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.
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 logarithmic function. And you will learn more about this special function in the University textbook.
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:
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.