needed plus the green rectangle.The two small areas on the sides of the triangle are not included in the integration.

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

## #30 2012-04-25 22:02:07

bob bundy
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### Re: Calculus - Area of a curve

hi amberzak and Stefy,

I think what amberzak is after is the explanation of why the reverse of differentiation is also the area under a curve.  Anyway, I hope so, because that's what you are going to get.

preamble.
Many calculus courses introduce integration as the reverse of differentiation.  Now, of course, that's going to be how you do your integration in practice.  You'll start by learning some rules for differentiation and then learn how to 'work backwards' when you are asked to integrate.  ie.  What function would differentiate to give me this?

However, for the best understanding of calculus, I think that definition is unhelpful.

I think it is best to think of integration as a summation process.  That's reallly what that integral sign means; it's a corrupted S for Sum.

And you can use integration to sum lots of things such as volumes, fuel used as a space rocket launches, moment about an axis leading to centres of gravity and so on.

But, at some stage, you need to know how to actually integrate something, and that's when it helps to know you can just reverse differentiation.

When I was at school my teacher told us that the proof that reverse of D and area under are the same, was too difficult for us.  It wasn't.  About 10 years later it was introduced to the A level course I was teaching at the time, so I had to find out.  Here's what I learnt.

Proof.

see the two graphs below.

f(x) is any differentiable function.

F(x) is made from f(x) as follows.

Calculate the area bounded by the y axis, the x axis, the curve f(x) and the vertical line through any point on the x axis.

This area will be the value of F(x) so you can construct the graph by plotting all points (x, F ) and joining them up.

Thus

So F is made by just summing the areas of all the infinitely thin rectangles with height f(x) and width dx.

Now, on the f(x) graph I have shaded an area in yellowy orange to indicate the area under f up to a vertical line x.

Then I have added a thin strip in red , width h (delta x) , for the next little bit of area to be added.

I've used h on the diagram because getting a delta x is a bit tricky on this diagram.

So how big is this extra area?  Well it is smaller than a rectangle width h and height f(x+h).

And it is bigger than a rectangle width h and height f(x)

Thus

or

so

so let delta x tend to zero

so the differential gets sandwiched between two values that approach each other and in the limit

So differentiation is the reverse of integration.

Bob

You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

## #31 2012-04-26 02:43:14

anonimnystefy
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### Re: Calculus - Area of a curve

Shouldn't it be less than or equal to in the inequalities?

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

## #32 2012-04-26 03:18:11

amberzak
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### Re: Calculus - Area of a curve

Thanks Bob. That was very interesting (and I actually understood what you were saying).

Don't think outside the box. Think there is no box

## #33 2012-04-26 04:08:58

bob bundy
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### Re: Calculus - Area of a curve

Shouldn't it be less than or equal to in the inequalities?

Yes it should.  And the diagram only works when f is increasing ..... so it's not a fully rigorous proof.

amberzak:  I'm really pleased you followed it.

Bob

You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

## #34 2012-04-26 05:10:33

anonimnystefy
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### Re: Calculus - Area of a curve

Hi Bob

Since the diagram has only clarification purposes,I don't think it affects the rigorousness of the proof.

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment