The FTC and Accumulation Functions — AP® Calculus AB Notes

So far a definite integral has been a single number: the area under $f$ from $a$ to $b$ . Now let the right end move. The area then depends on where you stop, so it is a function. Studying that function leads to the most important theorem in the course.

Accumulation functions

Fix a starting point $a$ and define

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

For each $x$ , $g(x)$ is the amount accumulated under $f$ from $a$ up to $x$ . This is an accumulation function. The variable inside is renamed $t$ so it does not clash with the $x$ in the limit. Two facts follow at once:

$g(a)$ $= \int_a^a f(t)\,dt$ $= 0$ : no width, no area.
Values of $g$ come from areas. If $f$ is a graph of lines and arcs, use geometry.

Worked example 1. $f$ is made of segments from $(0, 1)$ to $(2, 3)$ , from $(2, 3)$ to $(4, 3)$ and from $(4, 3)$ to $(6, 0)$ . Let $g(x)$ $= \int_0^x f(t)\,dt$ . Find $g(2)$ , $g(4)$ and $g(6)$ .

$g(2)$ : the trapezoid from 0 to 2, $\frac{1 + 3}{2} \cdot 2 = 4$ .

$g(4)$ : add the rectangle from 2 to 4, $3 \cdot 2 = 6$ . So $g(4) = 10$ .

$g(6)$ : add the triangle from 4 to 6, $\frac{1}{2}(2)(3) = 3$ . So $g(6) = 13$ .

How fast does the area grow?

Top: the graph of f(t) = 0.5t + 1 on [0, 6], with the area under it from 0 to x = 3 shaded; that area is g(3) = 5.25 and the height f(3) is 2.5. Bottom: the accumulation function g(x) = x squared over 4 + x, with the point (3, 5.25) and a dashed tangent line there of slope g prime of 3 = f(3) = 2.5. — **Figure 1.** The area under f up to x is g(x); the slope of g at x is the height f(x). LevlPrep original diagram.

Move $x$ to the right by a tiny amount $h$ . The area $g$ gains a thin strip: width $h$ , height about $f(x)$ . So

$g(x + h) - g(x) \approx f(x)\cdot h, \quad\text{so}\quad \frac{g(x + h) - g(x)}{h} \approx f(x).$

As $h \to 0$ the approximation becomes exact, because $f$ is continuous. The left side becomes the derivative $g'(x)$ . That is the Fundamental Theorem of Calculus (part 1):

If $f$ is continuous, then $\frac{d}{dx}\int_a^x f(t)\,dt$ $= f(x)$ .

In words: the rate at which area accumulates is the height of the graph. The figure shows it for $f(t) = 0.5t + 1$ : at $x = 3$ the height is 2.5, and the slope of $g$ there is 2.5.

A function $F$ with $F' = f$ is called an antiderivative of $f$ . So the theorem says: every accumulation function of a continuous $f$ is an antiderivative of $f$ . Even a function with no formula for its area, like $e^{-t^2}$ , has antiderivatives: $\int_0^x e^{-t^2}\,dt$ is one.

The lower limit $a$ does not appear in the answer. Changing it changes where $g$ is 0, which shifts the whole graph of $g$ up or down, but not its slope. Writing $f(x) - f(a)$ is a common mistake.

Reading derivatives of g from the graph of f

$g(x)$ is an area under $f$ .
$g'(x) = f(x)$ : read the height of $f$ .
$g''(x) = f'(x)$ : read the slope of $f$ . So $g$ is concave up where $f$ is increasing and concave down where $f$ is decreasing.

With the graph in Worked example 1: $g'(1) = f(1) = 2$ , and $g''(3) = f'(3) = 0$ because $f$ is flat on $(2, 4)$ .

A function as the upper limit: add the chain rule

If the upper limit is $u(x)$ instead of $x$ , then $\int_a^{u(x)} f(t)\,dt$ is $g(u(x))$ , a composition. The chain rule gives

$\frac{d}{dx}\int_a^{u(x)} f(t)\,dt = g'(u(x))\,u'(x) = f(u(x))\,u'(x).$

Worked example 2. Find $h'(x)$ for $h(x)$ $= \int_0^{x^3} e^{-t^2}\,dt$ .

Integrand: $f(t) = e^{-t^2}$ . Upper limit: $u = x^3$ , so $u' = 3x^2$ .

Evaluate the integrand at the upper limit: $f(x^3)$ $= e^{-(x^3)^2}$ $= e^{-x^6}$ .

Multiply by $u'$ : $h'(x) = 3x^2 e^{-x^6}$ .

Two things are easy to miss: the factor $3x^2$ , and putting $x^3$ (not $x$ ) into the integrand.

Accumulation with a starting amount

In context, an amount often equals a starting value plus an accumulation: $W(t)$ $= 200$ $+ \int_0^t r(s)\,ds$ liters of water. Then $W(0) = 200$ , and by the theorem $W'(t) = r(t)$ : the amount changes at the given rate, exactly as it should.

The theorem has a second part, about evaluating $\int_a^b f(x)\,dx$ with an antiderivative. It comes in topic 6.7, after you have learned more about accumulation functions and the properties of integrals.