We suggest that you first explore the Second Fundamental Theorem of Calculus via Dr. Jackson's GeoGebra activity before proceeding. https://www.geogebra.org/m/cn2khzjf

In the App[br][br]Start by typing in any formula for a function [i]f[/i]([i]x[/i]) in the input box. If you check the f(x) checkbox in the right window the graph of [i]f[/i]([i]x[/i]) will appear in the right window in blue. [br]The formula for f '(x) is displayed, along with the graph of [i]f[/i] '([i]x[/i]) in red in the left window.[br]For now, toggle off the graph of [i]f[/i]([i]x[/i]) to clear out most of the right window.[br][br]Choose a value for [i]a[/i] via the slider or input box in the left window. Similarly pick a value for [i]x[/i].[br]Start the value of [i]x[/i] the same as the value for a and slowly slide the slider for [i]x[/i] to the right. You will see area accumulating between the graph of [i]f[/i] '([i]x[/i]) and the [i]x[/i]-axis. Green areas accumulate positively and red areas accumulate negatively. The green area minus the red area is the value of the accumulation function for that value of [i]x[/i]. [br][br]Click the checkbox for ([i]x[/i], [i]A[/i]([i]x[/i])) in the right window to see this value graphed there. Move [i]x[/i] around via the slider to see it change. Now click on the checkbox for [i]A[/i]([i]x[/i]) to see the graph. Again move [i]x[/i] around to investigate. [br][br]Now deselect ([i]x[/i], [i]A[/i]([i]x[/i])) to hide that portion of the illustration. Does the graph of [i]A[/i]([i]x[/i]) look familiar?[br][br]Select the checkbox for[i] f[/i]([i]x[/i]). How does the graph of [i]A[/i]([i]x[/i]) compare to the graph of [i]f[/i]([i]x[/i])?[br][br]Does they look like vertical shifts of each other?[br]Select the checkbox for Shift in the right window. This will show how far apart the two graphs are vertically for a particular[i] x[/i]-value. Move this point on the graph of [i]f[/i]([i]x[/i]) around. Does the vertical distance stay constant? How does this distance compare to [i]f[/i]([i]x[/i])?[br][br]You should see that, yes, the graphs of [i]A[/i]([i]x[/i]) and [i]f[/i]([i]x[/i]) are vertical shifts of each other and that the amount of the vertical shift is [i]f[/i]([i]a[/i]).[br][br]What does this tell use about an alternate way to express [math]\int_{^a}^xf\left(t\right)dt[/math]?[br][br]

(Fundamental Theorem of Calculus Part 1)[br][br]If f is any function differentiable on the interval including a and b and any points between them, then[br][math]\int_a^xf'\left(x\right)=f\left(x\right)-f\left(a\right)[/math].[br][br]First differentiating and then integrating produces the original function, possibly with a vertical shift.[br][br]One of the consequences of this is that if we are integrating a function [i]g[/i]([i]x[/i]) and we can find a function [i]f[/i]([i]x[/i]) so that [i]g[/i]([i]x[/i]) = [i]f[/i] '([i]x[/i]) (i.e. we find any antiderivative [i]f[/i] for the original function [i]g[/i]), then we can find an exact value for the definite integral by finding the total change in the antiderivative over the interval:[br][math]\int_a^bg\left(x\right)dx=\int_a^bf'\left(x\right)dx=f\left(b\right)-f\left(a\right)=\Delta f[/math].