The Fundamental Theorem of Calculus: Part 1 vs Part 2

The two statements, side by side

Both halves of the Fundamental Theorem of Calculus (FTC) say the same big thing: differentiation and integration undo each other. They run that relationship in opposite directions, which is why they get used for completely different tasks.

The evaluation form (AP CED Topic 6.7, essential knowledge FUN-6.B.3) is the one you use to compute integrals. If $f$ is continuous on $[a,b]$ and $F$ is any antiderivative of $f$, then:

The accumulation form (Topic 6.4, FUN-6.B.2) runs the other way. It differentiates an integral whose upper limit is a variable. If $f$ is continuous on an interval containing $a$ and $g(x) = \int_a^x f(t)\,dt$, then:

Part 2: turning an integral into a number

Reach for the evaluation form whenever you need the value of a definite integral $\int_a^b f(x)\,dx$ and $f$ has an antiderivative you can find. This is the workhorse of Unit 6 and every integration application after it: area, average value, total change, volume.

The method is two steps. First, find an antiderivative $F$ of the integrand (Topic 6.8 basic rules, or substitution in Topic 6.9). Second, compute $F(b) - F(a)$, usually written with a bracket:

You never need the constant of integration here. If you used $F(x) + C$ instead of $F(x)$, the $+C$ appears in both $F(b)$ and $F(a)$ and cancels in the subtraction, so pick the simplest antiderivative. The theorem also requires $f$ continuous on the whole interval $[a,b]$: if the integrand blows up somewhere inside, this shortcut does not apply directly (that gap is what BC improper integrals in Topic 6.13 handle).

The evaluation form is not obvious on its own; it works because of the accumulation form. Since $F(x) = \int_a^x f(t)\,dt$ is an antiderivative of $f$ (FUN-6.B.2), and any two antiderivatives of the same function differ only by a constant, evaluating any antiderivative at $b$ and $a$ and subtracting recovers the accumulated area from $a$ to $b$. That is the bridge between the two parts: Part 1 guarantees the antiderivative exists, Part 2 puts it to work.

Part 1: the accumulation-function view

The accumulation form is about a specific kind of function: one defined by an integral with a variable upper limit.

Read $g(x)$ as the net area accumulated by $f$ from the fixed start $a$ out to $x$. The letter $t$ is just a placeholder (a dummy variable) for the integration; the real input is $x$, sitting in the upper limit. FUN-6.B.2 guarantees this $g$ is an antiderivative of $f$, and the accumulation form gives its derivative directly:

That is the whole point of Part 1: you can differentiate an integral without ever evaluating it. When the integrand has no elementary antiderivative, like $\sin(t^2)$ or $e^{-t^2}$, this is the only way to get the derivative at all.

In an applied setting $g(x)$ reads as a running total. If $f(t)$ is a rate (liters per minute, meters per second), then $g(x)$ is the accumulated amount from time $a$ to time $x$, and $g'(x) = f(x)$ just says the instantaneous rate of that running total is the original rate. That interpretation is exactly Topic 6.5, where you analyze the behavior of $g$ from the graph of $f$.

One extension is tested constantly: when the upper limit is itself a function $u(x)$ rather than plain $x$, the chain rule tacks on a factor of $u'(x)$.

If the variable sits in the lower limit instead, swap the limits first and pick up a negative sign, because reversing the limits of a definite integral negates it (Topic 6.6).

Deciding which part a problem wants

The fastest tell is the limits of integration. Look there before you do anything else.

Two constant limits, and the task is to produce a value (a number, an area, a total change, an average): that is the evaluation form. You will be finding an antiderivative and subtracting.

A variable in a limit, usually paired with a definition like $g(x)$ built from an integral with $x$ in the upper limit, then a question about $g'$, about where $g$ is increasing or has a maximum, or about the concavity of $g$: that is the accumulation form. Any time a new function is built out of an integral and you are asked to differentiate or analyze it, Part 1 is the tool.

• Numeric limits, asked for a value: evaluation form (Part 2). Antidifferentiate, then $F(b)-F(a)$.
• Variable upper limit, asked for a derivative: accumulation form (Part 1). Replace $t$ with the limit, times its derivative.
• Integrand you cannot antidifferentiate, but only a derivative is asked: accumulation form (Part 1).
• Function defined as an integral, asked about increasing/decreasing or extrema: accumulation form (Part 1), then standard Unit 5 analysis on $g'$.

Where students lose points

• Dropping the chain rule factor. With a variable upper limit like $x^2$, the answer is $f(x^2)\cdot 2x$, not just $f(x^2)$.
• Confusing the dummy variable. Inside $\int_a^x f(t)\,dt$ the integrand uses $t$; the $x$ appears only in the limit. Do not substitute $x$ for the integration variable.
• Subtracting in the wrong order. It is always $F(\text{upper}) - F(\text{lower})$, that is $F(b) - F(a)$. Flipping the order flips the sign.
• Ignoring continuity. The evaluation form needs $f$ continuous on $[a,b]$. A vertical asymptote inside the interval breaks it.
• Forgetting the sign when the variable is the lower limit: $\dfrac{d}{dx}\int_x^b f(t)\,dt = -f(x)$.