FTC and Definite Integrals for AP Calculus AB

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

1. The Second Fundamental Theorem of Calculus (FTC 2): Evaluating Definite Integrals★★☆☆☆⏱ 4 min

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 $F (x) ∣_{a}^{b}$ , 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.

📘 Definition

Second Fundamental Theorem of Calculus

$\int_{a}^{b} f (x) d x = F (b) - F (a)$

The theorem connecting antiderivatives to exact evaluation of definite integrals for continuous functions.

📐 Worked Example

Evaluate $\int_{1}^{4} (3 x^{2} - 2 x) d x$ .

1
First, find the antiderivative of each term using the power rule for antiderivatives: $\int x^{n} d x = \frac{x ^{n + 1}}{n + 1} + C$ for $n \neq = - 1$ .
2
Antiderivative of $3 x^{2}$ is $3 \cdot \frac{x ^{3}}{3} = x^{3}$ . Antiderivative of $- 2 x = - 2 x^{1/2}$ 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 $F (x) ∣_{a}^{b}$ 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.

2. The First Fundamental Theorem of Calculus (FTC 1): Derivatives of Accumulation Functions★★★☆☆⏱ 4 min

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 $g (x) = \int_{v (x)}^{u (x)} f (t) d t$ is:

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

📘 Definition

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 $\frac{d}{d x} [\int_{c o s x}^{x^{3}} e^{2 t} d t]$ .

1
Apply the extended FTC 1 rule for variable upper and lower bounds: the derivative equals $f (upper bound) \cdot \frac{d}{d x} (upper bound) - f (lower bound) \cdot \frac{d}{d x} (lower bound)$ .
2
Identify components: $f (t) = e^{2 t}$ , upper bound $u (x) = x^{3}$ , lower bound $v (x) = cos x$ .
3
Compute derivatives of the bounds: $u^{'} (x) = 3 x^{2}$ , $v^{'} (x) = - sin x$ .
4
Evaluate $f$ at both bounds: $f (u (x)) = e^{2 (x^{3})} = e^{2 x^{3}}$ , $f (v (x)) = e^{2 c o s x}$ .
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.

3. The Net Change Theorem: FTC for Contextual Problems★★★☆☆⏱ 3 min

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: $F (b) = F (a) + \int_{a}^{b} f (t) d t$ .

📐 Worked Example

The rate at which water flows into a tank is given by $r (t) = 2 t + 5 sin (\frac{π t}{12})$ 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.

4. AP Style Concept Check★★★☆☆⏱ 3 min

✓ Quick check

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

If $F (x) = \int_{0}^{x^{2}} 1 + t^{3} d t$ , what is the value of $F^{'} (2)$ ?
- A) $65$
- B) $465$
- C) $41$
- D) $265$
Reveal answer
1 —
Correct. Using FTC 1 with chain rule: $F^{'} (x) = 2 x 1 + x^{6}$ , so $F^{'} (2) = 465$ .
Evaluate $\int_{0}^{2} (2 x - 6 e^{x}) d x$ . Which result is correct?
- A) $10 - 6 e^{2} \approx - 34.17$
- B) $4 - 6 e^{2} \approx - 40.17$
- C) $- 6 e^{2} \approx - 44.17$
- D) $10 - 6 e \approx - 6.31$

5. Common Pitfalls

Wrong move:

When differentiating $g (x) = \int_{2}^{x^{2}} t^{3} + 1 d t$ , you write $g^{'} (x) = (x^{2})^{3} + 1$ and omit the chain rule term.

Why:

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 move:

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 move:

When evaluating $\int_{- 2}^{2} \frac{1}{x ^{2}} d x$ , you apply FTC 2 to get $- \frac{1}{x}_{- 2}^{2} = - 1$ and accept this as the final answer.

Why:

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 move:

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 move:

When computing a definite integral, you subtract the upper bound value from the lower bound value.

Why:

Students mix up the order of evaluation, especially when bounds are written in reverse order in the problem.

Correct move:

Always follow the notation convention $F (x) ∣_{a}^{b} = F (b) - F (a)$ , which means upper bound minus lower bound, no exceptions.

Wrong move:

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 $\int_{0}^{10} r (t) d t$ as your final answer.

Why:

Students confuse net change with total final amount.

Correct move:

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 move:

When differentiating $\int_{x}^{5} e^{s i n t} d t$ , you get $- e^{s i n x}$ and then remove the negative sign because you think it is a mistake.

Why:

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 move:

Flip the bounds to get $- \int_{5}^{x} e^{s i n t} d t$ , apply FTC, and keep the negative sign in your final derivative.

6. Quick Reference Cheatsheet

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

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
Differentiate variable-bound accumulation function
2022 · FRQ
Net change for real-world volume problem

Going deeper

unit overviewUnit 6: Integration and Accumulation of Change OverviewParent unit overview for this topic

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.