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

For: AP Calculus BC candidates sitting AP Calculus BC.

Covers: Definition of accumulation functions, both parts of the Fundamental Theorem of Calculus (FTC), derivative rules for accumulation functions with variable bounds, and analysis of accumulation functions for extrema and concavity, aligned to the AP CED.

You should already know: Derivative rules including the chain rule for composite functions. Definition of the definite integral as a limit of Riemann sums. Basic antiderivative formulas for common functions.

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 Fundamental Theorem of Calculus and accumulation functions?

The Fundamental Theorem of Calculus (FTC) is the core result that connects the two pillars of calculus: differentiation and integration, proving they are inverse operations. This topic makes up roughly 8–10% of the total AP Calculus BC exam, per the College Board AP CED, and is tested in both multiple-choice (MCQ) and free-response (FRQ) sections.

An accumulation function is a function defined as a definite integral with at least one variable bound, rather than two constant bounds. For example, $A(x)={\int }_a^{x}f(t)dt$ is an accumulation function that gives the net area accumulated under $f(t)$ from a constant lower bound $a$ to the upper bound $x$, which varies as $x$ changes.

FTC has two key parts: Part 1 gives the derivative of an accumulation function, and Part 2 gives a method to evaluate definite integrals using antiderivatives. This topic is central to almost all applied integration problems in BC, from net change of physical quantities to analyzing integrals with variable bounds on FRQ graph problems.

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

FTC Part 1 formalizes the inverse relationship between integration and differentiation. The formal statement is: if $f(t)$ is continuous on an interval $[a,b]$, and we define the accumulation function $A(x)={\int }_a^{x}f(t)dt$ for $a\le x\le b$, then $A^{\prime }(x)=f(x)$. The intuition here is simple: the rate of change of the total accumulated area up to $x$ is exactly the height of the function $f$ at $x$.

This rule extends easily to accumulation functions with composite variable bounds, using the chain rule. If the upper bound is a function $u(x)$ instead of just $x$, then: $$\frac{d}{dx}{\int }_a^{u(x)}f(t)dt=f(u(x))\cdot u^{\prime }(x)$$ For accumulation functions with variable bounds on both ends, we split the integral at a constant $a$, flip the lower bound integral to get a negative sign, then apply the chain rule to each bound: $$\frac{d}{dx}{\int }_{l(x)}^{u(x)}f(t)dt=f(u(x))u^{\prime }(x)-f(l(x))l^{\prime }(x)$$

Worked Example

Find $\frac{d}{dx}\left[{\int }_{\cos x}^{x^2}\ln (t^2+1)dt\right]$.

1. Identify the components: this is an accumulation function with variable lower bound $l(x)=\cos x$ and variable upper bound $u(x)=x^2$, with integrand $f(t)=\ln (t^2+1)$.
2. Compute derivatives of the bounds: $u^{\prime }(x)=2x$, $l^{\prime }(x)=-\sin x$.
3. Apply the two-bound derivative rule: $\frac{d}{dx}=f(u(x))u^{\prime }(x)-f(l(x))l^{\prime }(x)$.
4. Substitute and simplify: $$\ln \left((x^2{)}^2+1\right)(2x)-\ln \left((\cos x{)}^2+1\right)(-\sin x)=2x\ln (x^4+1)+\sin x\ln ({\cos }^{2}x+1)$$

Exam tip: Always label $f(t)$, the upper bound function, and lower bound function explicitly before taking the derivative. This prevents mixing up terms and avoids avoidable sign errors.

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

FTC Part 2 gives a straightforward way to calculate the exact value of a definite integral, avoiding the need to compute limits of Riemann sums. The formal statement is: if $f(x)$ is continuous on $[a,b]$, and $F(x)$ is any antiderivative of $f(x)$ (meaning $F^{\prime }(x)=f(x)$), then: $${\int }_a^{b}f(x)dx=F(b)-F(a)$$

The intuition here aligns with the net change interpretation of integration: the definite integral of a rate of change $F^{\prime }(x)$ from $a$ to $b$ is just the total change in $F(x)$ over that interval, which is $F(b)-F(a)$. The constant of integration we add for indefinite integrals cancels out between $F(b)$ and $F(a)$, so we do not need to include it for definite integral calculations. This works for all continuous integrands over closed intervals, and is the basis for almost every definite integral evaluation you will do on the AP exam.

Worked Example

Evaluate ${\int }_0^{\pi /2}\left(2\sin x+4x\cos (x^2)\right)dx$.

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 $2\sin x$ is $-2\cos x$. For $4x\cos (x^2)$, use u-substitution ($u=x^2$, $du=2xdx$) to get the antiderivative $2\sin (x^2)$. The full antiderivative is $F(x)=-2\cos x+2\sin (x^2)$.
3. Evaluate at the upper bound: $F\left(\frac{\pi }{2}\right)=-2\cos \left(\frac{\pi }{2}\right)+2\sin \left({\left(\frac{\pi }{2}\right)}^2\right)=0+2\sin \left(\frac{{\pi }^2}{4}\right)$.
4. Evaluate at the lower bound: $F(0)=-2\cos (0)+2\sin (0)=-2(1)+0=-2$.
5. Subtract lower from upper: ${\int }_0^{\pi /2}...dx=F\left(\frac{\pi }{2}\right)-F(0)=2\sin \left(\frac{{\pi }^2}{4}\right)-(-2)=2+2\sin \left(\frac{{\pi }^2}{4}\right)$.

Exam tip: Always write $F(b)-F(a)$ explicitly, do not reverse the order. Reversing bounds will flip the sign of your answer incorrectly, and AP graders deduct points for this mistake.

4. Analyzing Accumulation Functions

A very common exam question asks you to find intervals of increase/decrease, local extrema, intervals of concavity, or inflection points for an accumulation function, often given a graph of the integrand $f(t)$. Because of FTC Part 1, we can get all derivative information for the accumulation function directly from $f$, no integration required.

If $A(x)={\int }_a^{x}f(t)dt$, then by FTC 1, $A^{\prime }(x)=f(x)$. That means:

• $A(x)$ is increasing when $f(x)>0$, decreasing when $f(x)<0$
• Local extrema of $A(x)$ occur where $f(x)$ changes sign (critical points where $f(x)=0$ for continuous $f$)

For the second derivative, differentiate $A^{\prime }(x)=f(x)$ to get $A^{\prime \prime }(x)=f^{\prime }(x)$. That means:

• $A(x)$ is concave up when $f^{\prime }(x)>0$ (when the graph of $f$ is increasing)
• $A(x)$ is concave down when $f^{\prime }(x)<0$ (when the graph of $f$ is decreasing)
• Inflection points of $A(x)$ occur where $f$ changes slope (where $f^{\prime }$ changes sign)

Worked Example

The graph of $y=f(t)$ is piecewise linear with the following points: $(-2,3)$, $(0,3)$, $(1.5,0)$, $(2,-1)$, $(3.5,0)$, $(4,1)$. Let $A(x)={\int }_0^{x}f(t)dt$. Identify all local maxima of $A(x)$ on $(-2,4)$.

1. By FTC 1, $A^{\prime }(x)=f(x)$, so critical points occur where $f(x)=0$, at $x=1.5$ and $x=3.5$.
2. First derivative test at $x=1.5$: for $0<x<1.5$, $f(x)>0$ so $A^{\prime }(x)>0$, so $A(x)$ is increasing. For $1.5<x<3.5$, $f(x)<0$ so $A^{\prime }(x)<0$, so $A(x)$ is decreasing. This means $x=1.5$ is a local maximum.
3. First derivative test at $x=3.5$: for $1.5<x<3.5$, $f(x)<0$ so $A(x)$ is decreasing. For $x>3.5$, $f(x)>0$ so $A(x)$ is increasing. This means $x=3.5$ is a local minimum.
4. For $-2<x<0$, $f(x)=3>0$, so $A(x)$ is increasing across the entire interval, with no sign changes, so no extrema here.
5. Conclusion: the only local maximum of $A(x)$ on $(-2,4)$ is at $x=1.5$.

Exam tip: When analyzing an accumulation function from a graph of $f$, remember $A^{\prime }=f$ and $A^{\prime \prime }=f^{\prime }$. Do not confuse the concavity of $A$ with the sign of $f$ — concavity depends on the slope of $f$'s graph.

5. Common Pitfalls (and how to avoid them)

• Wrong move: When finding $\frac{d}{dx}{\int }_a^{u(x)}f(t)dt$, you write $f^{\prime }(u(x))u^{\prime }(x)$ instead of $f(u(x))u^{\prime }(x)$. Why: Students confuse taking the derivative of $f$ versus just evaluating $f$ at the bound, mixing up FTC 1 with regular chain rule. Correct move: Always remember FTC 1 gives the derivative of the integral equals $f$ evaluated at the bound, not derivative of $f$. Write the formula down explicitly before starting.
• Wrong move: When the lower bound is variable, you forget the negative sign: $\frac{d}{dx}{\int }_{v(x)}^{b}f(t)dt=f(v(x))v^{\prime }(x)$ instead of $-f(v(x))v^{\prime }(x)$. Why: Students don't rewrite the integral to flip bounds before applying the chain rule, and misremember the generalized formula. Correct move: Always flip the bound and add the negative sign when the variable is on the lower bound before taking the derivative.
• Wrong move: When evaluating ${\int }_a^{b}f(x)dx$ with FTC 2, you leave the constant of integration $C$ in, getting $F(b)-F(a)+C$ instead of $F(b)-F(a)$. Why: Students are used to adding $C$ for indefinite integrals and carry it over by habit. Correct move: Recognize that $C$ cancels out for definite integrals, so you can omit it entirely when evaluating $F(b)-F(a)$.
• Wrong move: When finding where an accumulation function $A(x)$ is concave up, you use the sign of $f(x)$ (the integrand) instead of the sign of $f^{\prime }(x)$. Why: Students forget the relationship $A^{\prime }=f$, $A^{\prime \prime }=f^{\prime }$, so they mix up first and second derivative information. Correct move: Label derivatives explicitly: $A^{\prime }(x)=f(x)$, so $A^{\prime \prime }(x)=\frac{d}{dx}{A}^{\prime }(x)=f^{\prime }(x)$. Use that label to check the sign for concavity.
• Wrong move: You evaluate $\frac{d}{dx}{\int }_0^{2x}\sin (t)dt$ as $\sin (2x)$ instead of $2\sin (2x)$. Why: Students forget the chain rule step for the derivative of the upper bound, only evaluate $f$ at the bound and omit the derivative of the bound. Correct move: Always compute the derivative of the bound function and multiply after evaluating $f$ at the bound, no exceptions.
• Wrong move: You apply FTC 2 to a discontinuous integrand over an interval containing a discontinuity, and directly evaluate $F(b)-F(a)$. Why: Students assume FTC 2 works for all integrals, but it requires continuity of $f$ on the entire closed interval between the bounds. Correct move: If $f$ has a discontinuity in $[a,b]$, split the integral into subintervals where $f$ is continuous, apply FTC 2 to each piece, and add the results.

6. Practice Questions (AP Calculus BC Style)

Question 1 (Multiple Choice)

If $f$ is a differentiable function such that $f(2)=5$, $f^{\prime }(2)=-3$, and $f(4)=1$, $f^{\prime }(4)=2$, what is $\frac{d}{dx}\left[{\int }_x^{2x}f(t)dt\right]$ evaluated at $x=2$? (A) $-8$ (B) $-3$ (C) $2$ (D) $7$

Worked Solution: We use the derivative rule for accumulation functions with two variable bounds: $\frac{d}{dx}{\int }_{l(x)}^{u(x)}f(t)dt=f(u(x))u^{\prime }(x)-f(l(x))l^{\prime }(x)$. Here, the upper bound is $u(x)=2x$, so $u^{\prime }(x)=2$, and the lower bound is $l(x)=x$, so $l^{\prime }(x)=1$. At $x=2$, $u(2)=4$ and $l(2)=2$. Substitute the given values of $f$: $f(4)\cdot 2-f(2)\cdot 1=(1)(2)-(5)(1)=2-5=-3$. The correct answer is B.

Question 2 (Free Response)

Let $f(t)=2t{e}^{-t^2}$ for all real $t$. Let $A(x)={\int }_0^{x}f(t)dt$. (a) Find $A^{\prime }(x)$ and $A^{\prime \prime }(x)$ in terms of $x$. (b) Find all $x$-coordinates of inflection points of $A(x)$ on $(-\infty ,\infty )$. Justify your answer. (c) Evaluate $A(2)$ in exact form.

Worked Solution: (a) By FTC Part 1, $A^{\prime }(x)=f(x)=2x{e}^{-x^2}$. Differentiate using the product rule to get $A^{\prime \prime }(x)$: $$A^{\prime \prime }(x)=2e^{-x^2}+2x(-2x{e}^{-x^2})=2e^{-x^2}(1-2x^2)$$

(b) Inflection points occur where $A^{\prime \prime }(x)$ changes sign. Set $A^{\prime \prime }(x)=0$: $2e^{-x^2}(1-2x^2)=0$. Since $e^{-x^2}$ is never zero for real $x$, we solve $1-2x^2=0$ to get $x=\pm \frac{\sqrt{2}}{2}$. Checking sign changes: for $x<-\frac{\sqrt{2}}{2}$, $A^{\prime \prime }(x)<0$; between $-\frac{\sqrt{2}}{2}$ and $\frac{\sqrt{2}}{2}$, $A^{\prime \prime }(x)>0$; for $x>\frac{\sqrt{2}}{2}$, $A^{\prime \prime }(x)<0$. $A^{\prime \prime }$ changes sign at both points, so the inflection points are $x=-\frac{\sqrt{2}}{2}$ and $x=\frac{\sqrt{2}}{2}$.

(c) By FTC Part 2, $A(2)=F(2)-F(0)$, where $F(x)$ is the antiderivative of $2t{e}^{-t^2}$. Using u-substitution ($u=-t^2$, $du=-2tdt$), we get $F(x)=-e^{-t^2}$. Evaluate: $$A(2)=(-e^{-(2{)}^2})-(-e^0)=1-e^{-4}=1-\frac{1}{e^4}$$

Question 3 (Application / Real-World Style)

The rate of change of the volume of water in a reservoir, in cubic meters per hour, over a 12-hour period is given by $r(t)=100\cos \left(\frac{\pi t}{6}\right)$, where $t$ is measured in hours starting at $t=0$. At $t=0$, the volume of water in the reservoir is 2000 cubic meters. Write the volume $V(t)$ as an accumulation function, then calculate the volume of water in the reservoir at $t=6$ hours, and interpret your result.

Worked Solution: The volume at time $t$ equals the initial volume plus the net accumulation of the rate of change from 0 to $t$, so: $$V(t)=2000+{\int }_0^{t}100\cos \left(\frac{\pi s}{6}\right)ds$$ To find $V(6)$, apply FTC Part 2. The antiderivative of $100\cos \left(\frac{\pi s}{6}\right)$ is $\frac{600}{\pi }\sin \left(\frac{\pi s}{6}\right)$. Evaluate from 0 to 6: $$\frac{600}{\pi }\sin \left(\frac{\pi \cdot 6}{6}\right)-\frac{600}{\pi }\sin (0)=\frac{600}{\pi }\sin (\pi )-0=0$$ So $V(6)=2000+0=2000$ cubic meters. In context, this means that after 6 hours, the net change in the volume of the reservoir over the first 6 hours is zero, so the volume is exactly the same as it was at the start of the observation period.

7. Quick Reference Cheatsheet

8. What's Next

This topic is the foundation for all further integration work in AP Calculus BC. Immediately after this, you will learn integration techniques (u-substitution, integration by parts, partial fractions) that rely entirely on the Fundamental Theorem of Calculus to evaluate definite and indefinite integrals. You will also apply accumulation functions to calculate net change of quantities over an interval, arc length of curves, and volumes of revolution, all of which depend on the inverse relationship between differentiation and integration established by FTC. Without mastering this chapter, you will not be able to correctly solve most FRQ problems involving these more advanced topics, which make up a large portion of the AP BC exam.

• U-substitution for definite integrals
• Integration by parts
• Net change from rates of change
• Area and volume applications