Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °

You are not logged in.

## #1 2012-04-25 18:46:58

bobbym

Online

### Tricky Area

This recently came up in another thread.

http://www.mathisfunforum.com/viewtopic.php?id=17680

Let's see if geogebra could help out.

1)Enter f(x) = 5+2x-x^2.

2)Enter g(x)=2.

3) Use the intersection tool to find the intersection of g(x) and f(x).

Immediately A and B are created. That answers a) but we are going after the big fish in the lake, b)!

How can we use geogebra to estimate that area?

4) We use the polygon tool and start at A and click on the parabola 3 points, C, D  and E finishing up with B to A. See the first drawing. Check the area of poly1, I get 9.81555.

Can we do better with a polygon that has more vertices?

5) We again use the polygon tool and make 20 points on f(x) as the vertices. See the second drawing.

Reading the area of poly1 I get 10.625.

How about 40 points? I tried but could only fit 38 points, see the third drawing. When I read off poly1 I get 10.6567. That must be a good approximation of the area we want. Turns out it is very close to the exact answer.

In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

## #2 2012-05-01 07:30:00

bobbym

Online

### Re: Tricky Area

Hi;

Another one came up. We need the area of the red shape in the first picture.

1) Input f(x) = x+1. A straight line will be drawn.

2) Input g(x)=6x-x^2-3. A parabola will be drawn.

3) Get the points of intersection of the two equations using the intersection tool. The points will be labelled A,B.

4) Go into options, labelling and check no new objects. Also hide the labels of A and B.

5) Use the polygon tool and carefully click on A and then place as many points as you on that curve until you finish at B and then click A to complete the polygon. I managed 19 points on the curve. See the second picture.

6) Read off the value of poly1, I got 4.48347. That is a good estimate of the area of the red area.

Now geogebra has an incredible command that does this for us.

7) Input integral[f,g,x(A),x(B)] and immediately our area is hsaded darker and in the algebra pane you will see a new value of 4.5. That is the exact answer. I am quite close. How did you do?

In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

## #3 2012-05-01 08:07:07

anonimnystefy
Real Member

Offline

### Re: Tricky Area

Hi bobbym

Why don't you zoom in? That way you have "more" space to put more points. I think you can get a 100 even.

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

## #4 2012-05-01 08:26:47

bobbym

Online

### Re: Tricky Area

Yes, I left that for the readers to get.

In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

## #5 2012-05-01 08:30:06

anonimnystefy
Real Member

Offline

### Re: Tricky Area

Ok then.

I am thinking about making another one of these.

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

## #6 2012-05-01 08:33:08

bobbym

Online

### Re: Tricky Area

For this type of problem, Geogebra is more powerful than M.

In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

## #7 2012-05-01 08:34:32

anonimnystefy
Real Member

Offline

### Re: Tricky Area

How so?

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

## #8 2012-05-01 08:58:18

bobbym

Online

### Re: Tricky Area

In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

## #9 2012-05-01 19:29:43

anonimnystefy
Real Member

Offline

### Re: Tricky Area

Do you mean the integral function?

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

## #10 2012-05-01 19:31:28

bobbym

Online

### Re: Tricky Area

Yep, M has nothing like it.

In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

## #11 2012-05-01 19:36:14

anonimnystefy
Real Member

Offline

### Re: Tricky Area

Integrate[f(x),{x,a,b}] ?

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

## #12 2012-05-01 19:37:20

bobbym

Online

### Re: Tricky Area

That does not get the integral between two curves. Geogebra does that for us, M does not.

In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

## #13 2012-05-01 19:38:07

anonimnystefy
Real Member

Offline

### Re: Tricky Area

Why not just enter the difference of the two functions?

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

## #14 2012-05-01 19:40:47

bobbym

Online

### Re: Tricky Area

I did not say you can not get it, but the thinking has already been done for us
in Geogebra.

In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

## #15 2012-05-01 19:44:58

anonimnystefy
Real Member

Offline

### Re: Tricky Area

I agree. Geogebra has many cool and useful functions for something that is generally thought of as a graphing software.

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

## #16 2012-05-01 19:49:17

bobbym

Online

### Re: Tricky Area

Take what amberzak is doing in the other threads. Geogebra does those in under a minute!

In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.