FTC and definite integrals — AP Calculus BC Study Guide

For: AP Calculus BC candidates sitting AP Calculus BC.

Covers: The Fundamental Theorem of Calculus (FTC) Part 1 (derivatives of accumulation functions), FTC Part 2 (evaluating definite integrals), the net change theorem, and application of FTC to contextual rate problems.

You should already know: Limits of Riemann sums for definite integrals, basic derivative and antiderivative rules, how to check continuity of a function.

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 BC 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 result that connects differentiation and integration, the two core operations of calculus. Before FTC, the definite integral of a function f over [a,b] was defined only as the limit of Riemann sums, a tedious calculation that is impractical for all but the simplest polynomial functions. FTC organizes this relationship into two key parts that allow us to evaluate definite integrals quickly and analyze the behavior of accumulation functions (functions defined as integrals with variable bounds). This topic is the core of Unit 6 (Integration and Accumulation of Change), which accounts for 17-20% of the total AP Calculus BC exam score; FTC and definite integrals themselves make up roughly 7-10% of the total exam. This topic appears in both multiple-choice (MCQ) and free-response (FRQ) sections, and is often combined with other topics like differential equations, area, and volume in multi-part FRQ questions.

2. Fundamental Theorem of Calculus Part 1: Derivatives of Accumulation Functions

The first part of FTC establishes the relationship between differentiation and accumulation of area under a curve. The formal statement is: If f is continuous on the interval [a,b], and we define the accumulation function F(x) as $$F(x)={\int }_a^{x}f(t)dt$$ for all x ∈ [a,b], then F is differentiable on (a,b), and F'(x) = f(x). Intuitively, this means that the instantaneous rate of change of the total area accumulated from a to x is exactly equal to the height of the function f at x. This rule extends to cases where the upper bound is a function of x, not just x itself, using the chain rule: if u(x) is a differentiable function, then $$\frac{d}{dx}\left({\int }_a^{u(x)}f(t)dt\right)=f\left(u(x)\right)\cdot u^{\prime }(x)$$ If the lower bound is variable and the upper bound is constant, we can use the definite integral property ${\int }_{l(x)}^{b}f(t)dt=-{\int }_b^{l(x)}f(t)dt$ to rewrite the integral before applying the rule. This is one of the most commonly tested FTC skills on the AP exam, especially in MCQ.

Worked Example

Find $\frac{d}{dx}\left({\int }_2^{x^2}\cos \left(\sqrt{t}\right)dt\right)$.

1. First, confirm the integrand is continuous on the interval between 2 and $x^2$ for all non-negative x, so FTC 1 applies. The lower bound is constant, and the upper bound is the differentiable function $u(x)=x^2$.
2. Identify the integrand: $f(t)=\cos \left(\sqrt{t}\right)$.
3. Apply the chain rule form of FTC 1: substitute $u(x)$ into $f(t)$ and multiply by $u^{\prime }(x)$. This gives $f(u(x))\cdot u^{\prime }(x)=\cos \left(\sqrt{x^2}\right)\cdot \frac{d}{dx}\left(x^2\right)$.
4. Simplify the result: $\sqrt{x^2}=|x|$, and since $\cos (|x|)=\cos (x)$ (cosine is an even function), and $\frac{d}{dx}(x^2)=2x$. The final derivative is $2x\cos (x)$.

Exam tip: If both the upper and lower bounds are variable, split the integral at any constant between the bounds, then apply FTC 1 to each piece separately; do not rely on memorization alone for this case.

3. Fundamental Theorem of Calculus Part 2: Evaluating Definite Integrals

The second part of FTC allows us to compute the value of a definite integral using any antiderivative of the integrand, eliminating the need to compute limits of Riemann sums. The formal statement is: If f is continuous on [a,b], and F is any antiderivative of f (meaning $F^{\prime }(x)=f(x)$ for all x in [a,b]), then $${\int }_a^{b}f(x)dx=F(b)-F(a)$$ Intuitively, this makes sense because the definite integral of the derivative of F (which is f) over [a,b] should equal the total change in F from a to b, which is exactly $F(b)-F(a)$. The constant of integration we use for indefinite integrals cancels out in the subtraction step: if we used $F(x)+C$ as the antiderivative, we get $(F(b)+C)-(F(a)+C)=F(b)-F(a)$, so we can ignore the constant for definite integrals. This is the primary method we use to evaluate definite integrals for the entire AP Calculus course.

Worked Example

Evaluate ${\int }_0^{\pi /4}\left({\sec }^{2}x-2x\right)dx$.

1. First, confirm that the integrand $f(x)={\sec }^{2}x-2x$ is continuous on $[0,\pi /4]$, so FTC 2 applies.
2. Find the general antiderivative of f(x): The antiderivative of ${\sec }^{2}x$ is $\tan x$, and the antiderivative of $-2x$ is $-x^2$. This gives $F(x)=\tan x-x^2$ (we drop the constant C as it will cancel).
3. Evaluate F at the upper bound $x=\pi /4$: $F\left(\frac{\pi }{4}\right)=\tan \left(\frac{\pi }{4}\right)-{\left(\frac{\pi }{4}\right)}^2=1-\frac{{\pi }^2}{16}$.
4. Evaluate F at the lower bound $x=0$: $F(0)=\tan (0)-0^2=0$.
5. Subtract lower bound from upper bound: $F\left(\frac{\pi }{4}\right)-F(0)=\left(1-\frac{{\pi }^2}{16}\right)-0=1-\frac{{\pi }^2}{16}$.

Exam tip: Always evaluate the upper bound first, then subtract the lower bound; reversing the order is the most common source of sign errors on definite integral problems.

4. The Net Change Theorem

The net change theorem is a direct, applied interpretation of FTC Part 2 that is extremely common in contextual FRQ problems. The theorem states that if $Q(t)$ is a differentiable quantity with rate of change $r(t)=Q^{\prime }(t)$, then the total net change in Q over the interval from t=a to t=b is $$\text{Net Change}=Q(b)-Q(a)={\int }_a^{b}r(t)dt$$ Net change counts positive and negative changes: if the rate is positive, Q increases, if the rate is negative, Q decreases, and the integral adds these up to get the overall change from start to finish. This is distinct from total change, which is the total amount of change regardless of direction (for example, total distance traveled by a particle versus net displacement). For net change, we integrate the rate directly; for total change, we integrate the absolute value of the rate. This distinction is tested frequently on FRQs.

Worked Example

A particle moves along the x-axis with velocity $v(t)=t^2-4t+3$ meters per second, for $0\le t\le 4$. Find the net displacement of the particle over the interval $[0,4]$.

1. Velocity is the rate of change of position, so by the net change theorem, net displacement equals the integral of velocity over the interval.
2. Find the antiderivative of v(t): $F(t)=\frac{t^3}{3}-2t^2+3t$.
3. Apply FTC 2: Net displacement = $F(4)-F(0)$.
4. Calculate values: $F(4)=\frac{64}{3}-2(16)+3(4)=\frac{64}{3}-32+12=\frac{4}{3}$, and $F(0)=0$.
5. The net displacement is $\frac{4}{3}$ meters.

Exam tip: Always include units in your final answer for contextual net change problems; AP graders routinely deduct 1 point for missing or incorrect units.

5. Common Pitfalls (and how to avoid them)

• Wrong move: When differentiating ${\int }_{g(x)}^{b}f(t)dt$, writing the derivative as $f(g(x))\cdot g^{\prime }(x)$ with no negative sign. Why: Confuses variable upper limits with variable lower limits, forgetting that reversing the order of integration changes the integral's sign. Correct move: Always rewrite ${\int }_{g(x)}^{b}f(t)dt=-{\int }_b^{g(x)}f(t)dt$ before applying FTC 1, and carry the negative through your calculation.
• Wrong move: When evaluating ${\int }_a^{b}f(x)dx$, calculating $F(a)-F(b)$ instead of $F(b)-F(a)$. Why: Rushes to plug in values without reading the bounds, confuses the statement of FTC 2. Correct move: Write "F(upper bound) minus F(lower bound)" explicitly on your paper before substituting any values.
• Wrong move: When differentiating ${\int }_0^{x}tf(t)dt$, writing the derivative as $xf(t)$ leaving t as the variable. Why: Confuses the dummy variable of integration with the variable bound, treats the variable of integration as a constant. Correct move: Replace every instance of the dummy variable t with the variable bound after applying FTC 1, so the derivative is $xf(x)$, not $xf(t)$.
• Wrong move: Adding a constant of integration (+C) to the final numerical answer of a definite integral. Why: Confuses definite integrals (which evaluate to a single number) with indefinite integrals (which are families of antiderivatives). Correct move: Drop the constant of integration when applying FTC 2, as it cancels out in the subtraction step.
• Wrong move: Applying FTC to evaluate ${\int }_{-1}^1\frac{1}{x}dx=\ln |1|-\ln |-1|=0$, ignoring the discontinuity at $x=0$. Why: Forgets that FTC only applies if the integrand is continuous on the entire interval between the two bounds. Correct move: Before applying FTC, check for discontinuities, vertical asymptotes, or other points of non-continuity in the interval; if any exist, the integral is improper and requires different evaluation methods.
• Wrong move: Interpreting the integral of velocity over time as total distance instead of net displacement. Why: Confuses net change (integral of the rate) with total change (integral of the absolute value of the rate). Correct move: Always check the question prompt: if it asks for net change or displacement, integrate the rate directly; if it asks for total change or total distance, integrate the absolute value of the rate.

6. Practice Questions (AP Calculus BC Style)

Question 1 (Multiple Choice)

If $h(x)={\int }_1^{e^x}\ln (t^2+1)dt$, what is $h^{\prime }(2)$? (A) $e^2\ln (e^4+1)$ (B) $\ln (e^4+1)$ (C) $e^2\ln (2e^2+1)$ (D) $e^2\ln (e^{2x}+1)$

Worked Solution: This problem tests application of FTC Part 1 with the chain rule for a variable upper bound. For any function of the form $h(x)={\int }_a^{u(x)}f(t)dt$, the derivative is $h^{\prime }(x)=f(u(x))\cdot u^{\prime }(x)$ by FTC. Here, $u(x)=e^x$, so $u^{\prime }(x)=e^x$, and $f(t)=\ln (t^2+1)$. Substituting $u(x)$ into $f$ gives $f(e^x)=\ln ((e^x{)}^2+1)=\ln (e^{2x}+1)$. Multiplying by $u^{\prime }(x)$ gives $h^{\prime }(x)=e^x\ln (e^{2x}+1)$. Evaluating at $x=2$ gives $h^{\prime }(2)=e^2\ln (e^4+1)$, which matches option A. Correct answer: $\boxed{A}$

Question 2 (Free Response)

Let $f(x)=2x+\sin x$, and let $g(y)={\int }_0^{y}f(x)dx$. (a) Find $g(\pi )$. (b) Find $g^{\prime }\left(\frac{\pi }{2}\right)$. (c) Find the average value of $f(x)$ on the interval $[0,\pi ]$.

Worked Solution: (a) First find an antiderivative of $f(x)$: $F(x)=x^2-\cos x$, since $\frac{d}{dx}(x^2-\cos x)=2x+\sin x=f(x)$. By FTC Part 2: $$g(\pi )=F(\pi )-F(0)=({\pi }^2-\cos \pi )-(0^2-\cos 0)=({\pi }^2+1)-(-1)={\pi }^2+2$$

(b) By FTC Part 1, $g^{\prime }(y)=f(y)$ for all y. Substitute $y=\frac{\pi }{2}$: $$g^{\prime }\left(\frac{\pi }{2}\right)=2\left(\frac{\pi }{2}\right)+\sin \left(\frac{\pi }{2}\right)=\pi +1$$

(c) The average value of a function on $[a,b]$ is given by $\frac{1}{b-a}{\int }_a^{b}f(x)dx$. We already know ${\int }_0^{\pi }f(x)dx={\pi }^2+2$ from part (a), so: $$\text{Average value}=\frac{1}{\pi -0}({\pi }^2+2)=\pi +\frac{2}{\pi }$$

Question 3 (Application / Real-World Style)

A bakery runs a 4-hour morning sale, and tracks the rate of cookie sales as $r(t)=-t^3+5t^2+10t$ cookies per hour, where t is the number of hours after the sale starts ($0\le t\le 4$). How many total cookies are sold during the entire 4-hour sale? Round your answer to the nearest whole number.

Worked Solution: Total cookies sold is the net change in the number of cookies sold over 4 hours, so by the net change theorem, we integrate the sales rate from $t=0$ to $t=4$. First find the antiderivative of $r(t)$: $$R(t)=-\frac{t^4}{4}+\frac{5t^3}{3}+5t^2$$ Apply FTC 2: $$R(4)-R(0)=\left(-\frac{256}{4}+\frac{5(64)}{3}+5(16)\right)-0=-64+\frac{320}{3}+80=16+\frac{320}{3}=\frac{368}{3}\approx 122.67$$ Rounding to the nearest whole number gives 123. In context, the bakery sells approximately 123 cookies over the entire 4-hour sale.

7. Quick Reference Cheatsheet

8. What's Next

Mastery of the FTC and definite integrals is the non-negotiable foundation for all remaining integration topics in AP Calculus BC. Immediately after this topic, you will learn u-substitution for definite and indefinite integrals, which relies entirely on FTC to evaluate final results after changing variables. Without mastering the rules and conventions of FTC, you will struggle to solve more complex integration problems including integration by parts, partial fraction decomposition, and improper integrals, all of which are heavily tested on the BC exam. FTC is also the core concept behind accumulation functions, which are used to solve separable differential equations and calculate area and volume of solids later in the course.

• U-Substitution for Definite Integrals
• Integration by Parts
• Improper Integrals
• Area Between Curves