FTC Part 2: Evaluating Definite Integrals

6 min read

Reference: Stewart §5.3

Textbook Reference


Primary source	OpenStax Calculus Volume 2, Section 1.3: “The Fundamental Theorem of Calculus” (Part 2, the Evaluation Theorem)
Direct link	https://openstax.org/books/calculus-volume-2/pages/1-3-the-fundamental-theorem-of-calculus
Original home	OpenStax Calculus Volume 1, Section 5.3: “The Fundamental Theorem of Calculus”
Original link	https://openstax.org/books/calculus-volume-1/pages/5-3-the-fundamental-theorem-of-calculus
Stewart	Calculus (Stewart), Section 5.3: “The Fundamental Theorem of Calculus”

OpenStax names Part 2 the Evaluation Theorem: if $f$ is continuous on $[a,b]$ and $F$ is any antiderivative of $f$, then $\int_a^b f(x)\, dx = F(b) - F(a)$. Both sources are free and openly licensed. This is assumed knowledge in second-semester calculus, restated here so it does not have to be hunted down.

From limits of sums to antiderivatives

Before the Fundamental Theorem, computing the area under a curve meant dividing the region into thin rectangles, adding up their areas, and taking a limit. Finding the area under a parabola took Archimedes a long argument by hand.

With the Fundamental Theorem, the same calculation is short. If you can find an antiderivative, you can evaluate any definite integral by subtraction. The theorem replaces the limit of a sum with a single difference of two values.

The payoff: Instead of computing limits of sums, find an antiderivative and subtract.

Prerequisite Map

Quick Reference

Property	Value
Concept	Fundamental Theorem of Calculus
Course	MATH161 (Calculus I)
Difficulty	Intermediate
Time	~20 minutes

Key Concepts

The Evaluation Theorem

If $f$ is continuous on $[a, b]$ and $F$ is any antiderivative of $f$ (meaning $F' = f$), then:

$$\boxed{\int_a^b f(x)\, dx = F(b) - F(a)}$$

Notation

We use the following shorthand for “evaluate at the bounds”:

$$F(x)\Big\vert _a^b = F(b) - F(a)$$

Other common notations: $[F(x)]_a^b$ or $\left.F(x)\right]_a^b$

So the theorem can be written:

$$\int_a^b f(x)\, dx = F(x)\Big\vert _a^b$$

The Procedure

Find an antiderivative $F(x)$ of the integrand $f(x)$
Evaluate $F$ at the upper limit: $F(b)$
Evaluate $F$ at the lower limit: $F(a)$
Subtract: $F(b) - F(a)$

Why Any Antiderivative Works

You might wonder: if $F(x) = \frac{x^3}{3}$ is an antiderivative of $x^2$, could we also use $F(x) = \frac{x^3}{3} + 7$?

Yes. Here is why it does not matter:

$$\left(\frac{x^3}{3} + 7\right)\Bigg\vert _0^1 = \left(\frac{1}{3} + 7\right) - \left(0 + 7\right) = \frac{1}{3} + 7 - 7 = \frac{1}{3}$$

The constant cancels. So always use the simplest antiderivative (with $C = 0$).

Quick self-check. Evaluate $\displaystyle\int_1^2 3x^2\, dx$ using FTC Part 2. Pick an antiderivative, then subtract.

Show answer

An antiderivative of $3x^2$ is $F(x) = x^3$. Then $\int_1^2 3x^2\, dx = F(2) - F(1) = 8 - 1 = 7$. Undo the derivative to get $F$, then read off the change $F(b) - F(a)$.

Physical Interpretation

If $v(t)$ is velocity and $s(t)$ is position with $s'(t) = v(t)$, then:

$$\int_a^b v(t)\, dt = s(b) - s(a) = \text{displacement}$$

The total change in position equals the integral of velocity. This connects area under the velocity curve to net distance traveled.

When FTC2 Fails

FTC2 requires continuity on $[a, b]$. Watch out for:

Division by zero within the interval
Logarithms of negative numbers
Square roots of negative numbers

If the integrand has a discontinuity in $[a, b]$, FTC2 does not apply directly.

Practice Problems

Level 1 Basic Polynomial

Evaluate $\displaystyle\int_0^2 x^3\, dx$.

Thought Process

Find an antiderivative of $x^3$: Use the power rule backwards. Add 1 to the exponent and divide by the new exponent: $F(x) = \frac{x^4}{4}$
Evaluate at upper limit (2): $F(2) = \frac{2^4}{4} = \frac{16}{4} = 4$
Evaluate at lower limit (0): $F(0) = \frac{0^4}{4} = 0$
Subtract: $4 - 0 = 4$

Show Answer

An antiderivative of $x^3$ is $F(x) = \frac{x^4}{4}$.

$$\int_0^2 x^3\, dx = \frac{x^4}{4}\Bigg\vert _0^2 = \frac{2^4}{4} - \frac{0^4}{4} = \frac{16}{4} - 0 = 4$$

Level 2 Trigonometric Integral

Find the area under the curve $y = \cos x$ from $x = 0$ to $x = \frac{\pi}{2}$.

Thought Process

Area under a curve from $a$ to $b$ is $\int_a^b f(x)\, dx$ (when $f(x) \geq 0$).

The antiderivative of $\cos x$ is $\sin x$.
Evaluate: $\sin(\frac{\pi}{2}) - \sin(0) = 1 - 0 = 1$

This makes geometric sense: the area under one “bump” of the cosine curve is exactly 1 square unit.

Show Answer

$$A = \int_0^{\pi/2} \cos x\, dx = \sin x\Big\vert _0^{\pi/2}$$

$$= \sin\left(\frac{\pi}{2}\right) - \sin(0) = 1 - 0 = 1$$

The area is $1$ square unit.

Level 3 Polynomial with Negative Region

Evaluate $\displaystyle\int_{-1}^{2} (x^3 - x)\, dx$ and interpret the result as a difference of areas.

Thought Process

Find the antiderivative:
- Antiderivative of $x^3$ is $\frac{x^4}{4}$
- Antiderivative of $x$ is $\frac{x^2}{2}$
- So $F(x) = \frac{x^4}{4} - \frac{x^2}{2}$
Evaluate at limits and subtract.
For interpretation: The function $x^3 - x = x(x-1)(x+1)$ has roots at $x = -1, 0, 1$. It is negative on $(-1, 0)$, positive on $(0, 1)$, and positive on $(1, 2)$. The integral gives the net signed area.

Show Answer

Antiderivative: $F(x) = \frac{x^4}{4} - \frac{x^2}{2}$

$$\int_{-1}^{2} (x^3 - x)\, dx = \left(\frac{x^4}{4} - \frac{x^2}{2}\right)\Bigg\vert _{-1}^{2}$$

At $x = 2$: $\frac{16}{4} - \frac{4}{2} = 4 - 2 = 2$

At $x = -1$: $\frac{1}{4} - \frac{1}{2} = \frac{1}{4} - \frac{2}{4} = -\frac{1}{4}$

$$= 2 - \left(-\frac{1}{4}\right) = 2 + \frac{1}{4} = \frac{9}{4}$$

Interpretation: The curve is below the $x$-axis on $(-1, 0)$ and above on $(0, 2)$. The integral $\frac{9}{4}$ represents: (area above axis) $-$ (area below axis).

Level 4 Recognizing an Invalid Application

A student computes:

$$\int_{-1}^{1} \frac{1}{x^2}\, dx = \left[-\frac{1}{x}\right]_{-1}^{1} = -1 - (1) = -2$$

Explain why this answer is wrong and what the actual situation is.

Thought Process

First, notice the red flag: $\frac{1}{x^2}$ is always positive (when defined), so the integral of a positive function should be positive, not $-2$.

Check the integrand on $[-1, 1]$: At $x = 0$, we have $\frac{1}{0^2}$, which is undefined. The function has a discontinuity (actually, a vertical asymptote) at $x = 0$.

FTC2 requires continuity on the entire interval. Since $f(x) = \frac{1}{x^2}$ is not continuous on $[-1, 1]$, FTC2 does not apply.

This is an improper integral situation that requires different techniques.

Show Answer

The error: The function $f(x) = \frac{1}{x^2}$ is not continuous on $[-1, 1]$. It has a vertical asymptote at $x = 0$:

$$\lim_{x \to 0} \frac{1}{x^2} = +\infty$$

Why the answer is wrong: FTC Part 2 requires $f$ to be continuous on $[a, b]$. Since $\frac{1}{x^2}$ is undefined at $x = 0$, the theorem does not apply.

The clue something is wrong: The integrand $\frac{1}{x^2} > 0$ for all $x \neq 0$, so any legitimate integral should be positive, not $-2$.

The reality: This integral is improper and must be evaluated using limits (covered in later sections). In fact, this particular improper integral diverges (equals $+\infty$).

Level 5 Net Change Application

Water flows into a tank at a rate of $r(t) = 3t^2 - 12t + 9$ gallons per minute, where $t$ is measured in minutes since noon.

Find the net change in water volume from $t = 0$ to $t = 4$ minutes.
At what time(s) is water flowing out of the tank?
What is the maximum amount of water that leaves the tank during $[0, 4]$?

Thought Process

(a) Net change in volume = $\int_0^4 r(t)\, dt$. Find the antiderivative and apply FTC2.

(b) Water flows out when $r(t) < 0$. Factor $r(t) = 3t^2 - 12t + 9 = 3(t^2 - 4t + 3) = 3(t-1)(t-3)$. The parabola opens up, so $r(t) < 0$ between the roots: $1 < t < 3$.

(c) Water leaving = negative of the integral over the interval where $r(t) < 0$. That is $-\int_1^3 r(t)\, dt$ (the negative makes it positive, representing water lost).

Show Answer

(a) Net change in volume:

$$\int_0^4 (3t^2 - 12t + 9)\, dt = \left[t^3 - 6t^2 + 9t\right]_0^4$$

At $t = 4$: $64 - 96 + 36 = 4$

At $t = 0$: $0 - 0 + 0 = 0$

Net change = 4 gallons (4 gallons more at $t=4$ than at $t=0$)

(b) When water flows out:

Factor: $r(t) = 3(t-1)(t-3)$

This is negative when $(t-1)$ and $(t-3)$ have opposite signs, which occurs when $1 < t < 3$.

Water flows out during $(1, 3)$, from minute 1 to minute 3.

(c) Maximum water lost:

The amount that flows out is:

$$-\int_1^3 r(t)\, dt = -\left[t^3 - 6t^2 + 9t\right]_1^3$$

At $t = 3$: $27 - 54 + 27 = 0$

At $t = 1$: $1 - 6 + 9 = 4$

$$= -(0 - 4) = 4$$

Maximum water lost = 4 gallons (during minutes 1-3)

Note: The same 4 gallons that left during minutes 1-3 came back during minutes 3-4, resulting in a net gain of 4 gallons over the full interval.

Common Misconceptions

Common misconception

FTC Part 2 applies to any integrand as long as an antiderivative can be found.

This is the concept-image-conflicts-definition error. The theorem requires $f$ to be continuous on the entire closed interval $[a, b]$. If $f$ has a discontinuity inside $[a, b]$, the formula $F(b) - F(a)$ yields a wrong or meaningless answer. For example, applying the antiderivative $-1/x$ to $\int_{-1}^{1} x^{-2}\,dx$ gives $-2$, but the integrand $x^{-2}$ is always positive, so the integral cannot be negative. The discontinuity at $x = 0$ invalidates the theorem entirely.

Mastery Checklist

I can state FTC Part 2 precisely
I can use antiderivatives to evaluate definite integrals
I understand the notation $F(x)\Big\vert _a^b = F(b) - F(a)$
I know that any antiderivative works (the constant cancels)
I can recognize when FTC2 does not apply (discontinuity in $[a,b]$)
I can interpret definite integrals as net change

Mental Model

The Bank Account Analogy: Think of $f(x)$ as the rate you deposit (or withdraw) money. The antiderivative $F(x)$ is your balance. To find how much your balance changed from time $a$ to time $b$, you do not add up every tiny transaction. Just check your balance at both times and subtract:

$$\text{Balance change} = F(b) - F(a) = \int_a^b f(x)\, dx$$

FTC2 says: To find total change, just compare endpoints.

Common Mistakes

Mistake	Correction
Subtracting in wrong order	Always upper minus lower: $F(b) - F(a)$
Forgetting to evaluate at both limits	Must compute $F(b)$ AND $F(a)$
Applying FTC2 when $f$ is discontinuous	Check continuity on $[a, b]$ first
Adding $+C$ to the antiderivative	Do not include $+C$ for definite integrals

Connections

Looking back:

Antiderivatives provide the tool to apply FTC2
FTC1 provides the theoretical foundation

Looking ahead:

Substitution in definite integrals (change limits when substituting)
Area between curves uses FTC2 repeatedly
Net change theorem: total change = integral of rate of change

Differentiation and integration are inverse processes: Together, FTC1 and FTC2 say that differentiation and integration undo each other:

Differentiate an integral → get back the integrand
Integrate a derivative → get back the original function (up to constant)

Previous	Up	Next
FTC Part 1 + Chain Rule	Skills Index	Substitution (Definite Integrals)

Last updated: 2026-01-22