Fundamental Theorem of Calculus and accumulation functions

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

1. Core Concepts: Accumulation Functions and FTC Overview★☆☆☆☆⏱ 2 min

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, tested in both multiple-choice and free-response sections.

📘 Definition

Accumulation Function

An accumulation function is a function defined as a definite integral with at least one variable bound, rather than two constant bounds. It gives the net area (or net change) accumulated under the integrand from a fixed point to a variable upper bound.

Example:

$A (x) = \int_{a}^{x} f (t) d t$ is an accumulation function that gives net accumulated area from constant $a$ to variable $x$ .

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

FTC Part 1 formalizes the inverse relationship between integration and differentiation. If $f (t)$ is continuous on an interval $[a, b]$ , and we define $A (x) = \int_{a}^{x} f (t) d t$ for $a \leq x \leq b$ , then $A^{'} (x) = f (x)$ . Intuitively, 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 $x$ , the derivative is:

\frac{d}{dx} \int_a^{u(x)} f(t) dt = f(u(x)) \cdot u'(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'(x) - f(l(x)) l'(x)

📐 Worked Example

Find $\frac{d}{d x} [\int_{c o s x}^{x^{2}} ln (t^{2} + 1) d t]$ .

1
Identify the components: variable lower bound $l (x) = cos x$ , variable upper bound $u (x) = x^{2}$ , integrand $f (t) = ln (t^{2} + 1)$ .
2
Compute derivatives of the bounds: $u^{'} (x) = 2 x$ , $l^{'} (x) = - sin x$ .
3
Apply the two-bound derivative rule: $\frac{d}{d x} = f (u (x)) u^{'} (x) - f (l (x)) l^{'} (x)$ .
4
Substitute and simplify:
5
$\ln\left((x^2)^2 + 1\right)(2x) - \ln\left((\cos x)^2 + 1\right)(-\sin x) = 2x \ln(x^4 + 1) + \sin x \ln(\cos^2 x + 1)$

✓ Quick check

Test your understanding of FTC Part 1:

If $f$ is a differentiable function such that $f (2) = 5$ , $f^{'} (2) = - 3$ , and $f (4) = 1$ , $f^{'} (4) = 2$ , what is $\frac{d}{d x} [\int_{x}^{2 x} f (t) d t]$ evaluated at $x = 2$ ?
- -8
- -3
- 2
- 7
Reveal answer
1 —
Using the two-bound rule: $f (2 x) \cdot 2 - f (x) \cdot 1$ . At $x = 2$ , this is $1 \cdot 2 - 5 \cdot 1 = - 3$ .

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

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 $F^{'} (x) = f (x)$ ), then:

\int_a^b f(x) dx = F(b) - F(a)

The intuition aligns with the net change interpretation of integration: the definite integral of a rate of change $F^{'} (x)$ 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.

📐 Worked Example

Evaluate $\int_{0}^{π /2} (2 sin x + 4 x cos (x^{2})) d x$ .

1
Confirm the integrand is continuous on $[0, π /2]$ , so FTC Part 2 applies.
2
Find the antiderivative term by term: the antiderivative of $2 sin x$ is $- 2 cos x$ . For $4 x cos (x^{2})$ , use u-substitution ( $u = x^{2}$ , $d u = 2 x d x$ ) 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:
4
$F\left(\frac{\pi}{2}\right) = -2 \cos\left(\frac{\pi}{2}\right) + 2 \sin\left(\left(\frac{\pi}{2}\right)^2\right) = 0 + 2 \sin\left(\frac{\pi^2}{4}\right)$
5
Evaluate at the lower bound: $F (0) = - 2 cos (0) + 2 sin (0) = - 2 (1) + 0 = - 2$ .
6
Subtract lower from upper to get the final result:
7
$F\left(\frac{\pi}{2}\right) - F(0) = 2 \sin\left(\frac{\pi^2}{4}\right) - (-2) = 2 + 2 \sin\left(\frac{\pi^2}{4}\right)$

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★★★☆☆⏱ 5 min

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 $A (x) = \int_{a}^{x} f (t) d t$ , then $A^{'} (x) = f (x)$ : $A (x)$ increases when $f (x) > 0$ , decreases when $f (x) < 0$ , and has local extrema where $f (x)$ changes sign.
For the second derivative, $A^{''} (x) = f^{'} (x)$ : $A (x)$ is concave up when $f (x)$ is increasing ( $f^{'} (x) > 0$ ), concave down when $f (x)$ is decreasing ( $f^{'} (x) < 0$ ), and has inflection points where $f (x)$ changes slope ( $f^{'} (x)$ changes sign).

📐 Worked Example

The graph of $y = f (t)$ is piecewise linear through $(- 2, 3)$ , $(0, 3)$ , $(1.5, 0)$ , $(2, - 1)$ , $(3.5, 0)$ , $(4, 1)$ . Let $A (x) = \int_{0}^{x} f (t) d t$ . Identify all local maxima of $A (x)$ on $(- 2, 4)$ .

1
By FTC 1, $A^{'} (x) = f (x)$ , 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 $A^{'} (x) > 0$ ( $A$ increasing). For $1.5 < x < 3.5$ , $f (x) < 0$ so $A^{'} (x) < 0$ ( $A$ decreasing). Thus $x = 1.5$ is a local maximum.
3
First derivative test at $x = 3.5$ : for $1.5 < x < 3.5$ , $f (x) < 0$ so $A$ decreasing. For $x > 3.5$ , $f (x) > 0$ so $A$ increasing. Thus $x = 3.5$ 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 $x = 1.5$ .

📐 Worked Example

Let $f (t) = 2 t e^{- t^{2}}$ for all real $t$ . Let $A (x) = \int_{0}^{x} f (t) d t$ . (a) Find $A^{'} (x)$ and $A^{''} (x)$ 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.

1
(a) By FTC Part 1, $A^{'} (x) = f (x) = 2 x e^{- x^{2}}$ . Differentiate using the product rule to get $A^{''} (x)$ :
2
$A''(x) = 2e^{-x^2} + 2x(-2x e^{-x^2}) = 2e^{-x^2}(1 - 2x^2)$
3
(b) Inflection points occur where $A^{''} (x)$ changes sign. Set $A^{''} (x) = 0$ : since $e^{- x^{2}}$ is never zero for real $x$ , solve $1 - 2 x^{2} = 0$ to get $x = \pm \frac{2}{2}$ . Checking sign changes confirms $A^{''}$ changes sign at both points, so these are the inflection points.
4
(c) By FTC Part 2, $A (2) = F (2) - F (0)$ , where $F (x)$ is the antiderivative of $2 t e^{- t^{2}}$ . Using u-substitution, we get $F (x) = - e^{- x^{2}}$ . Evaluate:
5
$A(2) = (-e^{-4}) - (-e^{0}) = 1 - e^{-4} = 1 - \frac{1}{e^4}$

📐 Worked Example

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 $r (t) = 100 cos (\frac{π t}{6})$ , 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.

1
The volume at time $t$ equals the initial volume plus the net accumulation of the rate of change from 0 to $t$ , so:
2
$V(t) = 2000 + \int_0^t 100 \cos\left(\frac{\pi s}{6}\right) ds$
3
To find $V (6)$ , apply FTC Part 2. The antiderivative of $100 cos (\frac{π s}{6})$ is $\frac{600}{π} sin (\frac{π s}{6})$ . Evaluate from 0 to 6:
4
$\frac{600}{\pi} \sin(\pi) - \frac{600}{\pi} \sin(0) = 0$
5
So $V (6) = 2000 + 0 = 2000$ cubic meters. In context, the net change in volume over the first 6 hours is zero, so the volume matches its initial value.

Exam tip:

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

5. Common Pitfalls

Wrong move:

When finding $\frac{d}{d x} \int_{a}^{u (x)} f (t) d t$ , you write $f^{'} (u (x)) u^{'} (x)$ instead of $f (u (x)) u^{'} (x)$ .

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: $\frac{d}{d x} \int_{v (x)}^{b} f (t) d t = f (v (x)) v^{'} (x)$ instead of $- f (v (x)) v^{'} (x)$ .

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 $\int_{a}^{b} f (x) d x$ 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 $f^{'} (x)$ .

Why:

Students forget the relationship $A^{'} = f$ , $A^{''} = f^{'}$ , so they mix up first and second derivative information.

Correct move:

Label derivatives explicitly: $A^{'} (x) = f (x)$ , so $A^{''} (x) = \frac{d}{d x} A^{'} (x) = f^{'} (x)$ . Use that label to check the sign for concavity.

Wrong move:

You evaluate $\frac{d}{d x} \int_{0}^{2 x} sin (t) d t$ as $sin (2 x)$ instead of $2 sin (2 x)$ .

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. Quick Reference Cheatsheet

Category	Formula	Notes
Accumulation Function	$A (x) = \int_{a}^{x} f (t) d t$	$f$ continuous on interval containing $a$ and $x$ ; gives net accumulated change/area from $a$ to $x$
FTC Part 1 (Constant Lower Bound)	If $A (x) = \int_{a}^{x} f (t) d t$ , then $A^{'} (x) = f (x)$	Requires $f$ continuous; works for any constant lower bound $a$
FTC Part 1 (General Variable Bounds)	$\frac{d}{d x} \int_{l (x)}^{u (x)} f (t) d t = f (u (x)) u^{'} (x) - f (l (x)) l^{'} (x)$	Negative sign comes from flipping the lower bound integral; always multiply by derivative of each bound
FTC Part 2 (Definite Integral Evaluation)	If $F^{'} (x) = f (x)$ and $f$ continuous on $[a, b]$ , then $\int_{a}^{b} f (x) d x = F (b) - F (a)$	Constant of integration $C$ cancels out, so it is omitted for definite integrals
First Derivative of Accumulation	$A^{'} (x) = f (x)$ for $A (x) = \int_{a}^{x} f (t) d t$	Used to find intervals of increase/decrease and local extrema of $A (x)$
Second Derivative of Accumulation	$A^{''} (x) = f^{'} (x)$ for $A (x) = \int_{a}^{x} f (t) d t$	Used to find intervals of concavity and inflection points of $A (x)$ ; do not confuse with $f (x)$
Derivative of Constant-Bound Integral	$\frac{d}{d x} \int_{a}^{b} f (t) d t = 0$	Common MCQ trick: an integral over constant bounds is a constant, so its derivative is zero

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
Derivative of variable bound integral
2022 · FRQ
Analyze accumulation from f graph

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. Mastery of FTC is required to correctly solve most FRQ problems involving these more advanced topics, which make up a large portion of the AP BC exam.