# FTC and Definite Integrals

> AP Calculus BC · Integration and Accumulation of Change
> Source: https://www.owlsprep.com/study/ap-calculus-bc-u6-ftc-and-definite-integrals/

This subtopic covers both parts of the Fundamental Theorem of Calculus (FTC), derivatives of accumulation functions, evaluation of definite integrals, and the Net Change Theorem, making up 7-10% of the AP Calculus BC exam.

**Prerequisites:** Limits of Riemann sums for definite integrals; Basic derivative and antiderivative rules; Checking continuity of functions

## Learning objectives

- Differentiate accumulation functions with variable bounds using FTC Part 1
- Evaluate definite integrals using FTC Part 2
- Apply the Net Change Theorem to contextual rate problems
- Identify common pitfalls when applying FTC rules

## FTC Part 1: Derivatives of Accumulation Functions

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.

**FTC Part 1 (Accumulation Functions)** — 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 $\frac{d}{dx} \left( \int_a^{u(x)} f(t) dt \right) = f(u(x)) \cdot u'(x)$. For a variable lower bound, reverse the bounds to get a negative sign before applying the rule.

*Notation:* F(x) = \int_a^x f(t) dt

**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, 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: $\sqrt{x^2} = |x|$, and cosine is even so $\cos(|x|) = \cos(x)$. The derivative of $x^2$ is $2x$.
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.

## FTC Part 2: Evaluating Definite Integrals

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.

**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: $\int_a^b f(x) dx = F(b) - F(a)$. The constant of integration for indefinite integrals cancels out during subtraction, so we can omit it for definite integrals.

**Worked example:** Evaluate $\int_0^{\pi/4} \left( \sec^2 x - 2x \right) dx$

1. Confirm the integrand $f(x) = \sec^2 x - 2x$ is continuous on $[0, \pi/4]$, 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 $x = \pi/4$:
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.

## The Net Change Theorem

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.

**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: $\text{Net Change} = Q(b) - Q(a) = \int_a^b r(t) dt$. 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 $v(t) = t^2 - 4t + 3$ meters per second, for $0 ≤ t ≤ 4$. 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.

## Additional AP-Style Worked Examples

**Worked example:** If $h(x) = \int_{1}^{e^x} \ln(t^2 + 1) dt$, what is $h'(2)$?<br>Options: (A) $e^2 \ln(e^4 + 1)$, (B) $\ln(e^4 + 1)$, (C) $e^2 \ln(2e^2 + 1)$, (D) $e^2 \ln(e^{2x} + 1)$

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

**Worked example:** Let $f(x) = 2x + \sin x$, and $g(y) = \int_0^y f(x) dx$. Find (a) $g(\pi)$, (b) $g'\left(\frac{\pi}{2}\right)$, (c) the average value of $f(x)$ on $[0, \pi]$.

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 $y = \frac{\pi}{2}$:
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 $\frac{1}{b-a} \int_a^b f(x) dx$. 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 $r(t) = -t^3 + 5t^2 + 10t$ cookies per hour ($0 ≤ t ≤ 4$). 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.

## Common pitfalls

- **Wrong:** When differentiating $\int_{g(x)}^b f(t) dt$, writing the derivative as $f(g(x)) \cdot g'(x)$ with no negative sign.
  - Why it fails: Confuses variable upper limits with variable lower limits, forgetting reversing integration order changes the integral's sign.
  - Correct: 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:** When evaluating $\int_a^b f(x) dx$, calculating $F(a) - F(b)$ instead of $F(b) - F(a)$.
  - Why it fails: Rushes to plug in values without reading the bounds, confuses the statement of FTC 2.
  - Correct: Write "F(upper bound) minus F(lower bound)" explicitly on your paper before substituting any values.
- **Wrong:** When differentiating $\int_{0}^{x} t f(t) dt$, writing the derivative as $x f(t)$ leaving $t$ as the variable.
  - Why it fails: Confuses the dummy variable of integration with the variable bound, treats the variable of integration as a constant.
  - Correct: Replace every instance of the dummy variable $t$ with the variable bound after applying FTC 1, so the derivative is $x f(x)$.
- **Wrong:** Adding a constant of integration (+C) to the final numerical answer of a definite integral.
  - Why it fails: Confuses definite integrals (which evaluate to a single number) with indefinite integrals (which are families of antiderivatives).
  - Correct: Drop the constant of integration when applying FTC 2, as it cancels out in the subtraction step.
- **Wrong:** Applying FTC to evaluate $\int_{-1}^1 \frac{1}{x} dx = \ln|1| - \ln|-1| = 0$, ignoring the discontinuity at $x=0$.
  - Why it fails: Forgets that FTC only applies if the integrand is continuous on the entire interval between the two bounds.
  - Correct: 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:** Interpreting the integral of velocity over time as total distance instead of net displacement.
  - Why it fails: Confuses net change (integral of the rate) with total change (integral of the absolute value of the rate).
  - Correct: 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.

## 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 |

## 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.

- [Integration by Parts](https://www.owlsprep.com/study/ap-calculus-bc-u6-integration-by-parts/)
- [Improper Integrals](https://www.owlsprep.com/study/ap-calculus-bc-u6-improper-integrals/)
- [Antiderivatives and indefinite integrals (basic rules)](https://www.owlsprep.com/study/ap-calculus-bc-u6-antiderivatives-and-indefinite-integrals/)

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