AP Calculus AB and BC

Accumulation Function vs Definite Integral

Q: Do I use FTC part 1 or part 2 on this question?

If a variable limit appears and the question wants a derivative, that is part 1, and the answer is the integrand evaluated at the upper limit, times the derivative of that limit. If the limits are constants and the question wants a value, that is part 2, and you need an antiderivative.

Q: Why is the inside variable written as t instead of x?

To keep it apart from the x on the limit. The integration variable is consumed by the integral, so \int_0^x f(t)\,dt and \int_0^x f(s)\,ds are the same function of x. Writing \int_0^x f(x)\,dx makes one letter do two jobs, which is where sign and substitution errors start.

Q: What if the variable sits in the lower limit instead?

Swap the limits and pick up a minus sign, so \frac{d}{dx}\int_x^b f(t)\,dt = -f(x). If both limits move, the derivative is the top one processed by the chain rule minus the bottom one processed the same way.

A definite integral with constant limits is a single number, so its derivative is zero. An accumulation function puts the variable in the upper limit, so it assigns an output to every input x, which is what makes it a function, and differentiating it hands back the integrand.

Definite integral

Use when: Both limits are constants, so the question wants a value: an area, a net change over a fixed interval, or an average.

Accumulation function

Use when: The upper limit contains the variable, and the question asks you to differentiate it, say where it increases, or locate its maximum.

Side by side

	Definite integral	Accumulation function
Written as	$\int_a^b f(t)\,dt$	$g(x) = \int_a^x f(t)\,dt$
What it is	A number	A function of $x$
Its derivative	$0$ , because a constant has no rate of change	$g'(x) = f(x)$ wherever $f$ is continuous, by the first fundamental theorem
What it depends on	Nothing: the letter $t$ inside is a placeholder	The upper limit $x$ , and nothing else
Typical question	Evaluate it, or read a total change off a rate graph	Where is $g$ increasing, and where does it reach its maximum

A definite integral $\int_a^b f(t)\,dt$ has constant limits, so it collapses to one number, and the letter $t$ inside is a placeholder: rename it $s$ and nothing at all changes. An accumulation function $g(x) = \int_a^x f(t)\,dt$ puts the variable on top, so each choice of $x$ names its own definite integral. Assigning exactly one number to each $x$ is what makes it a function, and functions can be differentiated.

\text{for } f \text{ continuous on an interval containing } a \text{ and } x: \quad \frac{d}{dx}\int_a^x f(t)\,dt = f(x) \quad\text{and}\quad \frac{d}{dx}\int_a^{u(x)} f(t)\,dt = f\bigl(u(x)\bigr)\,u'(x)

Once $g'(x) = f(x)$ is in hand, every first and second derivative fact transfers to the graph of $f$ . Where $f$ is positive, $g$ increases. Where $f$ crosses from positive to negative, $g$ has a relative maximum. Where $f$ is itself increasing, $g$ is concave up. And $g(a) = 0$ for any integrand, because the interval has shrunk to a point.

The derivative that gets set to zero

The error is writing $\frac{d}{dx}\int_0^x f(t)\,dt = 0$ on the grounds that an integral is just a number. The limits decide. With a constant on top the result genuinely is a constant and its derivative genuinely is $0$ ; with $x$ on top the derivative is $f(x)$ at every point where $f$ is continuous, and that statement is the content of FTC part 1. The continuity hypothesis is doing real work: for the step function equal to $0$ for $t < 1$ and $1$ for $t \ge 1$ , the accumulation function $g(x) = \int_0^x f(t)\,dt$ exists for every $x$ but has a corner at $x = 1$ , so $g'(1)$ does not exist.

Frequently asked questions

Do I use FTC part 1 or part 2 on this question?

If a variable limit appears and the question wants a derivative, that is part 1, and the answer is the integrand evaluated at the upper limit, times the derivative of that limit. If the limits are constants and the question wants a value, that is part 2, and you need an antiderivative.

Why is the inside variable written as $t$ instead of $x$ ?

To keep it apart from the $x$ on the limit. The integration variable is consumed by the integral, so $\int_0^x f(t)\,dt$ and $\int_0^x f(s)\,ds$ are the same function of $x$ . Writing $\int_0^x f(x)\,dx$ makes one letter do two jobs, which is where sign and substitution errors start.

What if the variable sits in the lower limit instead?

Swap the limits and pick up a minus sign, so $\frac{d}{dx}\int_x^b f(t)\,dt = -f(x)$ . If both limits move, the derivative is the top one processed by the chain rule minus the bottom one processed the same way.

In the CED: Unit 6: Integration and Accumulation