Fundamental Theorem of Calculus and accumulation functions

AP Calculus AB· AP Calculus AB CED — Integration and Accumulation of Change· 14 min read

1. Accumulation Functions★★☆☆☆⏱ 3 min

An accumulation function is a function whose value equals the net signed area accumulated under another function from a fixed starting bound to a variable input. The standard form is:

A(x) = \int_{a}^{x} f(t) dt

Here, $a$ is a fixed constant, $f$ is continuous on the interval containing $a$ and $x$ , and $t$ is a dummy variable: a placeholder that does not appear in the final output of $A (x)$ . Common variations include variable lower bounds or two variable bounds, which can be rewritten using the property $\int_{b}^{a} f (t) d t = - \int_{a}^{b} f (t) d t$ .

📘 Definition

Accumulation Function

A(x) = \int_{a}^{x} f(t) dt

A function defined by a definite integral with at least one variable limit of integration, whose value equals the net signed area accumulated under the integrand from the fixed bound to the variable bound.

Example:

g(x) = \int_{2}^{x} (t^2 - 3t) dt

📐 Worked Example

Define $g (x) = \int_{2}^{x} (t^{2} - 3 t) d t$ . (a) Calculate $g (4)$ ; (b) Describe the net signed area represented by $g (- 1)$ .

1
Substitute $x = 4$ to get the definite integral:
$g(4) = \int_{2}^{4} (t^2 - 3t) dt$
2
Find the antiderivative of the integrand:
$\frac{t^3}{3} - \frac{3t^2}{2}$
3
Evaluate antiderivative at upper bound minus lower bound:
$\left(\frac{64}{3} - 24\right) - \left(\frac{8}{3} - 6\right) = \frac{56}{3} - 18 = \frac{2}{3}$
4
For part (b), rewrite the integral using the bound flip property:
$g(-1) = \int_{2}^{-1} (t^2 - 3t) dt = -\int_{-1}^{2} (t^2 - 3t) dt$
5
Thus $g (- 1)$ is the negative of the net signed area between $t = - 1$ and $t = 2$ under $y = t^{2} - 3 t$ .

2. FTC Part 1: Derivative of Accumulation Functions★★★☆☆⏱ 4 min

FTC Part 1 formalizes the inverse relationship between integration and differentiation, telling us how to differentiate an accumulation function. The basic statement of the theorem is:

📘 Definition

Fundamental Theorem of Calculus Part 1 (Basic Form)

\frac{d}{dx} \int_{a}^{x} f(t) dt = f(x)

If $f$ is continuous on $[a, b]$ , and $A (x) = \int_{a}^{x} f (t) d t$ for $a \leq x \leq b$ , then $A^{'} (x) = f (x)$ for all $x \in (a, b)$ . Intuitively, the rate of change of accumulated area at $x$ equals the height of the original function at $x$ .

For AP exam purposes, we extend this rule to variable bounds using the chain rule, resulting in three common cases:

If $A (x) = \int_{a}^{u (x)} f (t) d t$ , then $A^{'} (x) = f (u (x)) \cdot u^{'} (x)$
If $A (x) = \int_{v (x)}^{a} f (t) d t$ , then $A^{'} (x) = - f (v (x)) \cdot v^{'} (x)$
If $A (x) = \int_{v (x)}^{u (x)} f (t) d t$ , then $A^{'} (x) = f (u (x)) u^{'} (x) - f (v (x)) v^{'} (x)$

📐 Worked Example

Find $\frac{d y}{d x}$ for $y = \int_{c o s x}^{2 x} e^{t^{2}} d t$ .

1
Apply the derivative rule for accumulation functions with two variable bounds.
2
Identify all components: $f (t) = e^{t^{2}}$ , upper bound $u (x) = 2 x$ , lower bound $v (x) = cos x$ .
3
Calculate each term: $f (u (x)) = e^{(2 x)^{2}} = e^{4 x^{2}}$ , $u^{'} (x) = 2$ ; $f (v (x)) = e^{c o s^{2} x}$ , $v^{'} (x) = - sin x$ .
4
Substitute into the formula:
$\frac{dy}{dx} = (e^{4x^2} \cdot 2) - \left(e^{\cos^2 x} \cdot (-\sin x)\right) = 2e^{4x^2} + \sin x \cdot e^{\cos^2 x}$

3. FTC Part 2: Evaluating Definite Integrals★★☆☆☆⏱ 3 min

FTC Part 2 gives a method to calculate the exact value of a definite integral using antiderivatives, instead of approximating with Riemann sums. The formal statement is:

📘 Definition

Fundamental Theorem of Calculus Part 2

\int_{a}^{b} f(x) dx = F(b) - F(a)

If $f$ is continuous on $[a, b]$ , and $F$ is any antiderivative of $f$ (meaning $F^{'} (x) = f (x)$ ), then the value of the definite integral equals the antiderivative evaluated at the upper bound minus the antiderivative evaluated at the lower bound. The constant of integration cancels out, so $+ C$ is never needed for definite integrals.

A key requirement for this theorem to hold is that $f$ is continuous over the entire interval $[a, b]$ — it does not apply to integrands with discontinuities inside the interval.

📐 Worked Example

Evaluate $\int_{0}^{π /2} (3 sin x + 2 cos 2 x) d x$ .

1
Confirm the integrand is continuous on $[0, π /2]$ , so FTC Part 2 applies.
2
Find the antiderivative term-by-term: antiderivative of $3 sin x$ is $- 3 cos x$ , antiderivative of $2 cos 2 x$ is $sin 2 x$ , so:
$F(x) = -3\cos x + \sin 2x$
3
Evaluate $F$ at the upper bound:
$F\left(\frac{\pi}{2}\right) = -3\cos\left(\frac{\pi}{2}\right) + \sin(\pi) = -3(0) + 0 = 0$
4
Evaluate $F$ at the lower bound, then subtract lower bound from upper bound:
$F(0) = -3\cos(0) + \sin(0) = -3(1) + 0 = -3 \\ \int_{0}^{\pi/2} \left(3\sin x + 2\cos 2x\right) dx = 0 - (-3) = 3$

4. AP Style Practice Problems★★★☆☆⏱ 4 min

✓ Quick check

Test your understanding with these AP-style problems:

If $h (x) = \int_{1}^{x^{2}} 3 1 + t^{2} d t$ , what is $h^{'} (2)$ ?
- $35$
- $4317$
- $2317$
- $235$
Reveal answer
1 —
Correct. Apply FTC Part 1 with chain rule: $h^{'} (x) = 3 1 + (x^{2})^{2} \cdot 2 x$ , so $h^{'} (2) = 4317$ .

📐 Worked Example

Let $f (t) = 4 - 2 t$ for all real $t$ , and define $A (x) = \int_{0}^{x} f (t) d t$ . (a) Find $A (x)$ as a function of $x$ . (b) Compute $A^{'} (x)$ and verify it matches FTC Part 1. (c) For what values of $x$ is $A (x)$ increasing? Justify your answer.

1
(a) Use FTC Part 2 to evaluate the definite integral:
$A(x) = \int_{0}^{x} (4 - 2t) dt = \left[4t - t^2\right]_{0}^{x} = 4x - x^2$
2
(b) Differentiate $A (x)$ directly:
$A'(x) = \frac{d}{dx}(4x - x^2) = 4 - 2x = f(x)$
3
This matches the FTC Part 1 result that the derivative of $\int_{a}^{x} f (t) d t$ is $f (x)$ .
4
(c) A function is increasing when its first derivative is positive. Solve:
$A'(x) = 4 - 2x > 0 \implies x < 2$
5
Thus $A (x)$ is increasing for all real $x < 2$ , justified by the first derivative test.

📐 Worked Example

A bakery's dough machine adds flour to a bowl at a rate of $r (t) = 3 t + 2$ ounces per minute, where $t$ is minutes after starting. At $t = 0$ , the bowl already has 8 ounces of flour. Write an accumulation function for total flour after $x$ minutes, and find the total after 5 minutes.

1
Total flour equals initial amount plus accumulated flour added, so the accumulation function is:
$F(x) = 8 + \int_{0}^{x} (3t + 2) dt$
2
Evaluate the definite integral from $t = 0$ to $t = 5$ :
$\int_{0}^{5} (3t + 2) dt = \left[\frac{3t^2}{2} + 2t\right]_{0}^{5} = 47.5$
3
Add the initial 8 ounces:
$F(5) = 8 + 47.5 = 55.5$
4
After 5 minutes, the mixing bowl contains 55.5 ounces of flour.

5. Common Pitfalls

Wrong move:

For $y = \int_{a}^{u (x)} f (t) d t$ , writing $y^{'} = f (u (x))$ and omitting the $u^{'} (x)$ chain rule term.

Why:

Students forget the upper bound is a function of $x$ , so they skip the chain rule step.

Correct move:

Always identify if the bound is a function of $x$ , and automatically multiply by its derivative every time you apply FTC Part 1.

Wrong move:

For $y = \int_{v (x)}^{a} f (t) d t$ , writing $y^{'} = f (v (x)) v^{'} (x)$ . No negative sign included.

Why:

Students forget flipping the limits of integration introduces a negative sign.

Correct move:

Rewrite any integral with a variable lower bound as $- \int_{a}^{v (x)} f (t) d t$ before taking the derivative, so the negative sign is explicit.

Wrong move:

Evaluating $\int_{a}^{b} f (x) d x$ as $F (a) - F (b)$ instead of $F (b) - F (a)$ .

Why:

Students mix up the order of subtraction when recalling FTC Part 2.

Correct move:

Always write "upper bound evaluation minus lower bound evaluation" in your work before plugging in values to enforce the correct order.

Wrong move:

Writing $A (x) = \int_{a}^{x} f (x) d x$ , reusing $x$ for both the bound and dummy integration variable.

Why:

Students reuse the output variable by habit, leading to confusion when taking derivatives.

Correct move:

Always use a different letter (e.g., $t$ , $s$ ) for the dummy integration variable when defining an accumulation function.

Wrong move:

Applying FTC Part 2 to evaluate $\int_{- 1}^{2} \frac{1}{x ^{2}} d x$ , getting $- \frac{3}{2}$ as a final answer.

Why:

Students forget FTC only applies when the integrand is continuous on the entire interval; $\frac{1}{x ^{2}}$ has an infinite discontinuity at $x = 0$ inside $[- 1, 2]$ .

Correct move:

Always check for discontinuities of the integrand inside the interval of integration before applying FTC.

Wrong move:

For $A (x) = \int_{1}^{4} f (t) d t$ , computing $A^{'} (x) = f (4) - f (1)$ .

Why:

Students confuse constant definite integrals with accumulation functions that have a variable bound.

Correct move:

If both bounds of integration are constant, the entire integral is a constant, so its derivative with respect to $x$ is 0.

6. Quick Reference Cheatsheet

Category	Formula	Notes
Basic Accumulation Function	$A (x) = \int_{a}^{x} f (t) d t$	$a$ constant, $t$ dummy variable, $f$ continuous
FTC Part 1 (Basic Form)	$\frac{d}{d x} \int_{a}^{x} f (t) d t = f (x)$	Only applies to continuous $f$
FTC Part 1 (Variable Upper Bound)	$\frac{d}{d x} \int_{a}^{u (x)} f (t) d t = f (u (x)) \cdot u^{'} (x)$	Chain rule term $u^{'} (x)$ always required
FTC Part 1 (Variable Lower Bound)	$\frac{d}{d x} \int_{v (x)}^{a} f (t) d t = - f (v (x)) \cdot v^{'} (x)$	Negative sign from flipping integral limits
FTC Part 1 (Two Variable Bounds)	$\frac{d}{d x} \int_{v (x)}^{u (x)} f (t) d t = f (u (x)) u^{'} (x) - f (v (x)) v^{'} (x)$	Combine upper and lower bound rules
FTC Part 2 (Evaluate Definite Integral)	$\int_{a}^{b} f (x) d x = F (b) - F (a), F^{'} (x) = f (x)$	Only applies if $f$ is continuous on $[a, b]$
Derivative of Constant Integral	$\frac{d}{d x} \int_{a}^{b} f (t) d t = 0$	Both bounds constant → integral is constant

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · MCQ
Differentiate two-variable-bound accumulation
2022 · FRQ
Accumulation for real-world rate problem

What's Next

This topic is the foundation for all remaining integration topics in AP Calculus AB. Next you will apply FTC and accumulation functions to solving net change problems in context, including motion along a line, population change, and fluid flow, where you will model total change from a given rate function. Mastery of FTC rules is required to correctly solve these applied problems, as well as later topics including areas between curves and volumes of revolution, all of which rely on FTC to set up and evaluate integrals. This topic also builds the core AP skill of connecting derivatives and integrals, which is central to almost every free-response question on the exam.