What is the connection between the area under a function curve and the antiderivative of that function?

[list=1][*]When the app is first started (or reset), the function [math]g\left(x\right)[/math] is shown (in [b][color=#ff0000]red[/color][/b]). [math]g[/math] is defined to be [math]g\left(x\right)=\int_a^xf\left(t\right)dt[/math]. This means that the value of [math]g[/math] is the same as the area under [math]f[/math] between a fixed point [math]t=a[/math] and another point [math]t=x[/math].[br][br][/*][*]By the FTC, [math]g'\left(x\right)=f\left(x\right)[/math]. Check the "Show f(t)" box to display [math]f[/math], the derivative of [math]g[/math] (in [b][color=#0000ff]blue[/color][/b]).[br][br][/*][*]You can drag the point on the black cursor on the function [math]g[/math] (red) to analyze its [i]derivative[/i] [math]f[/math] (blue), such as finding sign changes of [math]f[/math] at extrema of [math]g[/math].[br][br][/*][*]We can also use a function graph [math]f[/math] to analyze its [i]antiderivative [/i][math]g[/math].[br][br][/*][*]Check the "Show FTC" box. [b]Do not move point "[color=#ff0000]a[/color]" at this time[/b].[br][br][/*][*]Drag the point [b][color=#1e84cc]x[/color][/b] so that it coincides with [b][color=#ff0000]a[/color][/b]. What is the area under [math]f[/math] between [b][color=#ff0000]a[/color][/b] and [b][color=#1e84cc]x[/color][/b]? What is the value of [math]g[/math] at this point?[br][br][/*][*]Drag [b][color=#1e84cc]x[/color][/b] to various points left and right of [b][color=#ff0000]a[/color][/b]. [b]The signed area under [/b][math]f[/math][b] between [color=#ff0000]a[/color] and [color=#1e84cc]x[/color] is the [/b][math]y[/math][b]-value of [/b][math]g[/math][b] at [color=#1e84cc]x[/color][/b]. This is because [i]the area under the graph of a function's rate of change [/i]on an interval (for example, velocity) gives [i]the amount of change of the function[/i] (position) on that interval. For our example, we could say that [math]\int_a^xf\left(t\right)dt=\Delta g[/math]: the [i]area [/i]under [math]f[/math] from [b][color=#ff0000]a[/color][/b] to [b][color=#1e84cc]x[/color][/b] is equal to the [i]change in the value[/i] of [math]g[/math] from [b][color=#ff0000]a[/color][/b] to [b][color=#1e84cc]x[/color][/b].[br][br][/*][*]Now leave [b][color=#1e84cc]x[/color][/b] fixed, and move the point [b][color=#ff0000]a[/color][/b] along the x-axis. What do you observe? This happens because [i]changing the value of [b][color=#ff0000]a[/color][/b] changes the constant of integration of the function [/i][math]g[/math].[br][/*][/list][br][math]g(x)[/math] is an antiderivative of [math]f(x)[/math]. [math]g[/math]'s [math]y[/math]-value at any value of [b][color=#1e84cc]x[/color][/b] is equal to the area under [math]f[/math] between [b][color=#ff0000]a[/color][/b] and [b][color=#1e84cc]x[/color][/b], which is [math]\int_{a}^{x}f(t)dt[/math] . So both [b][color=#ff0000]a[/color][/b] and [b][color=#1e84cc]x[/color][/b] affect the value of [math]g(x)[/math]. In this sense, [math]g\left(x\right)=\int_a^xf\left(t\right)dt[/math] is a function of [b][i][color=#1e84cc]x[/color][/i][/b], and we write [math]g\left(x\right)=\int_a^xf(t)dt[/math]. Notice the use of the variable [math]t[/math] in the integral. Using the value of [b][color=#1e84cc]x[/color][/b] as a limit of integration gives it a different role than the variable of integration, so we have to use a different variable inside the integral. As we know, however, the variable name we choose makes no difference in a definite integral. We just can't use the same variable for two different purposes.[br][br]If we fix [b][color=#1e84cc]x[/color][/b] and change [b][color=#ff0000]a[/color][/b], we again change the area between them by a certain amount. (The amount is the area between the old and new positions of [b][color=#ff0000]a[/color][/b]). This amount added to [math]g(x)[/math] shifts the graph up or down. Thus, [b][color=#ff0000]a[/color][/b] determines the value of the "[math]+C[/math]" in the antiderivative.[br][br][br][math]g(x)[/math] is "an" antiderivative of [math]f(x)[/math], whose [math]+C[/math] value depends on [b][color=#ff0000]a[/color][/b]. So no matter what the value of [b][color=#ff0000]a[/color][/b] is, the [i]derivative[/i] of [math]g(x)[/math] is just [math]f(x)[/math], since the derivative of [u][i]any [/i][/u]value of [math]C[/math] is zero. This leads to the following equation, which is the Antiderivative Part of the Fundamental Theorem of Calculus:[br][br][math]\frac{\mathrm{d}}{\mathrm{d}x}g(x)=\frac{\mathrm{d}}{\mathrm{d}x}\int_a^xf(t)dt=f\left(x\right)[/math]