Accumulation Functions (College Board AP® Calculus AB)

Study Guide

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Written by: Roger B

Reviewed by: Dan Finlay

Accumulation functions

What is an accumulation function?

An accumulation function is a function that outputs values which
- represent an accumulation of change
- over an interval from a given starting point
  - to a variable endpoint
- E.g. if a strawberry harvester gathers half a kilogram of strawberries for each meter of strawberry plants (rate of change = 0.5 kilograms per meter)
  - Then the accumulation function from the harvester's starting point,
  - to a point $x$ meters from that starting point,
  - is simply $0.5 x$
- Substituting in a value for $x$ gives you the amount of strawberries harvested up to that point

How do I write an accumulation function as a definite integral?

Most often you will see accumulation functions written as definite integrals:
$g (x) = \int_{a}^{x} f (t) d t$
- $g$ here is the accumulation function
  - $g$ is a function of x
    - Its value changes as the value of $x$ changes
- $f$ is the associated rate of change function
- $a$ is the (fixed) starting point of the integral
- $x$ is the (variable) ending point of the integral
- $t$ is merely a 'dummy variable' used for evaluating the integral
  - any letter except x can be used for the dummy variable inside the integral
    - This is true even if $f$ is defined as a function of $x$
    - A dummy variable must still be used inside the integral
- $g (x)$ is calculating the accumulation of change as $t$ goes from $a$ to $x$

A graph showing an example of a rate of change function and its associated accumulation function

Note here that a definite integral is being used to define a new function of x
- If $f$ is a function of $x$ defined by $f (x)$
- Then $g$ defined by $g (x) = \int_{a}^{x} f (t) d t$ is another function of $x$

Worked Example

Let $f$ be the function defined by $f (x) = 2 x + \sin x$ .

Let $g$ be the function defined by $g (x) = \int_{0}^{x} f (t) d t$ .

(a) Show that $g (x) = x^{2} - \cos x + 1$ .

Answer:

We need to integrate $f$ with respect to $t$ , and then evaluate the definite integral between $0$ and $x$

$\begin{array}{rcl} \int_{0}^{x} f (t) d t & = & \int_{0}^{x} (2 t + \sin t) d t \\ = & {[t^{2} - \cos t]}_{0}^{x} \\ = & {(x)}^{2} - \cos (x) - ({(0)}^{2} - \cos (0)) \\ = & x^{2} - \cos x - (0 - 1) \\ = & x^{2} - \cos x + 1 \end{array}$

Therefore $g (x) = x^{2} - \cos x + 1$

(b) Let $r$ be a quantity for which $f (x)$ is the rate of change function. Explain why $r (x)$ is not necessarily equal to $g (x)$ .

Answer:

$g (x)$ is an accumulation of change, and only tells us how much $r (x)$ changes between $0$ and $x$

It will only be equal to $r (x)$ if $r (0) = 0$

$g (x)$ tells us the change in $r$ between $0$ and $x$

To find the value of $r (x)$ we need to add $g (x)$ to the value of $r$ when $x = 0$ , i.e. $r (x) = g (x) + r (0)$

Last updated: 8 October 2024

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