The Fundamental Theorem of Calculus Part I says [math]f'(x)=\frac{d}{dx}\left [\int_{a}^{x} {f'(t)}{d}t\right ][/math]. This can be a little difficult to navigate, with the [math]x[/math] in the limits of integration and the [math]t[/math] as the variable of integration. This app is intended to help make this concept more concrete using a real-life example.[br][br]On the left is a graph of a [i]rate of change[/i] (derivative). It represents the [b][i]rate[/i] [/b], [math]g'[/math], at which a pump moves water. In this example, the pump starts pumping at a rate of 5 gallons per minute, but continuously slows down as it runs, dropping to 2 gallons per minute somewhere around 9 minutes later. Time for a new pump, I guess![br][br]The amount of water that it pumps in a short time interval [math]\Delta t[/math] is found as the flow rate [math]g'[/math] times [math]\Delta t[/math]. (Since [math]g'[/math] is changing during [math]\Delta t[/math], we would hold constant some value of [math]g'[/math] within [math]\Delta t[/math]). You can think of these as the height ([math]g'[/math]) and the width ([math]\Delta t[/math]) of a rectangle on the left graph, and the amount pumped during [math]\Delta t[/math] as the area of this rectangle. So, the total amount pumped between times [math]t=a[/math] and [math]t=x[/math] is the sum of the areas of many of these narrow rectangles. As the number [math]n[/math] of rectangles increases, they get narrower, and the approximation of the area gets better. Thus we have a Riemann sum, [math]\lim_{n\rightarrow \infty} \sum_{i=1}^{n} {g'(t)\Delta}t=\int_{a}^{x}g'(t)dt[/math].[br][br]On the right is plotted a graph of an [i]accumulated quantity[/i] (integral). It represents the [b][i]amount[/i][/b] of water [math]g[/math] pumped at the rate [math]g'[/math] since time [math]t=a[/math]. Let's fix [math]t=a[/math] at some point in time that we want to start measuring the amount of water pumped. Now, we let [math]x[/math] be the (arbitrary) point in time at which we want to know the total amount of water pumped since [math]t=a[/math]. Thus, the total amount of water [math]g[/math] pumped between times [math]a[/math] and [math]x[/math] is really a function [math]g(x)-g(a)[/math], calculated by [math]{g(x)-g(a)}=\int_{a}^{x}{g'(t)}{d}t[/math]. This is what's graphed at the right - it's really [math]\Delta g[/math].[br][br]So we've established the relationship between [math]g'[/math] (the derivative) and [math]g[/math] (the integral). Now let's see how this relates to the Fundamental Theorem.[br][br]If you actually solve the "definite integral" [math]\int_{a}^{x}{g'(t)}{d}t[/math], you would first find the antiderivative of [math]g'[/math] (which is [math]g[/math]), then plug in [math]x[/math] and [math]a[/math] and take the difference: [math]\int_{a}^{x}{g'(t)}{d}t=g(x)-g(a)[/math]. Notice that the app's default is [math]g(a)=0[/math] so that what's graphed is just [math]g(x)[/math]. You can use the [math]g(a)[/math] slider to change the value of [math]g(a)[/math], which has the effect of shifting the graph vertically. Now let's take the derivative of [math]g(x)-g(a)[/math]. Since [math]g(a)[/math] is a constant, its derivative is zero. And the derivative of [math]g(x)[/math] is just [math]g'(x)[/math]. Thus, [math]g'(x)=\frac{d}{dx}\left [g(x)-g(a)\right ]=\frac{d}{dx}\left [\int_{a}^{x} {g'(t)}{d}t\right ][/math].[br][br]You can drag the "[math]a[/math]" and "[math]x[/math]" points on the [math]t[/math]-axis at the left to see the effects on [math]g(x)[/math]. You can also type a new [i]rate[/i] function for the [math]g'[/math] graph at the left, and see the effects on the [i]amount[/i] function [math]g[/math] on the right. (The software requires using "x" as the variable in the input box). Click the circular arrows at the top right of the left graph to reset the app.