FTC Part 2

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The Fundamental Theorem of Calculus, Part 2

Textbook Reference

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Key idea

Part 2 of the Fundamental Theorem of Calculus turns the hardest computation in integral calculus into subtraction. It says that to evaluate a definite integral, you find an antiderivative and subtract its values at the two endpoints. No rectangles, no limits of sums.

Recall the work it replaces. To find the area under $y = x^2$ from $0$ to $1$ as a limit of Riemann sums took a careful calculation with the sum-of-squares formula. Part 2 gives the same answer, $\frac{1}{3}$, in a single line: an antiderivative of $x^2$ is $\frac{x^3}{3}$, so the integral is $\frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3}$.

The statement is \[ \int_a^b f(x)\,dx = F(b) - F(a), \] where $F$ is any antiderivative of $f$, meaning any function with $F' = f$. It is genuinely surprising. The left side was defined by a complicated process involving every value of $f$ across the whole interval, yet the right side needs only two numbers, $F$ at the endpoints. The reason it works is Part 1: the “area so far” function is itself an antiderivative of $f$, and any two antiderivatives differ by a constant, so subtracting endpoint values of any antiderivative recovers the accumulated area. Part 1 builds the antiderivative; Part 2 spends it.

Prerequisite Check

Before this lesson, make sure you can do all of the following:

• State what an antiderivative is: a function $F$ with $F' = f$ (review Antiderivative Concept)
• Find a basic antiderivative, for instance an antiderivative of $x^2$ is $\frac{x^3}{3}$
• Understand that the accumulation function is an antiderivative (review FTC Part 1)
• Interpret $\int_a^b f(x)\,dx$ as a signed area
• Evaluate a function at two endpoints and subtract

If finding an antiderivative is shaky, review the antiderivative rules first. Part 2 is only as easy as your ability to reverse a derivative.

Quick Reference

The Fundamental Theorem of Calculus, Part 2 (the Evaluation Theorem). If $f$ is continuous on $[a, b]$, then \[ \int_a^b f(x)\,dx = F(b) - F(a), \] where $F$ is any antiderivative of $f$ (any function with $F' = f$).

Notation. Write $F(x)\big|_a^b = F(b) - F(a)$, so \[ \int_a^b f(x)\,dx = F(x)\Big|_a^b. \]

The procedure.

1. Find any antiderivative $F$ of the integrand $f$.
2. Evaluate $F$ at the upper limit $b$ and the lower limit $a$.
3. Subtract: $F(b) - F(a)$.

Any antiderivative works. The constant $C$ cancels in the subtraction, so use the simplest antiderivative.

Hypothesis. $f$ must be continuous on $[a, b]$. If $f$ has a discontinuity inside the interval, the theorem does not apply.

Key Concepts

1. The Evaluation Theorem

The central statement converts a definite integral into a subtraction of antiderivative values.

The Fundamental Theorem of Calculus, Part 2. If $f$ is continuous on $[a, b]$ and $F$ is any antiderivative of $f$, then \[ \int_a^b f(x)\,dx = F(b) - F(a). \] (Plain gloss: the total signed area equals the change in any antiderivative from $a$ to $b$.)

Why it holds: Part 1 says the accumulation function $g(x) = \int_a^x f(t)\,dt$ is an antiderivative of $f$. Any other antiderivative $F$ differs from $g$ by a constant, $F(x) = g(x) + C$. Since $g(a) = 0$, \[ F(b) - F(a) = [g(b) + C] - [g(a) + C] = g(b) - g(a) = g(b) = \int_a^b f(x)\,dx. \] The constant cancels, which is exactly why any antiderivative gives the right answer.

2. The First Evaluation, and Why Any Antiderivative Works

The simplest use is a polynomial integrand.

Example 1. Evaluate $\displaystyle\int_{-2}^{1} x^3\,dx$. (This is Stewart 4.3, Example 5.)

Goal. Find an antiderivative of $x^3$, then subtract endpoint values.

The function $x^3$ is continuous, and an antiderivative is $F(x) = \frac{x^4}{4}$. By Part 2, \[ \int_{-2}^{1} x^3\,dx = F(1) - F(-2) = \frac{1^4}{4} - \frac{(-2)^4}{4} = \frac{1}{4} - \frac{16}{4} = -\frac{15}{4}. \]

Boxed answer: $-\dfrac{15}{4}$.

Recap. The result is negative because more signed area lies below the axis (over $[-2, 0]$) than above it (over $[0, 1]$). You could use $\frac{x^4}{4} + 7$ or $\frac{x^4}{4} + C$ instead; the added constant cancels in the subtraction, so the simplest antiderivative is always the right choice.

3. Areas in One Line: The Contrast With Riemann Sums

The theorem’s power is clearest when you compare it with computing the same area as a limit of sums.

Example 2. Find the area under $y = x^2$ from $0$ to $1$. (This is Stewart 4.3, Example 6.)

Goal. Antiderivative, then subtract.

An antiderivative of $x^2$ is $F(x) = \frac{x^3}{3}$, so \[ A = \int_0^1 x^2\,dx = \frac{x^3}{3}\bigg|_0^1 = \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3}. \]

Boxed answer: the area is $\dfrac{1}{3}$.

Recap. This is the same $\frac{1}{3}$ that took a full limit-of-Riemann-sums computation (with the formula $\sum i^2 = \frac{n(n+1)(2n+1)}{6}$) to obtain directly. Part 2 reduces that entire derivation to one subtraction. That contrast is why the theorem is called fundamental: it makes area problems routine.

4. Trigonometric and Algebraic Integrands

The procedure is identical for any continuous integrand whose antiderivative you can write.

Example 3. Find the area under $y = \cos x$ from $x = 0$ to $x = b$, where $0 \leq b \leq \frac{\pi}{2}$. (This is Stewart 4.3, Example 7.)

Goal. An antiderivative of $\cos x$ is $\sin x$; evaluate and subtract.

\[ A = \int_0^b \cos x\,dx = \sin x\Big|_0^b = \sin b - \sin 0 = \sin b. \] In particular, taking $b = \frac{\pi}{2}$ gives area $\sin\frac{\pi}{2} = 1$.

Boxed answer: the area is $\sin b$; for $b = \frac{\pi}{2}$ it equals $1$.

Recap. Recognizing the antiderivative ($\sin x$ for $\cos x$) is the whole task. Building a table of antiderivatives is what makes Part 2 fast, just as a table of derivatives makes differentiation fast.

5. The Continuity Hypothesis Is Not Optional

Part 2 requires the integrand to be continuous on the entire interval. Skipping that check produces nonsense.

Example 4. What is wrong with the following calculation? (This is Stewart 4.3, Example 8.) \[ \int_{-1}^{3}\frac{1}{x^2}\,dx = -\frac{1}{x}\bigg|_{-1}^{3} = -\frac{1}{3} - 1 = -\frac{4}{3}. \]

Goal. Spot why the answer cannot be right, then identify the broken hypothesis.

The integrand $\frac{1}{x^2}$ is never negative, so any genuine area is nonnegative. A negative answer is impossible. The error is that $\frac{1}{x^2}$ has an infinite discontinuity at $x = 0$, which lies inside $[-1, 3]$. Part 2 requires continuity on the whole closed interval, so it does not apply here. In fact this integral does not exist as an ordinary integral.

Boxed answer: the calculation is invalid because $\frac{1}{x^2}$ is discontinuous at $x = 0 \in [-1, 3]$; the theorem cannot be used.

Important: check continuity across the whole interval before applying Part 2. Blindly evaluating an antiderivative at the endpoints when the integrand has a discontinuity inside the interval gives a wrong answer, sometimes one with the wrong sign. Always confirm $f$ is continuous on all of $[a, b]$ first.

6. Differentiation and Integration Are Inverse Processes

Putting the two parts together states the deep symmetry of calculus.

• Part 1: $\dfrac{d}{dx}\int_a^x f(t)\,dt = f(x)$. Integrate, then differentiate, and you return to $f$.
• Part 2: $\displaystyle\int_a^b F'(x)\,dx = F(b) - F(a)$. Differentiate $F$ to get $F'$, then integrate, and you return to $F$ (up to the constant $F(a)$).

Read together, the two parts say that integration and differentiation undo each other. This also gives Part 2 its practical reading as the net change theorem: the integral of a rate of change $F'$ over $[a, b]$ is the total change $F(b) - F(a)$. Integrating a velocity gives displacement; integrating a marginal cost gives total added cost. The theorem links the local (the rate, the derivative) to the global (the accumulated change, the integral).

Common Errors Summary

Leveled Practice

Level 1 -- Direct Application

Problem 1. Evaluate $\displaystyle\int_1^2 x^2\,dx$.

Problem 2. Evaluate $\displaystyle\int_0^{\pi}\sin x\,dx$.

Problem 3. Evaluate $\displaystyle\int_5^6 (x^2 + 2x - 4)\,dx$.

Level 2 -- Multiple Steps

Problem 4. Evaluate $\displaystyle\int_0^4 (\sqrt{x} - 1)\,dx$ and interpret it as a difference of areas.

Problem 5. Find the area of the region enclosed by $y = 4 - x^2$ and the $x$-axis.

Problem 6. Evaluate $\displaystyle\int_1^2 \frac{s^4 + 1}{s^2}\,ds$.

Level 3 -- Deeper Problems

Problem 7. If $f(1) = 12$, $f'$ is continuous, and $\displaystyle\int_1^4 f'(x)\,dx = 17$, find $f(4)$.

Problem 8. A particle has velocity $v(t) = t^2 - 4$ meters per second on $[0, 3]$. Find the displacement (net change in position) over $[0, 3]$.

Problem 9. Show that the area under $y = x^2$ on $[0, 1]$ found here, $\frac{1}{3}$, agrees with the Riemann-sum result, and explain in one sentence why Part 2 is preferred.

Common Misconceptions

Mastery Checklist

You have mastered this skill when you can do all of the following without referring to notes:

• State Part 2: for continuous $f$, $\int_a^b f(x)\,dx = F(b) - F(a)$ with $F' = f$
• Evaluate a definite integral by finding an antiderivative and subtracting endpoint values
• Explain why any antiderivative works (the constant cancels)
• Use the notation $F(x)\big|_a^b$ correctly, upper limit minus lower limit
• Rewrite an integrand into power-rule pieces before antidifferentiating
• Find the limits of integration from where a curve crosses the axis, for an area problem
• Check the continuity hypothesis and recognize when Part 2 cannot be used
• Read Part 2 as the net change theorem: the integral of a rate is the total change

Mental Model

Think of Part 2 as reading total change off the endpoints.

If $F$ records a quantity and $f = F'$ records its rate of change, then adding up all the little changes across $[a, b]$, the integral $\int_a^b f$, must equal the overall change $F(b) - F(a)$. You do not need to track the quantity continuously; you only need its value at the start and the end. That is the whole theorem.

Three things follow from the picture:

• Integration becomes subtraction. The definite integral, defined as a limit of sums, collapses to two evaluations of an antiderivative. The hard part is finding the antiderivative; once you have it, the integral is one subtraction.
• The constant does not matter. Any antiderivative differs from another by a constant, and that constant cancels when you subtract endpoint values. So you always pick the simplest antiderivative.
• It is the inverse of Part 1. Part 1 integrates to build an antiderivative; Part 2 differentiates-then-integrates back to recover total change. Together they say differentiation and integration undo each other, which is why net change (displacement from velocity, total cost from marginal cost) is computed by integration.

Connections

Within Integrals (Chapter 4)

• FTC Part 1 and the antiderivative concept: Part 1 guarantees an antiderivative exists (the accumulation function), which is exactly what Part 2 needs. Part 2 is only useful once you can find antiderivatives, so the antiderivative rules feed directly into it.
• Riemann sums and the definite integral: Part 2 evaluates the very integrals that the definite integral defines as limits of Riemann sums, replacing the hard limit with a subtraction. The $\frac{1}{3}$ area appears both ways.
• Evaluating definite integrals and the net change theorem: Part 2 is how you evaluate definite integrals, and read in reverse it is the net change theorem, $\int_a^b F'\,dx = F(b) - F(a)$.

Toward later calculus (MATH161 and beyond)

• Integration techniques: Because Part 2 reduces integration to finding antiderivatives, the rest of integral calculus, substitution, integration by parts, partial fractions, is the project of finding antiderivatives for harder integrands so that Part 2 can finish the job.
• Applications of integration: Area between curves, volumes, arc length, work, and average value all reduce to a definite integral evaluated by Part 2. The displacement-from-velocity reading (Problem 8) is the first of many net-change applications.
• Improper integrals: Example 4 shows Part 2 failing across a discontinuity. Integrals over intervals containing a discontinuity, or over infinite intervals, are handled separately as improper integrals in Calculus 2, defined through limits precisely because Part 2 does not apply directly.

Audience Notes

For students who find math intimidating: The recipe is short. Find a function whose derivative is the integrand (the antiderivative), plug in the top number, plug in the bottom number, and subtract. The only skill you need to build is reversing derivatives, which is just the derivative rules run backward. Always do top minus bottom, and always check the function has no break inside the interval.

For career-focused students: The net change reading is the everyday use: total displacement is the integral of velocity, total energy is the integral of power, accumulated rainfall is the integral of rainfall rate. Part 2 says you can get the total from the antiderivative’s endpoint values, which is how integrated sensor data and simulations report cumulative quantities.

For gifted and curious students: Compare evaluating $\int_0^1 x^2\,dx$ by Part 2 (one line) with the Riemann-sum derivation (a full limit calculation). Then appreciate that some continuous functions, like $e^{-x^2}$, have no elementary antiderivative, so Part 2 cannot evaluate $\int_0^1 e^{-x^2}\,dx$ in closed form even though the integral exists. That gap is what motivates numerical integration and special functions.

For PhD-track students: The clean form $\int_a^b f = F(b) - F(a)$ requires $f$ continuous, but the sharp statement is that it holds whenever $F$ is differentiable with $F' = f$ and $f$ is Riemann integrable. Extending it to absolutely continuous $F$ with $f$ only Lebesgue integrable is the Lebesgue differentiation and fundamental-theorem program, and the exact hypotheses (why absolute continuity, not mere continuity, is the right condition) are a central topic of real analysis.

Back to Integrals | Related: FTC Part 1 | Next: Definite Integral Evaluation