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Topic: What is a function #1 HS-Textbook 7th ed. : TRUE CALCULUS; without
the phony limit concept

Replies: 12   Last Post: Jul 13, 2013 2:38 AM

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 plutonium.archimedes@gmail.com Posts: 18,572 Registered: 3/31/08
integration of y=x, y=x^2, and y=1/x #7 HS-Textbook 7th ed.: TRUE
CALCULUS; without the phony limit concept

Posted: Jul 12, 2013 12:30 PM

integration of y=x, y=x^2, and y=1/x #7 HS-Textbook 7th ed.: TRUE
CALCULUS; without the phony limit concept

Now we use the 10-Grid system with its x-axis going from 0 to 0.1 then
0.2 then 0.3 on up to 10.0. Now it would be far easier to teach High
School students Calculus if I used whole numbers of 0, 1, 2, 3, etc
rather than 0, .1, .2, .3, etc. Because High School students have a
much rougher time of operating on fractions or decimals than they have
on whole numbers.

But the reason I want the High School student to stick with and apply
to fractions and decimals of 0, then 0.1 then 0.2 then 0.3 etc is
because Calculus is dependent on empty space between successive number
points. The student will not learn the True Calculus when we do
nothing but whole numbers and look at fractions and decimals as
inconvenience. Calculus is linked to finite versus infinity and the
infinity has two types, a macroinfinity, a large infinity and a
microinfinity, a small microscopic infinity which is the empty space
between the points of the graph. The fractions or decimals are
essential to understand the microinfinity.

Now let me do these three functions of y=x, y=x^2 and y=1/x of their
integrals and show the importance of empty space between points of the
graph, because without the empty space, the microinfinity we would not
be able to sum picketfences that have internal area. Fake Calculus,
that uses the phony limit concept, they are adding up line segments
which have no internal area.

For the function y = x we have this:

.    .    .    .    x
.    .    .    x   .
.    .    x   .    .
.    x   .    .    .
x   .    .    .    .
0  .1  .2  .3   .4

And we integrate from 0 to that of 0.3 so we add up the area in each
of those three cells. The first cell has the triangle of sides 0.1,
0.1 and its diagonal. The second cell has a square of side 0.1 and
another triangle the same as in the first cell, and finally the third
cell has a rectangle that is 0.1 by 0.2 or equivalently two of the
squares found in cell two, and another triangle the same as in cell
one and cell two. So if we add up all the squares of side 0.1 we have
three of those squares and if we add up the triangles we have three of
them and we combine two of them to form a square, and altogether we
have now 4 squares of side 0.1 and 1 triangle which is 1/2 the area of
a square. In total, we have 4.5 squares for a integral of 4.5, but now
we must realize that the squares involved are 0.1*0.1 which is equal
to 0.01 and there are four of them so we have 0.04 and then adding on
the single triangle whose area is 1/2 of 0.01 we have 0.005. Finally
we have 0.04 + 0.005 = 0.045.

Now we learned from the Antiderivative that the function y= x has an
integral of Int = 1/2x^2. You likely may have to review the
Antiderivative. Anyway, if we plug into that formula of 1/2x^2 for
that of 0.3 we have 1/2(0.3*0.3) = 1/2(0.09) = 0.045.

So, for the function y=x, its integral by picketfence summing is the
same as the integral by Antiderivative.

Now the function y= x^2 is a bit more complicated than y = x.

If the x-axis was this, then for the first four cells of y=x^2 looks
like this:

.    .    .    .    .

x
.    .    .    .    .
x
x
x   x    .    .    .
0  .1   .2  .3  .4   -->

x=0, y = 0
x=.1, y =.01
x=.2, y = .04
x=.3, y = .09
x=.4, y = .16

first cell has a tiny bit of area of 1/2(.1*.01) = .0005; second cell
has a triangle of 1/2(.1*.04) =.002; third cell has .04*.04 + 1/2(.1*.
05) = .0016 + .0025 = .0041; and fourth cell has
.09*.09 + 1/2(.1*.07) = .0081 + .0035 = approx .012. So the total area
is approximately .0005 + .002 + .0041 +.012 = .019.

Now earlier the Antiderivative technique gave us the integral of y=
x^2 as being Int = 1/3x^3. Now what happens if we plug in x=.4 into
that 1/3x^3 we have 1/3 (.4*.4*.4) = 1/3(.064) = .021. So it is a bit off and not
agreeing exactly with the summed cell area. The reason being is that we are in 10 Grid and as we go to 100 Grid or 1000 Grid or higher up we have the antiderivative come in closer agreement with cell sums.

The function y = 1/x looks like this:

.    x   .    .    .    .
.    .    .    .    .    .
.    .    .    .    .    .
.    .    .    .    .    .
.    .    .    .    .    .
.    .   x    .    .    .
.    .    .    .    .    .
???              x
.    .    .    .    .    .
? ?                    x
.    .    .    .    .    x
.    .    .    .    .    .
.    .    .    .    .    .
0  .1  .2  .3   .4  .5

I can graph this function in ascii art but better yet, I can list the results:

x= 0 y is undefined
x=.1, y= 10
x=.2, y= 5
x=.3, y =3.3
x=.4, y =2.5

Here there are only 3 cells involved since x=0 is undefined.
In the first cell we have a picketfence of (.1*5) + 1/2(.1*5) = .5+.25
= .75; second cell we have (.1*3.3) + 1/2(.1*1.7) = .33+ .085 = .415;
third cell we have (.1*2.5) + 1/2(.1*.8) = .25 + .04 = .29. For a
summation total of  .75 + .415 + .29 = 1.46

Now the function y= 1/x is a special function for it is the

Now let me talk about the integral of one more function, the broken function:

F(x) = 0 when x is even number and F(x) = 10 when x is odd number.

So the graph of this Sawtooth function in 10-Grid, 1st quadrant only,
looks like this:

.    x    .   x    .   x    .   x    .    x   . 10.0
.    .    .    .    .    .    .    .    .    .    .  9.9
.    .    .    .    .    .    .    .    .    .    .  9.8
.    .    .    .    .    .    .    .    .    .    .
.    .    .    .    .    .    .    .    .    .    .
.    .    .    .    .    .    .    .    .    .    .
.    .    .    .    .    .    .    .    .    .    .
.    .    .    .    .    .    .    .    .    .    .
.    .    .    .    .    .    .    .    .    .    .
.    .    .    .    .    .    .    .    .    .    .
.    .    .    .    .    .    .    .    .    .    .
out of scale for there should be 10 blocks of 10
.    .    .    .    .    .    .    .    .    .    .
.    .    .    .    .    .    .    .    .    .    .
.    .    .    .    .    .    .    .    .    .    .
.    .    .    .    .    .    .    .    .    .    .
.    .    .    .    .    .    .    .    .    .    .
.    .    .    .    .    .    .    .    .    .    .
.    .    .    .    .    .    .    .    .    .    .
.    .    .    .    .    .    .    .    .    .    .
.    .    .    .    .    .    .    .    .    .    .
.    .    .    .    .    .    .    .    .    .    .
x    .   x    .    x   .    x    .   x    .   x
0  .1   .2  .3  .4  .5   .6  .7  .8  .9  1.0  -->

And for the sawtooth function above, the integral is the summation of
those steep triangles in every one of those successive number points
of width 0.1 metric, for every cell has one steep triangle. So the
integral is the area of the triangle in the 0 to 0.1 interval added
with the triangle in 0.1 to 0.2 interval, and so on. In fact, the
integral of this sawtooth function is 1/2 of (10 by 10) or 50 square
units. For it is apparent that if we stack the triangles of one cell
with its successive cell we have a rectangle that is half the area of
10x10.

Many integrals will be the summation of triangles only, while many
others will be the summation of picketfences such as the y = x
function. Some functions will be the summation of pure rectangles
only for the integral, such as the function y= 3.

The functions, y= 1/x or y = x^2 will be picketfence integrals, since
each cell of 0.1 width will be partly a rectangle with a triangle
atop.

Archimedes Plutonium
http://www.iw.net/~a_plutonium
whole entire Universe is just one big atom
where dots of the electron-dot-cloud are galaxies