Accumulation Functions and the Fundamental Theorem, Part 1, calculus lesson

Picture water running into a tank at a rate that changes over time. The amount in the tank at time $x$ is the integral of the flow rate from the start up to $x$ . Ask how fast that amount is changing right now, and the obvious answer is the flow rate right now. That obvious answer is the first half of the fundamental theorem of calculus, and it ties integrals to derivatives.

Key terms

Accumulation function

Definition Let $f$ be integrable on every closed subinterval of an interval $I$ and let $a$ be a point of $I$ . The accumulation function of $f$ starting at $a$ is

g(x) = \int_a^x f(t) \, dt, \qquad x \in I.

The variable $x$ is in the bound, and $t$ is the dummy variable inside. Every such $g$ is continuous on $I$ , and $g(a) = 0$ .

In plain words Fix a starting point, move the right edge to $x$ and record the integral so far. As $x$ moves, the recorded number traces out a new function.

Example For $f(t) = t$ and $a = 0$ , the region from $0$ to $x \ge 0$ is a triangle of area $\frac{x^2}{2}$ .

$x$	$0$	$1$	$2$	$3$
$g(x)$	$0$	$0.5$	$2$	$4.5$

For $x < 0$ the reversed bounds rule gives

g(x) = -\int_x^0 t \, dt = -\left(-\frac{x^2}{2}\right) = \frac{x^2}{2}

too. So $g(x) = \frac{x^2}{2}$ for all $x$ .

Signed area and the shape of an accumulation function

Definition Let $g(x) = \int_a^x f(t) \, dt$ . For $x_1 < x_2$ in $I$ , interval additivity gives

g(x_2) - g(x_1) = \int_{x_1}^{x_2} f(t) \, dt.

So $g$ is nondecreasing on any interval where $f \ge 0$ and nonincreasing on any interval where $f \le 0$ . The value $g(x)$ itself is the signed area from $a$ to $x$ , with area below the axis counted negatively.

In plain words The accumulation function goes up while the integrand is above the axis and down while it is below. Its peaks occur where $f$ crosses from positive to negative.

Example Let $g(x) = \int_0^x \sin t \, dt$ on $[0, 2\pi]$ . Since $\sin t > 0$ on $(0, \pi)$ , $g$ increases there. Since $\sin t < 0$ on $(\pi, 2\pi)$ , $g$ decreases there. So $g$ is largest at $x = \pi$ . The negative hump on $[\pi, 2\pi]$ is the positive hump flipped, so it cancels exactly and $g(2\pi) = 0$ .

Derivative of an integral with variable upper bound

Definition (Fundamental theorem of calculus, part 1.) If $f$ is continuous on $[a, b]$ and

g(x) = \int_a^x f(t) \, dt \qquad (a \le x \le b),

then $g$ is continuous on $[a, b]$ , differentiable on $(a, b)$ , and

g'(x) = f(x) \qquad \text{for } a < x < b.

At the endpoints the matching one-sided derivatives also equal $f(a)$ and $f(b)$ . In particular every continuous function on an interval has an antiderivative there.

In plain words The rate at which the accumulated area grows is the height of the curve at the moving edge. Integrating and then differentiating gets you back where you started.

Example For $f(t) = t$ we found $g(x) = \frac{x^2}{2}$ , and $g'(x) = x = f(x)$ , as promised. The theorem also handles cases with no simple formula. For $g(x) = \int_0^x \sqrt{1 + t^3} \, dt$ with $x \ge 0$ ,

g'(x) = \sqrt{1 + x^3}.

Continuity at the evaluation point

Definition The theorem is local. If $f$ is integrable on $[a, b]$ and continuous at a particular point $c$ in $(a, b)$ , then $g(x) = \int_a^x f(t) \, dt$ is differentiable at $c$ and $g'(c) = f(c)$ . The reason is that for $h \neq 0$ small,

\frac{g(c + h) - g(c)}{h} = \frac{1}{h} \int_c^{c + h} f(t) \, dt,

which is the average of $f$ over a tiny interval at $c$ , and continuity at $c$ forces that average toward $f(c)$ . Without continuity at $c$ the conclusion can fail.

In plain words What matters is that $f$ is continuous at the point where you take the derivative. At a jump of $f$ , the accumulation function has a corner.

Example Let $f(t) = 0$ for $t < 0$ and $f(t) = 1$ for $t \ge 0$ , and $g(x) = \int_{-1}^{x} f(t) \, dt$ . Then $g(x) = 0$ for $x \le 0$ and $g(x) = x$ for $x > 0$ . The left derivative of $g$ at $0$ is $0$ and the right derivative is $1$ , so $g'(0)$ does not exist. At every $c \neq 0$ , $f$ is continuous and $g'(c) = f(c)$ .

Variable bounds and the chain rule

Definition Let $f$ be continuous on an interval $I$ , and let $u$ and $v$ be differentiable on an interval $J$ with all their values in $I$ . Then for every $x$ in $J$

\frac{d}{dx} \int_{u(x)}^{v(x)} f(t) \, dt = f\big(v(x)\big) \, v'(x) - f\big(u(x)\big) \, u'(x).

To see why, fix $a$ in $I$ , let $F(y) = \int_a^y f(t) \, dt$ , write the integral as $F(v(x)) - F(u(x))$ , and apply part 1 together with the chain rule.

In plain words Plug the upper bound into $f$ and multiply by its derivative. Then subtract the same thing for the lower bound.

Example For $\int_0^{x^2} 2t \, dt$ the formula gives $2x^2 \cdot 2x = 4x^3$ . A direct check agrees, since

\int_0^{x^2} 2t \, dt = (x^2)^2 = x^4,

whose derivative is $4x^3$ . A case with no formula is

\frac{d}{dx} \int_x^{x^3} e^{t^2} \, dt = e^{x^6} \cdot 3x^2 - e^{x^2}.

Check yourself

Find $\frac{d}{dx} \int_x^5 \sin(t^2) \, dt$ .

Show the answer

Reverse the bounds to get $-\int_5^x \sin(t^2) \, dt$ , whose derivative is $-\sin(x^2)$ .

Find $\frac{d}{dx} \int_0^{\sin x} \sqrt{1 + t^2} \, dt$ .

Show the answer

By the chain rule version, $\sqrt{1 + \sin^2 x} \cdot \cos x$ .

On which interval is $g(x) = \int_0^x (t - 1)(t - 3) \, dt$ decreasing?

Show the answer

By part 1, $g'(x) = (x - 1)(x - 3)$ , which is negative exactly for $1 < x < 3$ . So $g$ is decreasing on $[1, 3]$ .

Accumulation Functions and the Fundamental Theorem, Part 1

Theory, part 103

Practice, part 104

Key terms