Fundamental Theorem of Calculus and accumulation functions — AP Calculus AB Study Guide

For: AP Calculus AB candidates sitting AP Calculus AB.

Covers: Fundamental Theorem of Calculus (FTC) Part 1, FTC Part 2, definition of accumulation functions, differentiation of accumulation functions, chain rule for variable bounds, and evaluating definite integrals via antiderivatives.

You should already know: Antiderivatives of basic functions, the chain rule for derivatives, definite integrals as limits of Riemann sums.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Calculus AB style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Fundamental Theorem of Calculus and accumulation functions?

Fundamental Theorem of Calculus (FTC) and accumulation functions are the core connecting link between differential and integral calculus, and form 17–20% of the total AP Calculus AB exam weight per the College Board CED, appearing in both multiple-choice (MCQ) and free-response (FRQ) sections. An accumulation function is any function defined as a definite integral with at least one variable limit of integration: its output equals the net signed area accumulated under an integrand function from a fixed starting bound to a variable upper (or lower) bound. The FTC codifies the inverse relationship between differentiation and integration, resolving the two separate problems of finding tangent slopes and finding area under a curve into one unified framework. For AP candidates, this topic is the basis for almost all applied integration problems, including motion, net change, and volume, making it one of the highest-impact topics for exam success. Mastery of FTC rules is required to solve nearly every non-Riemann sum integration question on the exam.

2. Accumulation Functions

An accumulation function is a function whose value depends on the net area accumulated under another function up to a variable input. The standard form of a basic accumulation function is: $$A(x)={\int }_a^{x}f(t)dt$$ where $a$ is a fixed constant, $f$ is continuous on the interval containing $a$ and $x$, and $t$ is called a dummy variable — it acts only as a placeholder for integration and does not appear in the final output of $A(x)$.

Common variations include accumulation functions with a variable lower bound (and constant upper bound) or two variable bounds. We can always rewrite these variations using the definite integral property ${\int }_b^{a}f(t)dt=-{\int }_a^{b}f(t)dt$ to align with the standard form above. Intuitively, you can think of an accumulation function as a "running total" of net signed area starting at $x=a$ and ending at any input $x$.

Worked Example

Define $g(x)={\int }_2^x(t^2-3t)dt$. (a) Calculate $g(4)$; (b) Describe the net signed area represented by $g(-1)$.

1. For part (a), 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 the antiderivative at the upper bound minus the 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 $g(-1)={\int }_2^{-1}(t^2-3t)dt=-{\int }_{-1}^2(t^2-3t)dt$, so $g(-1)$ is the negative of the net signed area between $t=-1$ and $t=2$ under $y=t^2-3t$.

Exam tip: Always use a different variable for the dummy integration variable than the variable bound to avoid confusion — mixing up $x$ for $t$ leads to frequent derivative errors on later problems.

3. Fundamental Theorem of Calculus Part 1 (Derivative of Accumulation Functions)

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

If $f$ is continuous on $[a,b]$, and $A(x)={\int }_a^{x}f(t)dt$ for $a\le x\le b$, then $A^{\prime }(x)=f(x)$ for all $x\in (a,b)$.

Intuitively, the rate of change of the accumulated area at $x$ is exactly the height of the original function at $x$: if you increase $x$ by a tiny amount $dx$, the added area is roughly $f(x)dx$, so $\frac{dA}{dx}=f(x)$.

For AP exam purposes, we extend this rule to variable bounds using the chain rule:

• If $A(x)={\int }_a^{u(x)}f(t)dt$, then $A^{\prime }(x)=f(u(x))\cdot u^{\prime }(x)$
• If $A(x)={\int }_{v(x)}^{a}f(t)dt$, then $A^{\prime }(x)=-f(v(x))\cdot v^{\prime }(x)$
• If $A(x)={\int }_{v(x)}^{u(x)}f(t)dt$, then $A^{\prime }(x)=f(u(x))u^{\prime }(x)-f(v(x))v^{\prime }(x)$

Worked Example

Find $\frac{dy}{dx}$ for $y={\int }_{\cos x}^{2x}{e}^{t^2}dt$.

1. Apply the derivative rule for accumulation functions with two variable bounds: $\frac{d}{dx}{\int }_{v(x)}^{u(x)}f(t)dt=f(u(x))u^{\prime }(x)-f(v(x))v^{\prime }(x)$.
2. Identify components: $f(t)=e^{t^2}$, $u(x)=2x$, $v(x)=\cos x$.
3. Calculate individual terms: $f(u(x))=e^{(2x{)}^2}=e^{4x^2}$, $u^{\prime }(x)=2$; $f(v(x))=e^{{\cos }^{2}x}$, $v^{\prime }(x)=-\sin x$.
4. Substitute into the formula: $\frac{dy}{dx}=(e^{4x^2}\cdot 2)-\left(e^{{\cos }^{2}x}\cdot (-\sin x)\right)=2e^{4x^2}+\sin x\cdot e^{{\cos }^{2}x}$.

Exam tip: When the lower bound is variable, always rewrite the integral with the negative sign to flip bounds before taking the derivative — this makes the negative sign explicit and avoids forgotten signs.

4. Fundamental Theorem of Calculus Part 2 (Evaluating Definite Integrals)

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

If $f$ is continuous on $[a,b]$, and $F$ is any antiderivative of $f$ (meaning $F^{\prime }(x)=f(x)$), then: $${\int }_a^{b}f(x)dx=F(b)-F(a)$$

Intuitively, the total change in the accumulation function from $a$ to $b$ is exactly the net area under $f$ over that interval. The constant of integration from finding the antiderivative cancels out automatically: $(F(b)+C)-(F(a)+C)=F(b)-F(a)$, so we never need to include $+C$ 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 work for integrands with discontinuities inside the interval.

Worked Example

Evaluate ${\int }_0^{\pi /2}\left(3\sin x+2\cos 2x\right)dx$.

1. Confirm the integrand is continuous on $[0,\pi /2]$, so FTC Part 2 applies.
2. Find the antiderivative term-by-term: the antiderivative of $3\sin x$ is $-3\cos x$, and the antiderivative of $2\cos 2x$ is $\sin 2x$, 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 and subtract: $F(0)=-3\cos (0)+\sin (0)=-3(1)+0=-3$, so ${\int }_0^{\pi /2}\dots dx=0-(-3)=3$.

Exam tip: Always differentiate your antiderivative before plugging in bounds to check for errors — a 10-second check will catch common sign mistakes for sine and cosine.

5. Common Pitfalls (and how to avoid them)

• Wrong move: For $y={\int }_a^{u(x)}f(t)dt$, writing $y^{\prime }=f(u(x))$ and omitting the $u^{\prime }(x)$ chain rule term. Why: Students forget the upper bound is a function of $x$, not just $x$, 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 $y={\int }_{v(x)}^{a}f(t)dt$, writing $y^{\prime }=f(v(x))v^{\prime }(x)$. Why: Students forget flipping the limits of integration introduces a negative sign. Correct move: Rewrite any integral with a variable lower bound as $-{\int }_a^{v(x)}f(t)dt$ before taking the derivative, so the negative sign is explicit.
• Wrong move: Evaluating ${\int }_a^{b}f(x)dx$ 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: Using the variable $x$ for both the bound of integration and the dummy integrand variable, e.g., writing $A(x)={\int }_a^{x}f(x)dx$. Why: Students reuse the output variable by habit. 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 ${\int }_{-1}^2\frac{1}{x^2}dx$, getting $-\frac{3}{2}$ as a final answer. Why: Students forget FTC only applies when the integrand is continuous on the interval of integration; $\frac{1}{x^2}$ has an infinite discontinuity at $x=0$ inside the interval. Correct move: Always check for discontinuities of the integrand inside the interval of integration before applying FTC.
• Wrong move: For $A(x)={\int }_1^{4}f(t)dt$, computing $A^{\prime }(x)=f(4)-f(1)$. 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. Practice Questions (AP Calculus AB Style)

Question 1 (Multiple Choice)

If $h(x)={\int }_1^{x^2}\sqrt[3]{1+t^2}dt$, what is $h^{\prime }(2)$? A. $\sqrt[3]{5}$ B. $4\sqrt[3]{17}$ C. $2\sqrt[3]{17}$ D. $2\sqrt[3]{5}$

Worked Solution: We apply FTC Part 1 with chain rule for an accumulation function of the form $h(x)={\int }_a^{u(x)}f(t)dt$, which gives $h^{\prime }(x)=f(u(x))u^{\prime }(x)$. Here, $f(t)=\sqrt[3]{1+t^2}$, $u(x)=x^2$, so $u^{\prime }(x)=2x$. Substitute $x=2$: $u(2)=4$, so $f(u(2))=\sqrt[3]{1+4^2}=\sqrt[3]{17}$, and $u^{\prime }(2)=4$. Multiply to get $h^{\prime }(2)=4\sqrt[3]{17}$. Correct answer: B.

Question 2 (Free Response)

Let $f(t)=4-2t$ for all real $t$, and define the accumulation function $A(x)={\int }_0^{x}f(t)dt$. (a) Find $A(x)$ as a function of $x$. (b) Compute $A^{\prime }(x)$ and verify it matches the result from FTC Part 1. (c) For what values of $x$ is $A(x)$ increasing? Justify your answer.

Worked Solution: (a) Use FTC Part 2 to evaluate the definite integral: $$A(x)={\int }_0^x(4-2t)dt={\left[4t-t^2\right]}_0^x=(4x-x^2)-(0-0)=4x-x^2$$ (b) Differentiate $A(x)$ directly: $A^{\prime }(x)=\frac{d}{dx}(4x-x^2)=4-2x=f(x)$, which matches the result from FTC Part 1 that the derivative of ${\int }_a^{x}f(t)dt$ is $f(x)$. (c) A function is increasing when its first derivative is positive. We solve $A^{\prime }(x)=4-2x>0⟹x<2$. So $A(x)$ is increasing for all real $x<2$, justifying by the first derivative test for increasing functions.

Question 3 (Application / Real-World Style)

A bakery uses a dough machine that adds flour to a mixing bowl at a rate of $r(t)=3t+2$ ounces per minute, where $t$ is the number of minutes after the machine starts. At $t=0$, the bowl already has 8 ounces of flour. Write an accumulation function for the total ounces of flour in the bowl after $x$ minutes, and find how much flour is in the bowl after 5 minutes.

Worked Solution: The total amount of flour is the initial amount plus the accumulated flour added by the machine, so the accumulation function is: $$F(x)=8+{\int }_0^x(3t+2)dt$$ Evaluate the definite integral from $t=0$ to $t=5$: $${\int }_0^5(3t+2)dt={\left[\frac{3t^2}{2}+2t\right]}_0^5=\left(\frac{75}{2}+10\right)-0=47.5$$ Add the initial 8 ounces: $F(5)=8+47.5=55.5$ ounces. After 5 minutes of running, the mixing bowl contains 55.5 ounces of flour.

7. Quick Reference Cheatsheet

8. 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 use accumulation functions to model total change from a given rate function. Without mastering FTC rules and differentiation of accumulation functions, it will be impossible to correctly solve these applied problems, or later topics including finding areas between curves and volumes of revolution, which all 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.

• Net Change and Definite Integrals in Context
• Integration by Substitution
• Area Between Curves
• Volumes of Solids of Revolution