Here is an opportunity to place a patch on a function that is undefined for x=0 since it is division by 0. So one patch would be to use 10.01 since it is a infinity number for the 10 Grid, or we can use 10.001. Or perhaps we can go to the first cell that is defined and see how to extrapolate what x= 0 would be.
In the first defined cell of x=.1 to x =.2 we have the cell wall to the left as y=10 and for the rightmost wall we have y = 5, for a drop of 5 in that cell. So if we extrapolate that 5 unit drop in the first defined cell, what if we make the x=0 to x =.1 cell a drop of 5 units. That would mean, for x=0, y = 15 (an infinity number in 10 Grid). So there is flexibility in applying a patch to get rid of discontinuity of undefined y.
Now let us continue with the integral of y = 1/x, starting with x = 1 and moving rightwards from x= 1.
I can graph this function in ascii art rather badly as shown above, but better yet, I can list the results:
Starting at x = 1 and using 100 and 1000 Grid for answers x= 1 y =1 x=1.1, y= .909 x=1.2, y= .833 x=1.3, y =.769 x=1.4, y = .714 x=1.5, y =.666 x=1.6, y = .625
So our first cell is x = 1 to x = 1.1 and we plot the leftside wall of the cell with y=1 and the rightward wall of the cell at y=.909. So we have a rectangle of at least base .1 and height of .909 for area of .1x.909 = .090, but we still have a triangle area to deal with. The triangle is .1 base and height is 1-.909 = .091 so we have an area of that triangle as 1/2(.1x.091) = .004. So we add the area of rectangle with triangle, the picketfence, and get .090 + .004 = .094.
Now the function y= 1/x starting x = 1, is a special function for it is the logarithmic function (ln) as integral.
So we ask what is the ln x=1.1 and it is equal to .095 which is extremely close to .094.
Now let us go to the next cell of x=1.1 to x=1.2 and we have a leftside wall of .909 and a rightward wall of .833 which gives a rectangle of .1 wide and .833 high for an area of .1x.833 = .083. Now the area of triangle is .1 wide and height of .909 -.833 = .076, so the area of triangle is 1/2(.1x.076) = .003. So totaling rectangle and triangle of that picketfence we have a full area of .083+.003 = .086. Now we look up ln(1.2) and we get ln(1.2) = .182 but we must subtract the cell prior of .095 so we have .182-.095 = .087 which is close to .086
Now for the third and last cell we will discuss is from x= 1.2 to x=1.3, and the rectangle area is .1x.769 = .076 and the triangle area is 1/2(.1x(.833-.769))= 1/2(.1x.064)= .003. So the area of picketfence is .076+.003 = .079. Here again the ln(1.3) = .262 but we must subract the prior picketfence of .182 and we end up with .262 - .182 = .08 which is close to .079.
So we see here that we can manually calculate the integral area and check it against the integral function of the area.
Now, the only decent search for AP posts on Google, is a author search which eliminates hate spam:
However, tonight, it appears that all Google archive searches are absent. So that Drexel is the only author search around. The only problem with Drexel is that it is math only and the bulk of my posts are other sciences. Maybe the out of order at Google is temporary, maybe not.