# The Accumulation Function and the FTC

- Author:
- Ken Schwartz

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

- When the app is first started (or reset), the function
is shown (in **red**).is defined to be . This means that the value of is the same as the area under between a fixed point and another point . - By the FTC,
. Check the "Show f(t)" box to display , the derivative of (in **blue**). - You can drag the point on the black cursor on the function
(red) to analyze its *derivative*(blue), such as finding sign changes of at extrema of . - We can also use a function graph
to analyze its *antiderivative*. - Check the "Show FTC" box.
**Do not move point "a" at this time**. - Drag the point
**x**so that it coincides with**a**. What is the area underbetween **a**and**x**? What is the value ofat this point? - Drag
**x**to various points left and right of**a**.**The signed area under****between a and x is the****-value of****at x**. This is because*the area under the graph of a function's rate of change*on an interval (for example, velocity) gives*the amount of change of the function*(position) on that interval. For our example, we could say that: the *area*underfrom **a**to**x**is equal to the*change in the value*offrom **a**to**x**. - Now leave
**x**fixed, and move the point**a**along the x-axis. What do you observe? This happens because*changing the value of***a**changes the constant of integration of the function.

**x**is equal to the area under

**a**and

**x**, which is

**a**and

**x**affect the value of

**, and we write**

*x***x**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. If we fix

**x**and change

**a**, we again change the area between them by a certain amount. (The amount is the area between the old and new positions of

**a**). This amount added to

**a**determines the value of the "

**a**. So no matter what the value of

**a**is, the

*derivative*of

__value of__

*any*