# FTC and Definite Integrals for AP Calculus AB

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

This module covers both parts of the Fundamental Theorem of Calculus (FTC), evaluating definite integrals via antiderivatives, differentiating accumulation functions, and the inverse relationship between differentiation and integration, all heavily tested on the AP Calculus AB exam.

**Prerequisites:** How to compute antiderivatives of basic functions; How to apply the chain rule for derivatives; What a definite integral represents as a limit of Riemann sums

## Learning objectives

- Evaluate definite integrals using the Second Fundamental Theorem of Calculus
- Differentiate accumulation functions using the First Fundamental Theorem of Calculus
- Apply the Net Change Theorem to solve contextual rate of change problems
- Avoid common errors when applying FTC rules to AP exam questions

## The Second Fundamental Theorem of Calculus (FTC 2): Evaluating Definite Integrals

FTC 2 gives the method for computing exact values of definite integrals using antiderivatives, eliminating the need to compute limits of infinite Riemann sums. The theorem applies when $f(x)$ is continuous on the closed interval $[a,b]$, and $F(x)$ is any antiderivative of $f(x)$ (meaning $F'(x) = f(x)$ for all $x$ in $[a,b]$).

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

The standard notation for this evaluation is $\left. F(x) \right|_a^b$, which explicitly means 'evaluate $F$ at the upper bound $b$, subtract the value of $F$ at the lower bound $a$.' The constant of integration $+C$ can be omitted entirely for definite integrals, because it cancels out when subtracting $F(a)$ from $F(b)$. FTC 2 can only be used if $f$ is continuous over the entire interval of integration; discontinuities like vertical asymptotes invalidate direct application of the theorem.

**Second Fundamental Theorem of Calculus** — The theorem connecting antiderivatives to exact evaluation of definite integrals for continuous functions.

*Notation:* $\int_a^b f(x) dx = F(b) - F(a)$

**Worked example:** Evaluate $\int_1^4 (3x^2 - 2\sqrt{x}) dx$.

1. First, find the antiderivative of each term using the power rule for antiderivatives: $\int x^n dx = \frac{x^{n+1}}{n+1} + C$ for $n \neq -1$.
2. Antiderivative of $3x^2$ is $3 \cdot \frac{x^3}{3} = x^3$. Antiderivative of $-2\sqrt{x} = -2x^{1/2}$ is $-2 \cdot \frac{x^{3/2}}{3/2} = -\frac{4}{3}x^{3/2}$.
3. Combine terms to get $F(x) = x^3 - \frac{4}{3}x^{3/2}$, omitting $+C$.
4. Evaluate at upper bound $x=4$:
5. $$F(4) = 4^3 - \frac{4}{3}(4)^{3/2} = 64 - \frac{32}{3} = \frac{160}{3}$$
6. Evaluate at lower bound $x=1$:
7. $$F(1) = 1^3 - \frac{4}{3}(1)^{3/2} = 1 - \frac{4}{3} = -\frac{1}{3}$$
8. Subtract lower bound from upper bound to get the final result:
9. $$\int_1^4 (3x^2 - 2\sqrt{x}) dx = F(4) - F(1) = \frac{160}{3} - \left(-\frac{1}{3}\right) = \frac{161}{3}$$

> **Exam tip:** Always write the evaluation as $\left. F(x) \right|_a^b$ explicitly to avoid mixing up the order of bounds. Reversing the bounds will flip the sign of your answer, which is a common avoidable error.

## The First Fundamental Theorem of Calculus (FTC 1): Derivatives of Accumulation Functions

FTC 1 formalizes the inverse relationship between integration and differentiation, and is heavily tested on AP Calculus AB multiple-choice questions involving accumulation functions (functions defined as integrals). The base case for a constant lower bound and variable upper bound $x$ is:

$$\frac{d}{dx} \left[ \int_a^x f(t) dt \right] = f(x)$$

When the upper bound is a function of $x$ ($u(x)$) rather than just $x$, we apply the chain rule to get the general form. If both bounds are functions of $x$, we split the integral at a constant to get the extended rule. The extended rule for $g(x) = \int_{v(x)}^{u(x)} f(t) dt$ is:

$$g'(x) = f(u(x))u'(x) - f(v(x))v'(x)$$

**Accumulation Function** — A function defined as a definite integral where one or both bounds are functions of the independent variable, representing accumulated area under a curve.

**Worked example:** Find $\frac{d}{dx} \left[ \int_{\cos x}^{x^3} e^{2t} dt \right]$.

1. Apply the extended FTC 1 rule for variable upper and lower bounds: the derivative equals $f(\text{upper bound}) \cdot \frac{d}{dx}(\text{upper bound}) - f(\text{lower bound}) \cdot \frac{d}{dx}(\text{lower bound})$.
2. Identify components: $f(t) = e^{2t}$, upper bound $u(x) = x^3$, lower bound $v(x) = \cos x$.
3. Compute derivatives of the bounds: $u'(x) = 3x^2$, $v'(x) = -\sin x$.
4. Evaluate $f$ at both bounds: $f(u(x)) = e^{2(x^3)} = e^{2x^3}$, $f(v(x)) = e^{2\cos x}$.
5. Substitute into the rule and simplify:
6. $$\frac{d}{dx} \left[ \int_{\cos x}^{x^3} e^{2t} dt \right] = (e^{2x^3})(3x^2) - (e^{2\cos x})(-\sin x) = 3x^2 e^{2x^3} + e^{2\cos x} \sin x$$

> **Exam tip:** When the variable of differentiation is in the lower bound, do not drop the required negative sign from splitting the integral. Double-check the sign of the lower bound term before submitting your final answer.

## The Net Change Theorem: FTC for Contextual Problems

The Net Change Theorem is a direct application of FTC 2 to real-world problems, where we integrate a rate of change to find total accumulated change over an interval. It states that the net change in a quantity $F(t)$ from $t=a$ to $t=b$ is equal to the definite integral of its rate of change $f(t) = F'(t)$ over that interval:

$$\text{Net Change} = F(b) - F(a) = \int_a^b f(t) dt$$

This theorem is the foundation for all contextual integration problems on AP Calculus AB, including particle displacement, population change, drug amount in the bloodstream, and total profit from marginal profit. Net change counts positive and negative change against each other (e.g., moving left vs. right for a particle), unlike total distance which requires integrating the absolute value of the rate. If asked for the final amount of a quantity at $t=b$ given an initial amount $F(a)$, rearrange the formula to get: $F(b) = F(a) + \int_a^b f(t) dt$.

**Worked example:** The rate at which water flows into a tank is given by $r(t) = 2t + 5\sin\left(\frac{\pi t}{12}\right)$ gallons per minute, for $0 \leq t \leq 20$ minutes. If the tank initially has 30 gallons of water at $t=0$, how much water is in the tank at $t=10$ minutes?

1. Let $W(t)$ be the total amount of water in the tank at time $t$. We know $W(0) = 30$ and $W'(t) = r(t)$, so by the Net Change Theorem:
2. $$W(10) = W(0) + \int_0^{10} r(t) dt$$
3. Find the antiderivative of $r(t)$:
4. $$\int \left(2t + 5\sin\left(\frac{\pi t}{12}\right)\right) dt = t^2 - \frac{60}{\pi} \cos\left(\frac{\pi t}{12}\right)$$
5. Evaluate the definite integral:
6. $$\left[ 10^2 - \frac{60}{\pi} \cos\left(\frac{10\pi}{12}\right) \right] - \left[ 0^2 - \frac{60}{\pi} \cos(0) \right] = 100 - \frac{60}{\pi}\left(-\frac{\sqrt{3}}{2}\right) + \frac{60}{\pi}(1) \approx 135.64$$
7. Add the initial amount to get the final total:
8. $$W(10) = 30 + 135.64 \approx 165.64$$
9. The tank has approximately 165.64 gallons of water at $t=10$.

> **Exam tip:** When asked for the final amount of a quantity, the definite integral only gives the net change, not the final amount. Always add the net change to the given initial amount to get the correct answer.

## AP Style Concept Check

**Check your understanding**

Test your understanding of FTC rules with these AP-style questions:

1. If $F(x) = \int_{0}^{x^2} \sqrt{1 + t^3} dt$, what is the value of $F'(2)$?

   - A) $\sqrt{65}$
   - B) $4\sqrt{65}$
   - C) $\sqrt{41}$
   - D) $2\sqrt{65}$

   *Answer:* B) $4\sqrt{65}$

   *Why:* Correct. Using FTC 1 with chain rule: $F'(x) = 2x\sqrt{1 + x^6}$, so $F'(2) = 4\sqrt{65}$.

2. Evaluate $\int_0^2 (2x - 6e^x) dx$. Which result is correct?

   - A) $10 - 6e^2 \approx -34.17$
   - B) $4 - 6e^2 \approx -40.17$
   - C) $-6e^2 \approx -44.17$
   - D) $10 - 6e \approx -6.31$

   *Answer:* A) $10 - 6e^2 \approx -34.17$

   *Why:* Correct. Antiderivative is $x^2 - 6e^x$, evaluate from 0 to 2: $(4 - 6e^2) - (0 - 6e^0) = 10 - 6e^2$.

## Common pitfalls

- **Wrong:** When differentiating $g(x) = \int_2^{x^2} \sqrt{t^3 + 1} dt$, you write $g'(x) = \sqrt{(x^2)^3 + 1}$ and omit the chain rule term.
  - Why it fails: Students remember FTC gives the integrand evaluated at the upper bound, but forget the chain rule when the bound is a function of $x$.
  - Correct: Always check if the upper or lower bound is a function of $x$; if it is, multiply by the derivative of the bound every time.
- **Wrong:** When evaluating $\int_{-2}^2 \frac{1}{x^2} dx$, you apply FTC 2 to get $\left. -\frac{1}{x} \right|_{-2}^2 = -1$ and accept this as the final answer.
  - Why it fails: Students forget FTC 2 requires $f$ to be continuous over the entire interval of integration, and this function has a discontinuity at $x=0$ inside the interval.
  - Correct: Always check for discontinuities of the integrand inside the interval before applying FTC 2; AP Calculus AB will only ask you to apply FTC to continuous functions.
- **Wrong:** When computing a definite integral, you subtract the upper bound value from the lower bound value.
  - Why it fails: Students mix up the order of evaluation, especially when bounds are written in reverse order in the problem.
  - Correct: Always follow the notation convention $\left. F(x) \right|_a^b = F(b) - F(a)$, which means upper bound minus lower bound, no exceptions.
- **Wrong:** When finding the final population at $t=10$ given an initial population of 1000 and a growth rate $r(t)$, you report the value of $\int_0^{10} r(t) dt$ as your final answer.
  - Why it fails: Students confuse net change with total final amount.
  - Correct: Remember that the integral of a rate gives only net change; add the net change to the initial amount to get the final total.
- **Wrong:** When differentiating $\int_{x}^5 e^{\sin t} dt$, you get $-e^{\sin x}$ and then remove the negative sign because you think it is a mistake.
  - Why it fails: Students are used to the constant lower bound case and forget the negative sign is required when the variable is in the lower bound.
  - Correct: Flip the bounds to get $-\int_5^x e^{\sin t} dt$, apply FTC, and keep the negative sign in your final derivative.

## Cheatsheet

| Category | Formula | Notes |
| --- | --- | --- |
| FTC 2 (Evaluate Definite Integral) | $\displaystyle \int_a^b f(x) dx = F(b) - F(a), \quad F'(x) = f(x)$ | $f$ must be continuous on $[a,b]$; constant of integration cancels, no $+C$ needed |
| FTC 1 (Constant Lower Bound, Variable Upper Bound) | $\displaystyle \frac{d}{dx} \left[ \int_a^{u(x)} f(t) dt \right] = f(u(x)) \cdot u'(x)$ | If $u(x) = x$, $u'(x) = 1$, so derivative simplifies directly to $f(x)$ |
| FTC 1 (Variable Upper and Lower Bounds) | $\displaystyle \frac{d}{dx} \left[ \int_{v(x)}^{u(x)} f(t) dt \right] = f(u(x))u'(x) - f(v(x))v'(x)$ | The negative sign on the lower bound term is required, do not drop it |
| Net Change Theorem | $\text{Net Change of } F(t) = F(b) - F(a) = \int_a^b F'(t) dt$ | Gives net change (accounts for positive/negative change), not total change of a quantity |
| Final Quantity from Rate | $\displaystyle F(b) = F(a) + \int_a^b F'(t) dt$ | Use for all contextual problems asking for total amount at time $b$ |
| Standard Evaluation Notation | $\displaystyle \left. F(x) \right\|_a^b = F(b) - F(a)$ | By convention, upper bound value always minus lower bound value; reversing bounds flips the sign |

## What's next

Mastery of the Fundamental Theorem of Calculus is the prerequisite for almost all remaining topics in AP Calculus AB. Immediately after this topic, you will apply FTC to compute areas between curves, solve integration by substitution problems, and analyze contextual problems involving motion, population growth, and accumulated change. Without solid command of both parts of FTC, you will struggle with nearly every applied integration problem that appears on the FRQ section, which accounts for a large share of your total exam score. FTC is also the core result that unifies differentiation and integration, the two big ideas of calculus tested across the entire exam.

- [Integration by Substitution](https://www.owlsprep.com/study/ap-calculus-ab-u6-integration-by-substitution/)
- [Antiderivatives and indefinite integrals (basic rules)](https://www.owlsprep.com/study/ap-calculus-ab-u6-antiderivatives-and-indefinite-integrals/)
- [Integration with long division and completing the square](https://www.owlsprep.com/study/ap-calculus-ab-u6-integration-with-long-division-and/)

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