FTC and Definite Integrals

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

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

The first part of FTC connects differentiation and accumulation of area under a curve. It establishes that the derivative of an accumulation function (a function defined by an integral with a variable bound) is exactly the original integrand.

📘 Definition

FTC Part 1 (Accumulation Functions)

F(x) = \int_a^x f(t) dt

If f is continuous on [a, b], then for all x ∈ [a,b], F is differentiable on (a,b) and F'(x) = f(x). For a variable upper bound u(x), the chain rule extension gives $\frac{d}{d x} (\int_{a}^{u (x)} f (t) d t) = f (u (x)) \cdot u^{'} (x)$ . For a variable lower bound, reverse the bounds to get a negative sign before applying the rule.

📐 Worked Example

Find $\frac{d}{d x} (\int_{2}^{x^{2}} cos (t) d t)$

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, upper bound is the differentiable function $u (x) = x^{2}$ .
2
Identify the integrand $f (t)$ :
3
$f(t) = \cos\left(\sqrt{t}\right)$
4
Apply the chain rule form of FTC 1: substitute $u (x)$ into $f (t)$ and multiply by the derivative of $u (x)$ :
5
$f(u(x)) \cdot u'(x) = \cos\left(\sqrt{x^2}\right) \cdot \frac{d}{dx}\left(x^2\right)$
6
Simplify: $x^{2} = ∣ x ∣$ , and cosine is even so $cos (∣ x ∣) = cos (x)$ . The derivative of $x^{2}$ is $2 x$ .
7
$2x \cos(x)$

Exam tip:

If both bounds are variable, split the integral at any constant between the bounds, then apply FTC 1 to each piece. Do not rely on memorization alone for this case.

2. FTC Part 2: Evaluating Definite Integrals★★☆☆☆⏱ 4 min

The second part of FTC lets us compute definite integrals using antiderivatives, eliminating the need for tedious Riemann sum limit calculations. It is the primary method for evaluating definite integrals in AP Calculus.

📘 Definition

FTC Part 2 (Definite Integral Evaluation)

If f is continuous on [a, b], and F is any antiderivative of f ( $F^{'} (x) = f (x)$ ), then: $\int_{a}^{b} f (x) d x = F (b) - F (a)$ . The constant of integration for indefinite integrals cancels out during subtraction, so we can omit it for definite integrals.

📐 Worked Example

Evaluate $\int_{0}^{π /4} (sec^{2} x - 2 x) d x$

1
Confirm the integrand $f (x) = sec^{2} x - 2 x$ is continuous on $[0, π /4]$ , so FTC 2 applies.
2
Find the antiderivative of $f (x)$ , omitting the constant of integration:
3
$F(x) = \tan x - x^2$
4
Evaluate $F$ at the upper bound $x = π /4$ :
5
$F\left(\frac{\pi}{4}\right) = \tan\left(\frac{\pi}{4}\right) - \left(\frac{\pi}{4}\right)^2 = 1 - \frac{\pi^2}{16}$
6
Evaluate $F$ at the lower bound $x = 0$ :
7
$F(0) = \tan(0) - 0^2 = 0$
8
Subtract lower bound from upper bound to get the final result:
9
$\left(1 - \frac{\pi^2}{16}\right) - 0 = 1 - \frac{\pi^2}{16}$

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.

3. The Net Change Theorem★★★☆☆⏱ 3 min

The Net Change Theorem is a direct applied interpretation of FTC Part 2, and it is extremely common in AP free-response questions involving rates of change.

📘 Definition

Net Change Theorem

If $Q (t)$ is a differentiable quantity with rate of change $r (t) = Q^{'} (t)$ , then the total net change in $Q$ over $[a, b]$ is: $Net Change = Q (b) - Q (a) = \int_{a}^{b} r (t) d t$ . Net change accounts for positive and negative changes; total change (total distance, for example) requires integrating the absolute value of the rate.

📐 Worked Example

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

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 $v (t)$ :
3
$F(t) = \frac{t^3}{3} - 2t^2 + 3t$
4
Apply FTC 2: Net displacement = $F (4) - F (0)$
5
Calculate values: $F (4) = \frac{64}{3} - 32 + 12 = \frac{4}{3}$ , $F (0) = 0$ .
6
Final result: 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.

4. Additional AP-Style Worked Examples★★★☆☆⏱ 3 min

📐 Worked Example

If $h (x) = \int_{1}^{e^{x}} ln (t^{2} + 1) d t$ , what is $h^{'} (2)$ ?
Options: (A) $e^{2} ln (e^{4} + 1)$ , (B) $ln (e^{4} + 1)$ , (C) $e^{2} ln (2 e^{2} + 1)$ , (D) $e^{2} ln (e^{2 x} + 1)$

1
This problem tests FTC Part 1 with the chain rule. For $h (x) = \int_{a}^{u (x)} f (t) d t$ , $h^{'} (x) = f (u (x)) u^{'} (x)$ . Here, $u (x) = e^{x}$ , so $u^{'} (x) = e^{x}$ , and $f (t) = ln (t^{2} + 1)$ .
2
Substitute $u (x)$ into $f$ : $f (e^{x}) = ln ((e^{x})^{2} + 1) = ln (e^{2 x} + 1)$
3
Multiply by $u^{'} (x)$ : $h^{'} (x) = e^{x} ln (e^{2 x} + 1)$
4
Evaluate at $x = 2$ : $h^{'} (2) = e^{2} ln (e^{4} + 1)$ , which matches option A. Correct answer: A

📐 Worked Example

Let $f (x) = 2 x + sin x$ , and $g (y) = \int_{0}^{y} f (x) d x$ . Find (a) $g (π)$ , (b) $g^{'} (\frac{π}{2})$ , (c) the average value of $f (x)$ on $[0, π]$ .

1
(a) Find an antiderivative of $f (x)$ :
2
$F(x) = x^2 - \cos x$
3
Apply FTC Part 2:
4
$g(\pi) = F(\pi) - F(0) = (\pi^2 - \cos \pi) - (0 - \cos 0) = \pi^2 + 2$
5
(b) By FTC Part 1, $g^{'} (y) = f (y)$ . Substitute $y = \frac{π}{2}$ :
6
$g'\left(\frac{\pi}{2}\right) = 2\left(\frac{\pi}{2}\right) + \sin\left(\frac{\pi}{2}\right) = \pi + 1$
7
(c) Average value formula is $\frac{1}{b - a} \int_{a}^{b} f (x) d x$ . Use the result from part (a):
8
$\text{Average value} = \frac{1}{\pi} (\pi^2 + 2) = \pi + \frac{2}{\pi}$

📐 Worked Example

A bakery runs a 4-hour morning sale, with cookie sales rate $r (t) = - t^{3} + 5 t^{2} + 10 t$ cookies per hour ( $0 \leq t \leq 4$ ). How many total cookies are sold during the entire sale? Round to the nearest whole number.

1
Total cookies sold is the net change in total cookies, so by the Net Change Theorem, we integrate the sales rate from $t = 0$ to $t = 4$ .
2
Find the antiderivative of $r (t)$ :
3
$R(t) = -\frac{t^4}{4} + \frac{5t^3}{3} + 5t^2$
4
Apply FTC 2:
5
$R(4) - R(0) = -64 + \frac{320}{3} + 80 = \frac{368}{3} ≈ 122.67$
6
Round to the nearest whole number: approximately 123 cookies are sold over the 4-hour sale.

5. Common Pitfalls

Wrong move:

When differentiating $\int_{g (x)}^{b} f (t) d t$ , writing the derivative as $f (g (x)) \cdot g^{'} (x)$ with no negative sign.

Why:

Confuses variable upper limits with variable lower limits, forgetting reversing integration order changes the integral's sign.

Correct move:

Always rewrite $\int_{g (x)}^{b} f (t) d t = - \int_{b}^{g (x)} f (t) d t$ before applying FTC 1, and carry the negative through your calculation.

Wrong move:

When evaluating $\int_{a}^{b} f (x) d x$ , 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 $\int_{0}^{x} t f (t) d t$ , writing the derivative as $x f (t)$ 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 $x f (x)$ .

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 $\int_{- 1}^{1} \frac{1}{x} d x = ln ∣1∣ - ln ∣ - 1∣ = 0$ , 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. Quick Reference Cheatsheet

Category	Formula	Notes
FTC Part 1 (constant lower bound)	\frac{d}{dx} \int_a^x f(t) dt = f(x)	Applies if f is continuous on $[a, x]$ ; t is a dummy variable
FTC Part 1 (variable upper bound)	\frac{d}{dx} \int_a^{u(x)} f(t) dt = f(u(x)) u'(x)	Always multiply by derivative of upper bound
FTC Part 1 (variable lower bound)	\frac{d}{dx} \int_{l(x)}^b f(t) dt = -f(l(x)) l'(x)	Flip bounds to get the negative sign
FTC Part 2 (evaluate definite integral)	\int_a^b f(x) dx = F(b) - F(a), \quad F'(x) = f(x)	Requires f continuous on entire $[a, b]$ ; no constant needed
Net Change Theorem	\int_a^b Q'(t) dt = Q(b) - Q(a)	Integral of a rate equals net change of Q
Two variable bounds	\frac{d}{dx} \int_{l(x)}^{u(x)} f(t) dt = f(u(x))u'(x) - f(l(x))l'(x)	Split at any constant to derive this rule

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 accumulation function
2022 · FRQ
Net change from rate function
2021 · MCQ
Evaluate definite integral via FTC

What's Next

Mastery of FTC and definite integrals is a non-negotiable foundation for all remaining integration topics in AP Calculus BC. Immediately after this topic, you will learn u-substitution, which relies entirely on FTC to evaluate final results after changing variables. Without solid mastery of FTC rules and conventions, you will struggle with more complex topics like integration by parts, partial fractions, and improper integrals, all heavily tested on the BC exam. FTC is also core to accumulation functions used for differential equations, area, and volume later in the course.