P.Etc: Integral as Inverse of Tangent

Integral as Inverse of Tangent

The area under a curve, however the curve may go as long as it doesn't double-back then it's good for figuring the area between it and the $x$-axis, and is derived similarly to the tangent, as follows. The area between a function and the $y=0$ line ($x$-axis) is named the integral of that function.

First we define the area between a curve and the line which is the zero-ordinate for all $x$, on an interval of abscissa, to be some function, $F$, of the interval, $[x_1, x_2]$, and the curve, as $F(x_1, x_2, f)$. The change in $F$ for a given change $b$ in the interval defined, chosen at the upper end of the interval ($x_2$). So, to use the Greek letter Delta (for the root of difference $\delta \iota \alpha \phi \omicron \rho \alpha$ ) to symbolize the difference of the quantity: $$ \Delta F = (b)\times(f(x_2+b)) $$ This is the formula for an area which is a rectangle of width $b$ and height $f$, and as long as this increment is not at a point where the difference of $f(x_2+b)$, the height of the addendum area, and $f(x_2)$, the height for the adjacent $b$-width portion, is infinite then this construction is meaningful.

So, the change in the area per change in interval is proportional to that change, ($\Delta x = b$), and the difference between $f(x)$ and $y=0$ at $x=x_2$ (which is simply $f(x_2+b)$). Below, the area between the low-point and zero is constant and left out.

Plot of a curvacious function, on the abscissic interval, zero to one (ordinate range, or measurement range on the y-axis, is zero to 1.87),
with a sub-interval of shaded area, where zero is less-than x-sub-1, is less than x-sub-2, is less than one.

Details: A plot of a portion of a well known function which is up to fifth-power of x polynomial (a sum of three terms of the form coefficient times $x^n$) on the interval $[x_1, x_2]$.

The area between function $f$ and $y=0$ (the line with zero slope and zero $y$-intercept) over the bounded yet imprecisely defined interval, $[x_1, x_2]$, is the filled region.

Plot of function integral area filled with a foliage-shade of green, over an arbitrary sub-interval, x-1 to x-2, with a narrow, eggplant-filled rectangle, with the same height as curve, at the right-edge of the sub-interval, point x-2, with a rectangle-width parameter b.

A plot of the same function, but with a change to the area illustrated as a rectangle b-wide and f(x_2)-high.

If one were to add up $N$ such rectangles, whose sum of widths was the same quantity found in the interval, $[x_1, x_2]$, and then that sum is compared with the very similar sum, all things being equal except the total number of rectangles, in which the index maximum, $m$, is a doubling of $N$ along with the widths being halved, $(x_2-x_1)=(N)(N^{-1})(x_2-x_1)=(2N)(2N)^{-1}(x_2-x_1)$, one would find the sums are different. The sum of rectangle areas, with more rectangles over the given interval, is absolutely more accurate, but relatively it's insignificant, as the size of the neglected part contributing to the change in $F$ per change in $x_2$ is about $b^2\ll b \ll 1$. $$ \frac{\Delta F(x)}{\Delta x} = f(x) $$

The difference in area acquired by extending the interval boundary by $b$ is simply the height of the function at $(x_2+b)$ times the width, the differential parameter $b$. The declaration that the area is some analytical quantity, calculable from some infinite number of contiguous rectangles, without adding more constraints than exist, we can say the quantity of such area is, in general, a function of the following parameters.

$f(x)$, the curve making the top boundary,
$x_1$, the line going from the point $(x_1, 0)$ to point $(x_1, f(x_1))$, as a section of the entire vertical line $(x_1, y)$, making the left boundary,
$x_2$, defining the right boundary.

Where the fourth bounding line, the $x$-axis, is constant. So the area which is bounded between $f(x)$ and the $x$-axis, over the interval, $[x_1, x_2]$, can be solved by the differential equation, $D_x F = fb$, which is identical to the tangent formula for $f$, except now the tangent is $f(x)$ (compare to $D_x f(x)$ tangent formula). We can write, with a change of variable $x_2\to x$ while keeping $x_1$ constant, we write: $$ \lim\limits_{b\ll1}\frac{F(x_1, x_2+b)-F(x_1, x_2)}{b} $$ $$ =D_{x_2} F(x_1, x_2) = f(x_2) $$ So for arbitrarily small $b=\Delta x$, the ratio becomes the derivative of $F$ at $x_2$, and is equal to $f(x_2)$. This immediately tells us that the area function, $F$, is the anti-derivative of the function $f$.

The area which could be thought of as a (finite) series of the function in subject as kernel to a summation operating from the interval lower bound in subject to the intervals upper bound all terms multiplied by some differential parameter $k$, as the uniform width of the rectangle heights.

The $k$-width rectangles add up to some area equivalent to the area between the function $f(x)$ and the $x$-axis (the abscissa), for the given interval $[a, b]$. So, the change in that total area as a function of the upper bound of the interval $b$ is the height of the the curve at $b$ so $F(x+b) - F(x) = b f(x) $

For historical placement of this notion, the anti-derivative of $x^{-1}$ was solved by Gottfried Leibniz in 1676, first publisher of the formalism of tangents [1].

The example polynomial used above, $f$, is a truncated sine series, of four times the argument, to sixth degree in the Maclaurin series. And the common jargon today refers to the relationship between the integral and the derivative of a function, the Fundamental theorem of calculus.

[1] Scriba, Christoph J. (1963). The inverse method of tangents: A dialogue between Leibniz and Newton. Archive for History of Exact Sciences 2 (2):113-137.

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