Let me first give what this Antiderivative rule or technique (sometimes called the Power formula) is. Let me give it to the High School student, so that they know up front what he or she is aiming for.
It is all a matter of geometry of angles involved with coordinate points and the area under the graph.
Antiderivative Technique (Power ?formula)
(1) for the derivative of a function x^n the derivative is n(x)^(n-1).
(2) for the integral the antiderivative works backward. So for x^n, the antiderivative is (1/(n +1)(x^(n+1))
The High School student is aiming for that rule listed above.
The High School student is going to work with the function y=x to derive that rule.
So graph the function in 10 Grid of integers only which looks like this of a diagonal that cuts the graph in half at a 45 degree angle:
So the derivative is the slope or the dy/dx and since the dy equals the dx, we have a derivative of 1. So how do we make a rule which starts from y = x and ends up with a derivative denoted as y', so that y'= 1? How do we get from y = x to y'=1?
How do we obtain the rule of n(x)^(n-1) starting from x^n, where the n in our case is n=1 for y = x ?
Well, y=x is really that of y=x^1 and what we want is x^1 be that of 1. So we take the exponent of 1 to be a coefficient, and then we take the function of x^1 and subtract 1 from the exponent, of x^(1-1) giving us x^0 and in math, x^0 =1.
So we have 1 * x^0 which is 1*1 or just 1.
So for y=x the derivative is 1(x)^(1-1) which is x^0 which is 1.
For the integral in the graph we easily add up 1/2 squares and whole squares to find the area under the graph. So for x= 3 we have 3 full whole squares and then we have 3 half squares for a total of 4.5 squares area. And the antiderivative formula for integral is 1/2x^2, so here the x=3 and if we plug into that formula we see we have 1/2(3*3) = 1/2(9) = 4.5 square area.
We instantly see that the graph of the function y= x is an expanding 1/2 of an ever expanding square with a 1 by 1 square, then a 2 by 2 square then a 3 by 3 square etc etc. So the Integral formula for y = x is that of 1/2x^2 because we all know from geometry that a right triangle area is 1/2 base times height and our right triangles have bases and heights that are equal to each other, so we have 1/2x^2. So how in mathematics do we go from y=x to a function of Int = 1/2x^2? (I abbreviate the derivative function as y' and I abbreviate the integral function as Int.)
We do it by having the power formula for x^n, the coefficient be 1/(n +1). So that for y=x^1, the coefficient is 1/(1+1) which is 1/2. And then the remainder of the power formula of x^n we increase the exponent by 1. In integration we increase the exponent by 1 whereas in differentiation we decrease it by 1.
For y=x the integral we have the coefficient as (1/(1+1)), and we have x^(1+1) which when combined is 1/2x^2. Now to see if that is correct we take the derivative of that to see if it lands us back to the original function y= x. So we have 2(1/2)x^(2-1) which surely is x.
Now, the bright High School student is going to wonder why I used y=x and not used y=x^2 for the antiderivative. And the answer is complicated for the function y=x^2 is far more complicated in using 0.1 and 0.2 than is y=x which does not involve fractions. But also, because True Calculus has no curves, only tiny straightline segments, and in the 10-Grid, my tiny straightline segments are only approximately the derivative and integral for y = x^2, whereas in the function y=x, it is a overall straightline function. The function y=x^2 is not an overall straightline function but rather a function of a combined tiny straightline segments. That is an important statement in True Calculus. Some functions are overall straightline functions such as y=3 or y=x, but many functions are overall a combination of tiny straightline segments such as y = x^2 or y= 1/x or y = sine x or y = cosine x.
Calculus without the limit concept has no curves, but rather has tiny straightline segments and when those segments are 1*10^-603 of a distance, the approximations become equality. At that small of a distance of 10^-603 we have a 10^603 regular polygon equalling a circle, where no curves exist in Euclidean Geometry.
Calculus with the limit concept would have you believe that the integral is the summation of line segments which have no internal area. And where the derivative is always running into trouble because its neighboring points are an infinity set of points that lead to what is described in math as pathological functions like the Weierstrass function or the function y = sin(1/x).
In True Calculus, the derivative itself connects a point with its immediate neighboring point on the leftside of the graph and its immediate neighboring point on the rightside of the graph. In the phony calculus of limit concept, the derivative was never a part of the function, but in True Calculus, the derivative is a part of the function itself for it connects successive points, and since it is a straightline segment then there are no curved lines in Calculus, nor in Euclidean Geometry.
Now what do I teach next? Do I teach the integral? Or do I teach the importance of the angle involved in Calculus. What I mean is that the derivative is the angle involved between successive points of the graph, so that in the function y=x, the derivative is a 45 degree angle between successive points of the graph. So how is 45 degrees expressed as y' = 1 for the function y=x? So, what is the logical order for the next discussion? Do I do the integral and integration, or do I talk about the huge importance that coordinate points of a graph in the Cartesian Coordinate System of Euclidean Geometry allows the Calculus to exist because of fixed angles? Let me ponder which is the better logical order. --
More than 90 percent of AP's posts are missing in the Google newsgroups author search archive from May 2012 to May 2013. Drexel University's Math Forum has done a far better job and many of those missing Google posts can be seen here: