Relationship between Area and the Antiderivative

Author:
cpeverley
Topic:
Area, Calculus
The Fundamental Theorem of Calculus, part (a) [pg. 319]: The function G defined by G(x) = the integral from a to x of f(t) dt, for x in [a,b] is an antiderivative of f on [a,b] This half of the FTC identifies a function G(x) as the area under the curve of a function f(t) between some constant a and x. This means that changes in the area under f(t) have an impact on the function (y) values of G(x). In the example below, the graph of f is an increasing linear function. If we're looking at positive x values starting at 0, then as x increases the area accumulates. You can move the point C (which is on the x-axis) to change the upper bound of the integral, and thus the area under f. According to the FTC, the function G(x) determined by the area under f as a function of x is AN antiderivative of f. It's hard to graph a "+C" function, so I arbitrarily decided that the y-intercept is -6 (marked E). Find the point H. If you've been moving around the point C, H has been marking it's path in green. The point H has coordinates (x,y), where x is the x-coordinate of C (which you control) and y is calculated by adding the area under the function of f up until x to -6. You can see that calculation around the coordinate (2,10). As you move C around, H will mark all points on a curve. This is the function G(x), an antiderivative of f with G(0)=-6. What function is that? Does that make sense with what you know about integrating? IGNORE THE OBJECTS SIDEBAR AND THE MENU, AND JUST LOOK AT THE PICTURE.