What does the Fundamental Theorem of Calculus say?

It links derivatives and integrals. If f is continuous, the derivative of the integral from a fixed number to x of f is f(x). And a definite integral equals the antiderivative at the upper limit minus the antiderivative at the lower limit.

How do I take the derivative of an integral with x in the limit?

Substitute the upper limit into the integrand and multiply by the derivative of that limit. If both limits depend on x, subtract the same expression for the lower limit.

What is an accumulation function?

A function defined by an integral with a variable limit. It adds up signed area from a fixed starting point to x.

How do I find the maximum amount in a rate in and rate out problem?

Find where the rate in minus the rate out changes sign, then compare the amount at those times with the amounts at the endpoints.

The Fundamental Theorem and accumulation for AP Calculus AB

Updated November 9, 2026

An integral of a rate gives the total change in the amount. That single idea, written in a few different ways, drives much of the free-response section of the AP Calculus AB exam.

This guide covers both parts of the Fundamental Theorem of Calculus, functions defined by integrals, and the rate in and rate out problems built from them.

The two parts of the theorem

The derivative of an accumulation function. If $f$ is continuous, and $g(x) = \int_a^x f(t)\,dt$ , then $g'(x) = f(x)$ .
Evaluating a definite integral. If $F' = f$ and $f$ is continuous on $[a, b]$ , then $\int_a^b f(x)\,dx = F(b) - F(a)$ .
The net change form. If $F'$ is continuous on $[a, b]$ , then $F(b) = F(a) + \int_a^b F'(x)\,dx$ . Starting amount plus total change gives the ending amount.

Evaluating an integral

Evaluate

\int_1^4 \frac{x + 2}{\sqrt{x}}\,dx

Split the fraction: $\frac{x + 2}{\sqrt{x}} = x^{1/2} + 2x^{-1/2}$ . An antiderivative is $F(x) = \frac{2}{3}x^{3/2} + 4x^{1/2}$ . So $F(4) - F(1) = \left(\frac{16}{3} + 8\right) - \left(\frac{2}{3} + 4\right) = \frac{26}{3}$ .

Answer: $\frac{26}{3}$

Net change without a formula

The function

f

is differentiable,

f(2) = 7

, and

\int_2^5 f'(x)\,dx = -3

. Find

f(5)

$f(5) = f(2) + \int_2^5 f'(x)\,dx = 7 + (-3) = 4$ .

Answer: $4$

Derivatives of integrals, with the chain rule

If the upper limit is a function of $x$ , plug it into the integrand and multiply by its derivative: $\frac{d}{dx}\int_a^{u(x)} f(t)\,dt = f(u(x))\,u'(x)$ . This requires $f$ to be continuous and $u$ to be differentiable.

One variable limit

Find

G'(x)

G(x) = \int_2^{x^3} \sqrt{t^2 + 5}\,dt

Substitute $x^3$ for $t$ and multiply by the derivative of $x^3$ : $G'(x) = \sqrt{x^6 + 5} \cdot 3x^2$ .

Answer: $3x^2\sqrt{x^6 + 5}$

Two variable limits

Find

H'(x)

H(x) = \int_x^{3x} (e^{-t} + t)\,dt

The upper limit contributes $(e^{-3x} + 3x)(3)$ , and the lower limit contributes $(e^{-x} + x)(1)$ , which you subtract.

So $H'(x) = 3e^{-3x} + 9x - e^{-x} - x = 3e^{-3x} + 8x - e^{-x}$ .

Answer: $3e^{-3x} + 8x - e^{-x}$

Accumulation functions from a graph

Let $g(x) = \int_a^x f(t)\,dt$ . Then $g(a) = 0$ , and $g$ adds signed area as $x$ moves right from $a$ . Moving left from $a$ flips the sign, because $\int_a^x f(t)\,dt = -\int_x^a f(t)\,dt$ . Since $g' = f$ , everything in the guide to reading the graph of a derivative applies here with $f$ in place of $f'$ .

y = f(t)

The graph of f. The piece from t = 2 to t = 6 is a semicircle of radius 2.

Reading an accumulation function

Let

K(x) = \int_0^x f(t)\,dt

on the interval

[-2, 8]

, where

f

is the function graphed above. (a) Find

K(-2)

and

K(2)

. (b) Find

K'(1)

and

K''(1)

. (c) On the open interval

(-2, 8)

, where does

K

have a relative maximum? (d) Find the absolute minimum value of

K

[-2, 8]

(a) $K(2)$ is the area of a triangle with base $2$ and height $2$ , so $K(2) = 2$ . For $K(-2)$ , the area from $-2$ to $0$ is $4$ , and moving left flips the sign, so $K(-2) = -4$ .

(b) $K'(1) = f(1) = 1$ . Then $K''(1) = f'(1)$ , the slope of the line through $(0, 2)$ and $(2, 0)$ , which is $-1$ .

(d) The candidates are the endpoints and the places where $f$ changes sign, $x = 2$ and $x = 6$ . The semicircle has area $\frac{1}{2}\pi(2)^2 = 2\pi$ below the axis, so $K(6) = 2 - 2\pi \approx -4.283$ . The last triangle has area $\frac{1}{2}(2)(3) = 3$ , so $K(8) = 5 - 2\pi \approx -1.283$ . Compared with $K(-2) = -4$ and $K(2) = 2$ , the smallest is $K(6)$ .

Answer: (a) $K(-2) = -4$ , $K(2) = 2$ . (b) $K'(1) = 1$ , $K''(1) = -1$ . (c) $x = 2$ . (d) $2 - 2\pi$ at $x = 6$ .

Rate in and rate out

When an amount changes because of two rates, the net rate is the rate in minus the rate out. The amount at time $t$ is the starting amount plus the integral of the net rate.

Amount at a time: $Q(T) = Q(0) + \int_0^T (\text{in}(t) - \text{out}(t))\,dt$ .
Increasing or decreasing at a time: compare the rates at that moment. If rate in is larger, the amount is increasing.
Maximum or minimum amount: find where the net rate changes sign, then compare the amount there with the amounts at the endpoints.

A(t)

, boxes in

S(t)

, boxes outRates in boxes per hour. The number of boxes rises while the arrival rate is above the shipping rate, then falls.

With a calculator

Boxes arrive at a warehouse at a rate of

A(t) = 30 + 18\sin\left(\frac{t}{2}\right)

boxes per hour, and are shipped out at a rate of

S(t) = 22 + 2.5t

boxes per hour, on the interval

[0, 8]

. At

t = 0

there are

120

boxes. (a) Is the number of boxes increasing or decreasing at

t = 3

? (b) How many boxes are there at

t = 8

? (c) At what time is the number of boxes greatest, and how many are there then?

(a) $A(3) - S(3) \approx 18.455$ , which is positive, so the number of boxes is increasing.

(b) Let $W(t)$ be the number of boxes at time $t$ . Then $W(8) = 120 + \int_0^8 (A(t) - S(t))\,dt \approx 120 + 43.531 = 163.531$ boxes.

(c) $W'(t) = A(t) - S(t)$ . On $[0, 8]$ it's zero only at $t \approx 5.603$ , where it changes from positive to negative.

$W(0) = 120$
$W(5.603) = 120 + \int_0^{5.603} (A(t) - S(t))\,dt \approx 195.520$
$W(8) \approx 163.531$

The greatest value is at $t \approx 5.603$ . For how to enter these integrals, see how to use Desmos on the exam.

Answer: (a) Increasing. (b) $163.531$ boxes. (c) About $195.520$ boxes at $t \approx 5.603$ hours.

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Questions

What does the Fundamental Theorem of Calculus say?: It links derivatives and integrals. If f is continuous, the derivative of the integral from a fixed number to x of f is f(x). And a definite integral equals the antiderivative at the upper limit minus the antiderivative at the lower limit.
How do I take the derivative of an integral with x in the limit?: Substitute the upper limit into the integrand and multiply by the derivative of that limit. If both limits depend on x, subtract the same expression for the lower limit.
What is an accumulation function?: A function defined by an integral with a variable limit. It adds up signed area from a fixed starting point to x.
How do I find the maximum amount in a rate in and rate out problem?: Find where the rate in minus the rate out changes sign, then compare the amount at those times with the amounts at the endpoints.