Math Core

So far each definite integral has produced a single number. Now let the upper limit move. The result is a function that keeps a running total of accumulated area, and it is one of the most tested ideas on the AP Calculus exam.

A running total

Picture filling a pool with a hose whose flow rate changes. At any moment you might ask, "How much water has come in since I started?" The answer depends on when you ask. That running total is an accumulation function.

Definition

Accumulation function

If $f$ is continuous and $a$ is a fixed number, the function

g(x) = \int_a^x f(t)\,dt

is an accumulation function. Its value $g(x)$ is the signed area between the graph of $f$ and the $t$ -axis from $t = a$ to $t = x$ .

Two notes on the notation:

The letter $t$ is a dummy variable. It only labels the horizontal axis while you sweep across; the answer depends on $x$ , not on $t$ . Writing $\int_a^x f(x)\,dx$ would use $x$ for two different jobs, so $t$ is used instead.
$g(a) = \displaystyle\int_a^a f(t)\,dt = 0$ . No width means no area.

If $x$ is to the left of $a$ , you sweep backward, and the area counts with the opposite sign: $g(x) = -\displaystyle\int_x^a f(t)\,dt$ .

Reading values from a graph

The graph of $f$ below is made of line segments on $[-2, 6]$ . Throughout this lesson, let $g(x) = \displaystyle\int_0^x f(t)\,dt$ .

The graph of f on [−2, 6]. It rises from (−2, 0) to (0, 2), stays at 2 until x = 2, falls to −2 at x = 4 (crossing zero at x = 3), then stays at −2.Open in grapher →

Worked example: Values of an accumulation function

Find $g(2)$ , $g(3)$ , $g(4)$ , $g(6)$ and $g(-2)$ .

Start at $x = 0$ and add up area as you move right.

$g(2)$ : the rectangle from 0 to 2 has area $2 \times 2 = 4$ . So $g(2) = 4$ .
$g(3)$ : add the triangle from 2 to 3 (base 1, height 2), area 1. So $g(3) = 5$ .
$g(4)$ : the triangle from 3 to 4 lies below the axis with area 1. So $g(4) = 5 - 1 = 4$ .
$g(6)$ : the rectangle from 4 to 6 lies below the axis with area $2 \times 2 = 4$ . So $g(6) = 4 - 4 = 0$ .

For $g(-2)$ , sweep left from 0. The triangle from $-2$ to 0 has area $\dfrac{1}{2}(2)(2) = 2$ and lies above the axis, but you're moving backward, so $g(-2) = -2$ .

The derivative of an accumulation function

How fast does $g$ grow? When $x$ moves a tiny bit to the right, by $\Delta x$ , the new sliver of area is almost a rectangle with height $f(x)$ and width $\Delta x$ . So

\frac{g(x + \Delta x) - g(x)}{\Delta x} \approx \frac{f(x)\,\Delta x}{\Delta x} = f(x).

As $\Delta x \to 0$ this becomes exact.

The accumulation function's rate is the integrand

If $g(x) = \displaystyle\int_a^x f(t)\,dt$ and $f$ is continuous, then

g'(x) = f(x).

The rate at which area accumulates is the current height of the graph. This is the first part of the Fundamental Theorem of Calculus, which the next lessons develop further.

That one fact lets you analyze $g$ with everything you learned about derivatives, using the graph of $f$ as the graph of $g'$ .

Feature of $f = g'$	Conclusion about $g$
$f(x) \gt 0$	$g$ is increasing
$f(x) \lt 0$	$g$ is decreasing
$f$ changes from positive to negative at $c$	$g$ has a relative maximum at $c$
$f$ changes from negative to positive at $c$	$g$ has a relative minimum at $c$
$f$ is increasing	$g$ is concave up
$f$ is decreasing	$g$ is concave down
$f$ has a relative extremum at $c$	$g$ has a point of inflection at $c$

Worked example: Analyzing g from the graph of f

For the graph above, on $[-2, 6]$ , find where $g$ is increasing, where $g$ has a relative maximum, where $g$ is concave down, and the absolute minimum value of $g$ .

$f \gt 0$ on $(-2, 3)$ , so $g$ is increasing on $(-2, 3)$ . It's decreasing on $(3, 6)$ .
$f$ changes from positive to negative at $x = 3$ , so $g$ has a relative maximum at $x = 3$ , where $g(3) = 5$ .
$f$ is decreasing on $(2, 4)$ , so $g$ is concave down on $(2, 4)$ . (It's concave up on $(-2, 0)$ where $f$ is increasing, and linear on $(0, 2)$ and $(4, 6)$ where $f$ is constant.)
For the absolute minimum, compare the endpoints and critical points: $g(-2) = -2$ , $g(3) = 5$ , $g(6) = 0$ . The absolute minimum value is $-2$ , at $x = -2$ .

Common mistake

The graph you're shown is the graph of $f$ , not $g$ . A relative maximum of $f$ is not a relative maximum of $g$ ; it's an inflection point of $g$ . The extrema of $g$ happen where $f$ crosses the axis.

Accumulation in context

In applications you usually start with some amount and add the accumulated change:

\text{amount at time } t = \text{starting amount} + \int_0^t (\text{rate})\,ds.

Worked example: Water in a tank

At time $t = 0$ a tank holds 40 gallons. Water flows in or out at a rate of $r(t) = 6 - 2t$ gallons per hour for $0 \le t \le 5$ (positive means in). Let $W(t) = 40 + \displaystyle\int_0^t r(s)\,ds$ . At what time is the amount of water greatest, and how much is there then?

$W'(t) = r(t) = 6 - 2t$ , which is positive before $t = 3$ and negative after. So $W$ has its maximum at $t = 3$ . The area under $r$ from 0 to 3 is a triangle with base 3 and height 6, area 9. So $W(3) = 40 + 9 = 49$ gallons.

Tip

To find where an accumulation function is largest or smallest, don't compute the function everywhere. Find where the integrand changes sign, then compare those candidates with the endpoints (the Candidates Test).

Practice

For Problems 2–6, use the graph of $f$ shown in the lesson and $g(x) = \displaystyle\int_0^x f(t)\,dt$ .

Practice 1

Let $h(x) = \displaystyle\int_0^x (4 - t)\,dt$ . Use geometry to find $h(6)$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 2

Find $g(5)$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 3

Find $g'(1)$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 4

Let $F(x) = \displaystyle\int_2^x f(t)\,dt$ . Find $F(0)$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 5

On which interval is the graph of $g$ concave up?

Practice 6

Which of the following statements is true?

Practice 7

A tank holds 40 gallons at $t = 0$ , and water flows at a rate of $r(t) = 6 - 2t$ gallons per hour for $0 \le t \le 5$ . Find the maximum amount of water in the tank on $0 \le t \le 5$ , in gallons.

Enter a number. Fractions like 3/4 and sqrt(2) are OK.