FTC and definite integrals — AP Calculus AB Study Guide

For: AP Calculus AB candidates sitting AP Calculus AB.

Covers: The First Fundamental Theorem of Calculus (FTC 1), Second Fundamental Theorem of Calculus (FTC 2), evaluating definite integrals via antiderivatives, derivatives of accumulation functions, and the inverse relationship between integration and differentiation.

You should already know: How to compute antiderivatives of basic functions. How to apply the chain rule for derivatives. What a definite integral represents as a limit 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 FTC and definite integrals?

The Fundamental Theorem of Calculus (FTC) is the unifying core result that connects the two foundational ideas of calculus: differentiation (measurement of instantaneous rate of change) and integration (measurement of accumulated change and area under a curve). For AP Calculus AB, this topic is the heart of Unit 6: Integration and Accumulation of Change, carrying roughly 15-18% of the total exam score per the CED, and appears in both multiple-choice (MCQ) and free-response (FRQ) sections every exam cycle. A definite integral of a function $f(x)$ over the interval $[a,b]$, written ${\int }_a^{b}f(x)dx$, is formally defined as the limit of Riemann sums, but that approach is only practical for simple shapes or approximation questions. FTC eliminates the need for infinite summation by linking integration to differentiation, giving efficient, exact methods for both evaluating definite integrals and differentiating functions defined as integrals. This topic is not just a set of formulas—it establishes that integration and differentiation are inverse operations, a fact that powers almost all applied calculus problems on the AP exam.

2. The Second Fundamental Theorem of Calculus (FTC 2): Evaluating Definite Integrals

FTC 2 gives the method for computing exact values of definite integrals using antiderivatives, replacing the need for infinite Riemann sums. The theorem states: If $f(x)$ is continuous on the closed interval $[a,b]$, and $F(x)$ is any antiderivative of $f(x)$ (meaning $F^{\prime }(x)=f(x)$ for all $x$ in $[a,b]$), then: $${\int }_a^{b}f(x)dx=F(b)-F(a)$$ The standard notation for this evaluation is ${F(x)|}_a^b$, which explicitly means "evaluate $F$ at the upper bound $b$, subtract the value of $F$ at the lower bound $a$." A key simplification here is that we do not need to add a constant of integration $+C$ for definite integrals: the constant cancels out when we subtract $F(a)$ from $F(b)$, so it can be omitted entirely. The only requirement for FTC 2 to apply is that $f$ is continuous over the entire interval of integration; if there is a discontinuity (like a vertical asymptote) inside the interval, FTC 2 cannot be used directly.

Worked Example

Evaluate ${\int }_1^4(3x^2-2\sqrt{x})dx$.

1. First, find the antiderivative of each term using the power rule for antiderivatives ($\int x^{n}dx=\frac{x^{n+1}}{n+1}+C$ for $n\ne -1$):
   • Antiderivative of $3x^2$: $3\cdot \frac{x^3}{3}=x^3$
   • Antiderivative of $-2\sqrt{x}=-2x^{1/2}$: $-2\cdot \frac{x^{3/2}}{3/2}=-\frac{4}{3}{x}^{3/2}$
2. Combine terms to get $F(x)=x^3-\frac{4}{3}{x}^{3/2}$ (no $+C$ needed).
3. Evaluate at the upper bound $x=4$: $F(4)=4^3-\frac{4}{3}(4{)}^{3/2}=64-\frac{32}{3}=\frac{160}{3}$.
4. Evaluate at the lower bound $x=1$: $F(1)=1^3-\frac{4}{3}(1{)}^{3/2}=1-\frac{4}{3}=-\frac{1}{3}$.
5. Subtract lower bound from upper bound: ${\int }_1^4(3x^2-2\sqrt{x})dx=F(4)-F(1)=\frac{160}{3}-\left(-\frac{1}{3}\right)=\frac{161}{3}$.

Exam tip: Always write the evaluation as ${F(x)|}_a^b$ explicitly to avoid mixing up the order of bounds. Reversing the bounds will flip the sign of your answer, which is a common avoidable error.

3. The First Fundamental Theorem of Calculus (FTC 1): Derivatives of Accumulation Functions

FTC 1 formalizes the inverse relationship between integration and differentiation, and is heavily tested on multiple-choice questions involving functions defined as integrals (called accumulation functions). The base case of FTC 1 states that if $f(t)$ is continuous on an interval containing constant $a$, then: $$\frac{d}{dx}\left[{\int }_a^{x}f(t)dt\right]=f(x)$$ Geometrically, this makes sense: the derivative of the total area under $f(t)$ from $a$ to $x$ is just the height of the function at $x$, which is exactly $f(x)$. When the upper bound is a function of $x$ (not just $x$ itself), we add the chain rule to get the general form: if $g(x)={\int }_a^{u(x)}f(t)dt$, then $g^{\prime }(x)=f(u(x))\cdot u^{\prime }(x)$. If both bounds are functions of $x$, we split the integral at a constant to get: $g^{\prime }(x)=f(u(x))u^{\prime }(x)-f(v(x))v^{\prime }(x)$ for $g(x)={\int }_{v(x)}^{u(x)}f(t)dt$.

Worked Example

Find $\frac{d}{dx}\left[{\int }_{\cos x}^{x^3}{e}^{2t}dt\right]$.

1. Apply the extended FTC 1 rule for variable upper and lower bounds: the derivative equals $f(\text{upper bound})\cdot \frac{d}{dx}(\text{upper bound})-f(\text{lower bound})\cdot \frac{d}{dx}(\text{lower bound})$.
2. Identify components: $f(t)=e^{2t}$, upper bound $u(x)=x^3$, lower bound $v(x)=\cos x$.
3. Compute derivatives of the bounds: $u^{\prime }(x)=3x^2$, $v^{\prime }(x)=-\sin x$.
4. Evaluate $f$ at the bounds: $f(u(x))=e^{2(x^3)}=e^{2x^3}$, $f(v(x))=e^{2\cos x}$.
5. Substitute into the formula: $\frac{d}{dx}\left[{\int }_{\cos x}^{x^3}{e}^{2t}dt\right]=(e^{2x^3})(3x^2)-(e^{2\cos x})(-\sin x)=3x^2{e}^{2x^3}+e^{2\cos x}\sin x$.

Exam tip: When the variable of differentiation is in the lower bound, do not drop the required negative sign from splitting the integral. Double-check the sign of the lower bound term before submitting your final answer.

4. The Net Change Theorem: FTC for Contextual Problems

The Net Change Theorem is a direct application of FTC 2 to real-world problems where we integrate a rate of change to find total accumulated change. The theorem states that the net change in a quantity $F(t)$ from time $t=a$ to $t=b$ is equal to the definite integral of its rate of change $f(t)=F^{\prime }(t)$ over that interval: $$\text{Net Change}=F(b)-F(a)={\int }_a^{b}f(t)dt$$ This is the foundation for all contextual integration problems on AP Calculus AB, including displacement of a moving particle, net change in population, net change in the amount of a drug in the bloodstream, and total profit from marginal profit. A critical distinction: net change counts positive and negative change (e.g., moving left vs. right for a particle) against each other, unlike total change which integrates the absolute value of the rate. If you need the final amount of a quantity at time $b$, given the initial amount $F(a)$, you just rearrange the formula to get $F(b)=F(a)+{\int }_a^{b}f(t)dt$.

Worked Example

The rate at which water flows into a tank is given by $r(t)=2t+5\sin \left(\frac{\pi t}{12}\right)$ gallons per minute, for $0\le t\le 20$ minutes. If the tank initially has 30 gallons of water at $t=0$, how much water is in the tank at $t=10$ minutes?

1. Let $W(t)$ be the total amount of water in the tank at time $t$. We know $W(0)=30$, $W^{\prime }(t)=r(t)$, so by the Net Change Theorem: $W(10)=W(0)+{\int }_0^{10}r(t)dt$.
2. Find the antiderivative of $r(t)$: $\int \left(2t+5\sin \left(\frac{\pi t}{12}\right)\right)dt=t^2-\frac{60}{\pi }\cos \left(\frac{\pi t}{12}\right)$.
3. Evaluate the definite integral: $\left[10^2-\frac{60}{\pi }\cos \left(\frac{10\pi }{12}\right)\right]-\left[0^2-\frac{60}{\pi }\cos (0)\right]=100-\frac{60}{\pi }\left(-\frac{\sqrt{3}}{2}\right)+\frac{60}{\pi }(1)\approx 135.64$.
4. Add the initial amount: $W(10)=30+135.64\approx 165.64$ gallons.

Exam tip: When asked for the final amount of a quantity, the definite integral only gives the net change, not the final amount. Always add the net change to the given initial amount to get the correct answer.

5. Common Pitfalls (and how to avoid them)

• Wrong move: When differentiating $g(x)={\int }_2^{x^2}\sqrt{t^3+1}dt$, you write $g^{\prime }(x)=\sqrt{(x^2{)}^3+1}$ and leave out the chain rule term. Why: Students remember FTC gives the integrand evaluated at the upper bound, but forget the chain rule when the bound is a function of $x$. Correct move: Always check if the upper or lower bound is a function of $x$; if it is, multiply by the derivative of the bound every time.
• Wrong move: When evaluating ${\int }_{-2}^2\frac{1}{x^2}dx$, you apply FTC 2 to get ${-\frac{1}{x}|}_{-2}^2=-1$, and accept this as the final answer. Why: Students forget FTC 2 requires $f$ to be continuous over the entire interval of integration, and this function has a discontinuity at $x=0$ inside the interval. Correct move: Always check for discontinuities of the integrand inside the interval before applying FTC 2; AP Calculus AB will only ask you to apply FTC to continuous functions.
• Wrong move: When computing a definite integral, you subtract the upper bound value from the lower bound value. Why: Students mix up the order of evaluation, especially when bounds are written in reverse order in the problem. Correct move: Always follow the notation convention ${F(x)|}_a^b=F(b)-F(a)$, which means upper bound minus lower bound, no exceptions.
• Wrong move: When finding the final population at $t=10$ given an initial population of 1000 and a growth rate $r(t)$, you report the value of ${\int }_0^{10}r(t)dt$ as your final answer. Why: Students confuse net change with total final amount. Correct move: Remember that the integral of a rate gives only net change; add the net change to the initial amount to get the final total.
• Wrong move: When differentiating ${\int }_x^5{e}^{\sin t}dt$, you get $-e^{\sin x}$ and then remove the negative sign because you think it is a mistake. Why: Students are used to the constant lower bound case and forget the negative sign is required when the variable is in the lower bound. Correct move: Flip the bounds to get $-{\int }_5^x{e}^{\sin t}dt$, apply FTC, and keep the negative sign in your final derivative.

6. Practice Questions (AP Calculus AB Style)

Question 1 (Multiple Choice)

If $F(x)={\int }_0^{x^2}\sqrt{1+t^3}dt$, what is the value of $F^{\prime }(2)$? A) $\sqrt{65}$ B) $4\sqrt{65}$ C) $\sqrt{41}$ D) $2\sqrt{65}$

Worked Solution: We use the First Fundamental Theorem of Calculus with the chain rule for accumulation functions: for $F(x)={\int }_a^{u(x)}f(t)dt$, $F^{\prime }(x)=f(u(x))\cdot u^{\prime }(x)$. Here, $f(t)=\sqrt{1+t^3}$, $u(x)=x^2$, so $u^{\prime }(x)=2x$. Substituting $u(x)=x^2$ into $f(t)$ gives $f(x^2)=\sqrt{1+(x^2{)}^3}=\sqrt{1+x^6}$. Multiply by $u^{\prime }(x)$ to get $F^{\prime }(x)=2x\sqrt{1+x^6}$. Evaluate at $x=2$: $F^{\prime }(2)=2(2)\sqrt{1+2^6}=4\sqrt{1+64}=4\sqrt{65}$. The correct answer is B.

Question 2 (Free Response)

Let $f(x)=2x-6e^x$. (a) Evaluate ${\int }_0^{2}f(x)dx$. Show your work. (b) Find $\frac{d}{dx}\left[{\int }_1^{\sin x}f(t)dt\right]$. Simplify your answer. (c) Let $g(x)={\int }_{x^2}^{4}f(t)dt$. Find $g^{\prime }(1)$.

Worked Solution: (a) First, find the antiderivative of $f(x)=2x-6e^x$: $F(x)=x^2-6e^x$. Evaluate from 0 to 2: $${\int }_0^2(2x-6e^x)dx={(x^2-6e^x)|}_0^2=(4-6e^2)-(0-6e^0)=10-6e^2\approx -34.17$$ (b) Apply FTC 1 with chain rule: $\frac{d}{dx}{\int }_1^{u(x)}f(t)dt=f(u(x))u^{\prime }(x)$, where $u(x)=\sin x$, $u^{\prime }(x)=\cos x$: $$f(\sin x)\cos x=(2\sin x-6e^{\sin x})\cos x=2\sin x\cos x-6\cos x{e}^{\sin x}$$ (c) Rewrite $g(x)=-{\int }_4^{x^2}f(t)dt$, then apply FTC 1 and chain rule: $g^{\prime }(x)=-f(x^2)\cdot 2x=-2x(2x^2-6e^{x^2})$. Evaluate at $x=1$: $$g^{\prime }(1)=-2(1)(2(1{)}^2-6e^1)=-4+12e\approx 28.62$$

Question 3 (Application / Real-World Style)

The velocity of a particle moving along the x-axis is given by $v(t)=t^2-4t+3$ meters per second, for $0\le t\le 4$ seconds. If the particle is at position $x=2$ meters at $t=0$, what is the position of the particle at $t=4$ seconds, and what is the net displacement over $[0,4]$? Include units in your answer.

Worked Solution: By the Net Change Theorem, final position equals initial position plus the integral of velocity (the rate of change of position) over the interval. We start with $x(0)=2$, so: $$x(4)=x(0)+{\int }_0^{4}v(t)dt=2+{\int }_0^4(t^2-4t+3)dt$$ Compute the antiderivative and evaluate: $\int (t^2-4t+3)dt=\frac{t^3}{3}-2t^2+3t$, so the definite integral becomes: $$\left(\frac{64}{3}-32+12\right)-0=\frac{4}{3}\approx 1.33$$ Add the initial position: $x(4)=2+\frac{4}{3}=\frac{10}{3}\approx 3.33$ meters. The net displacement is equal to the definite integral, so it is $\frac{4}{3}\approx 1.33$ meters. This means the particle ends up 1.33 meters to the right of its starting position after 4 seconds.

7. Quick Reference Cheatsheet

8. What's Next

Mastery of the Fundamental Theorem of Calculus is the prerequisite for almost all remaining topics in AP Calculus AB. Immediately after this topic, you will apply FTC to compute areas between curves, solve integration by substitution problems, and analyze contextual problems involving motion, population growth, and accumulated change. Without solid command of both parts of FTC, you will struggle with nearly every applied integration problem that appears on the FRQ section, which accounts for a large share of your total exam score. FTC is also the core result that unifies differentiation and integration, the two big ideas of calculus tested across the entire exam. Follow-on topics to study next are: Integration by substitution Area between two curves Integration applications for motion