Chapter 14 — The Fundamental Theorem of Calculus

14.1 Two Problems That Turned Out to Be One

For thirteen chapters you have studied two problems that look nothing alike.

The tangent problem asks: given a curve, how steep is it at a point? Its answer is the derivative — a local object, built from the behavior of a function in an infinitesimally small neighborhood. You compute it by letting a difference quotient shrink to zero.

The area problem asks: given a curve, how much area lies beneath it? Its answer is the definite integral — a global object, built by chopping an interval into many pieces and summing their contributions. You compute it by letting a Riemann sum's mesh shrink to zero.

These problems were studied for two thousand years, and almost nobody suspected they were related. Archimedes computed areas; Fermat and Descartes computed tangents; the two communities barely spoke. Then, within a decade of each other, Isaac Newton and Gottfried Leibniz saw something astonishing: the two problems are inverses of one another. Finding areas and finding slopes are the same activity run in opposite directions.

That statement is the Fundamental Theorem of Calculus (FTC), and it is, without exaggeration, the most important theorem in mathematics. It is "fundamental" in three distinct senses at once, and this whole chapter is the story of why.

The Key Insight. Differentiation and integration are inverse operations. Differentiating an accumulation of area gives back the height of the curve; integrating a rate of change recovers the total change. FTC welds the two halves of calculus into a single structure — and hands you a computational shortcut that turns nearly every integral from an infinite limit into a one-line subtraction.

This chapter has two halves. FTC Part 1 explains why every continuous function has an antiderivative and what that antiderivative looks like. FTC Part 2 — the part you will use ten thousand times — explains how to evaluate a definite integral without ever computing a Riemann sum.

14.2 The Accumulation Function

Everything starts with one idea: freeze the lower limit of an integral and let the upper limit move.

Fix a continuous function $f$ and a starting point $a$. Define a new function $F$ by

$$F(x) = \int_a^x f(t)\,dt.$$

We rename the integration variable $t$ so it cannot be confused with the upper limit $x$; the value of the integral does not depend on what we call the dummy variable. We call $F$ the accumulation function of $f$ from $a$, because $F(x)$ is the running total of signed area swept out from $a$ up to $x$.

Think of $f$ as a rate and $F$ as the accumulated amount. If $f(t)$ is the rate at which water flows into a tank at time $t$, then $F(x)$ is the total volume of water that has flowed in by time $x$. If $f(t)$ is your velocity, $F(x)$ is your displacement. The accumulation function is the heart of FTC because it is both an integral (by definition) and, as we are about to discover, an antiderivative.

Geometric Intuition. Picture the graph of $f$. The accumulation function $F(x)$ is the shaded area between the curve and the horizontal axis, from $a$ on the left to a movable wall at $x$ on the right. As you slide the wall to the right, area pours in; $F$ grows. Where $f$ is positive, $F$ rises; where $f$ is negative, $F$ falls (you are subtracting signed area); where $f = 0$, $F$ momentarily levels off. Keep that picture — it is the entire proof of FTC Part 1 in visual form.

Check Your Understanding. Let $F(x) = \int_0^x f(t)\,dt$ where $f$ is positive on $(0,3)$ and negative on $(3,5)$. On which interval is $F$ increasing, and where does $F$ attain its maximum?

Answer

$F$ increases where the integrand is positive, namely on $(0,3)$, and decreases on $(3,5)$. So $F$ attains its maximum at $x = 3$ — exactly where $f$ changes from positive to negative. Notice this is the first-derivative test applied to $F$, with $F' = f$. That is FTC Part 1 arriving a section early.

Watching the accumulation grow. To make the accumulation function concrete, take $f(t) = t$ — a straight line through the origin — and accumulate from $a = 0$. Because the region under $f$ from $0$ to $x$ is a triangle of base $x$ and height $x$, geometry alone gives $F(x) = \tfrac12 x^2$. Tabulate it:

Now look at the right-hand column as a function of $x$: $0,\ 0.5,\ 2,\ 4.5,\ 8$. Differentiate $F(x) = \tfrac12 x^2$ and you get $F'(x) = x = f(x)$ — the integrand, exactly as FTC Part 1 will promise. The accumulation of a linear rate is a quadratic amount, and the derivative of that amount is the rate you started with. Every accumulation function works this way; this is the simplest case where you can check it by pure geometry.

14.2½ Signed Area: Why the Accumulation Can Go Down

The accumulation function tracks signed area: regions where $f$ lies below the axis count negatively. This is not a technicality — it is what makes FTC consistent with the Net Change Theorem you will meet in §14.7. If you drive forward then backward, the "area" under your velocity curve must be allowed to cancel, or the math would never report that you ended where you started.

Concretely, with $f(t) = \cos t$ accumulated from $0$:

$$F(x) = \int_0^x \cos t\,dt = \sin x.$$

As $x$ runs from $0$ to $\pi/2$, $\cos t > 0$ and $F = \sin x$ climbs from $0$ to $1$. As $x$ runs from $\pi/2$ to $\pi$, $\cos t < 0$ and $F$ descends from $1$ back toward $0$ — the negative area is being subtracted off. And indeed $F'(x) = \cos x = f(x)$ throughout: $F$ rises exactly where $f>0$ and falls exactly where $f<0$. The accumulation function is a perfect bookkeeper of signed area.

14.3 FTC Part 1: The Derivative of an Accumulation

Here is the first half of the theorem.

Fundamental Theorem of Calculus, Part 1. If $f$ is continuous on an interval containing $a$, and $F(x) = \int_a^x f(t)\,dt$, then $F$ is differentiable and $$F'(x) = f(x).$$

In words: the derivative of the accumulation function is the integrand. Differentiating "undoes" integrating. This single line guarantees that every continuous function has an antiderivative — namely its own accumulation function. That is a remarkable existence theorem hiding inside a computational statement.

The key idea before the details. Increase the upper limit by a tiny amount $h$. The extra area you pick up, $F(x+h) - F(x)$, is a thin sliver above the interval $[x, x+h]$. That sliver is almost a rectangle of width $h$ and height $f(x)$, so its area is approximately $f(x)\cdot h$. Divide by $h$ and the height $f(x)$ is what survives. Let $h \to 0$ and the approximation becomes exact.

Watch it happen symbolically:

$$F(x+h) - F(x) = \int_a^{x+h} f(t)\,dt - \int_a^{x} f(t)\,dt = \int_x^{x+h} f(t)\,dt.$$

The two accumulations share all the area up to $x$; their difference is precisely the sliver from $x$ to $x+h$. So

$$\frac{F(x+h) - F(x)}{h} = \frac{1}{h}\int_x^{x+h} f(t)\,dt = \text{(average value of } f \text{ on } [x, x+h]).$$

As $h \to 0$, that average value is taken over a vanishing interval clamped against the point $x$, and by continuity of $f$ it converges to $f(x)$. Therefore $F'(x) = f(x)$. The picture and the algebra say the same thing.

Math Major Sidebar — The rigorous proof via the Mean Value Theorem for Integrals. The phrase "average value converges to $f(x)$" can be made airtight. The Mean Value Theorem for Integrals (Section 13.11) says: if $f$ is continuous on $[x, x+h]$, there exists a point $c$ between $x$ and $x+h$ with $\int_x^{x+h} f(t)\,dt = f(c)\cdot h$. Substituting, $$\frac{F(x+h)-F(x)}{h} = \frac{f(c)\,h}{h} = f(c).$$ As $h \to 0$, the point $c$ — trapped between $x$ and $x+h$ — is squeezed to $x$, and continuity gives $f(c) \to f(x)$. Hence $F'(x) = \lim_{h\to 0} f(c) = f(x)$. The proof handles $h < 0$ identically, with $c$ between $x+h$ and $x$. Continuity is doing all the work: drop it, and $f$ could jump inside every sliver and the limit could fail.

Warning. FTC Part 1 requires continuity of $f$. If $f$ has a jump discontinuity at $x$, the accumulation function $F$ still exists and is even continuous, but it has a corner at $x$ — its left and right derivatives are the two one-sided limits of $f$, which disagree. So $F'(x)$ fails to exist exactly where $f$ jumps. The theorem is sharp: continuity in, differentiability out.

14.4 FTC Part 2: The Evaluation Theorem

Part 1 tells you that antiderivatives exist. Part 2 tells you how to use any antiderivative to evaluate an integral — and this is the workhorse of all of applied calculus.

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

The infinite limit of Riemann sums collapses into a single subtraction. You no longer chop $[a,b]$ into a million rectangles; you find one antiderivative and evaluate it at the two endpoints.

Why it follows from Part 1. Let $G(x) = \int_a^x f(t)\,dt$ be the accumulation function. By Part 1, $G$ is an antiderivative of $f$. Now suppose $F$ is any other antiderivative. Two antiderivatives of the same function differ by a constant (a consequence of the Mean Value Theorem, Chapter 9), so $F(x) = G(x) + C$ for some constant $C$. Then

$$F(b) - F(a) = \big(G(b)+C\big) - \big(G(a)+C\big) = G(b) - G(a) = \int_a^b f\,dt - \int_a^a f\,dt = \int_a^b f\,dt - 0.$$

The constant $C$ cancels — which is why it does not matter which antiderivative you pick. Every "$+C$" you ever wrote in an indefinite integral evaporates the moment you subtract endpoint values.

Common Pitfall. Students often forget that FTC Part 2 demands continuity on the whole interval $[a,b]$. Writing $\int_{-1}^{1} \frac{1}{x^2}\,dx = \big[-\tfrac{1}{x}\big]_{-1}^{1} = -1 - 1 = -2$ is a disaster: the answer is negative, yet $1/x^2 > 0$ everywhere. The integrand blows up at $x = 0$ inside the interval, so it is not continuous on $[-1,1]$ and FTC simply does not apply. (This integral is in fact divergent — a case for improper integrals in Chapter 17.) Always check the integrand is continuous across the entire interval before applying the evaluation bar.

14.5 Notation: The Evaluation Bar

The standard shorthand for "evaluate the antiderivative at the endpoints and subtract" is the vertical bar:

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

Read $F(x)\big|_a^b$ as "$F$ evaluated from $a$ to $b$": plug in the top, plug in the bottom, subtract bottom from top. The bracket form $[\,\cdot\,]_a^b$ is convenient when $F$ is a long expression.

14.6 Worked Examples, Graduated

FTC Part 2 reduces every definite integral to two skills: finding an antiderivative, and arithmetic. Here is a ladder from routine to subtle.

Example 1 (power rule). $\displaystyle\int_0^1 x^2\,dx$. An antiderivative of $x^2$ is $x^3/3$. So $$\int_0^1 x^2\,dx = \frac{x^3}{3}\bigg|_0^1 = \frac{1}{3} - 0 = \frac{1}{3}.$$ Compare the labor: in Chapter 13 we got $1/3$ by taking the limit of $\sum (i/n)^2 \cdot (1/n)$. FTC gets it in one line.

Example 2 (trigonometric). $\displaystyle\int_0^\pi \sin x\,dx$. An antiderivative of $\sin x$ is $-\cos x$. So $$\int_0^\pi \sin x\,dx = -\cos x\Big|_0^\pi = -\cos\pi - (-\cos 0) = -(-1) - (-1) = 2.$$ The area under one hump of the sine curve is exactly $2$ — a clean number that no amount of staring at the graph would reveal.

Example 3 (logarithmic antiderivative). $\displaystyle\int_1^e \frac{1}{x}\,dx = \ln x\Big|_1^e = \ln e - \ln 1 = 1 - 0 = 1.$ This is the defining property of $e$: the number whose natural logarithm is exactly $1$ is the upper limit that makes the area under $1/x$ equal to $1$.

Example 4 (signed area / symmetry). $\displaystyle\int_{-1}^{1} x^3\,dx = \frac{x^4}{4}\bigg|_{-1}^{1} = \frac14 - \frac14 = 0.$ The negative area on $[-1,0]$ exactly cancels the positive area on $[0,1]$ because $x^3$ is odd. FTC and symmetry agree.

Example 5 (rational, leading to an arctangent). $\displaystyle\int_0^1 \frac{1}{1+x^2}\,dx = \arctan x\Big|_0^1 = \frac{\pi}{4} - 0 = \frac{\pi}{4}.$ A purely algebraic integrand produces $\pi$. The appearance of $\pi$ from a rational function is a first hint of the deep links between integration and geometry.

Check Your Understanding. Evaluate $\displaystyle\int_0^4 \sqrt{x}\,dx$.

Answer

Write $\sqrt{x} = x^{1/2}$; an antiderivative is $\tfrac{2}{3}x^{3/2}$. Then $\tfrac{2}{3}x^{3/2}\big|_0^4 = \tfrac{2}{3}\cdot 4^{3/2} = \tfrac{2}{3}\cdot 8 = \tfrac{16}{3}$.

Example 6 (antiderivative not yet in our toolkit). $\displaystyle\int_0^2 2x\,e^{x^2}\,dx$. The integrand is a chain-rule derivative in disguise: $\frac{d}{dx}e^{x^2} = 2x e^{x^2}$. So an antiderivative is $e^{x^2}$, and $$\int_0^2 2x\,e^{x^2}\,dx = e^{x^2}\Big|_0^2 = e^4 - 1.$$ Recognizing such hidden derivatives is the skill of $u$-substitution, which we systematize in Chapter 15.

14.7 The Net Change Theorem

Rewrite FTC Part 2 with $f$ renamed as a derivative. If $F' = f$, then $f = F'$ and the theorem reads

$$\int_a^b F'(x)\,dx = F(b) - F(a).$$

This is the Net Change Theorem, and it is FTC told from the other direction: the integral of a rate of change is the net change. Whenever you know how fast something is changing and you want to know how much it changed in total, you integrate the rate.

This interpretation is why FTC is the engine of every quantitative science:

• Integrate velocity $v(t) = s'(t)$ over $[t_1, t_2]$ → net displacement $s(t_2) - s(t_1)$.
• Integrate marginal cost $C'(q)$ over $[q_1, q_2]$ → total added cost of producing those units.
• Integrate a population growth rate $P'(t)$ → net change in population.
• Integrate a probability density $p(x) = F'(x)$ over $[a,b]$ → the probability $F(b) - F(a)$ that the value lands in $[a,b]$.

Worked example. A particle moves with velocity $v(t) = 3t^2 - 4$ m/s on $[0,2]$. Net displacement is $$\int_0^2 (3t^2 - 4)\,dt = \big[t^3 - 4t\big]_0^2 = (8 - 8) - 0 = 0\ \text{m}.$$ The particle returns to its starting position. (Note: net displacement is zero, but total distance traveled is $\int_0^2 |v(t)|\,dt$, which is positive — the particle went backward then forward. Distinguishing the two is a classic exam trap.)

14.8 Differentiating Integrals with Variable Limits

FTC Part 1 says $\frac{d}{dx}\int_a^x f(t)\,dt = f(x)$. What if the upper limit is not $x$ but a function of $x$? Then FTC Part 1 combines with the chain rule.

If $G(x) = \displaystyle\int_a^{u(x)} f(t)\,dt$, set the inner accumulation $\Phi(u) = \int_a^u f(t)\,dt$ so that $G(x) = \Phi(u(x))$. By FTC Part 1, $\Phi'(u) = f(u)$, and the chain rule gives

$$G'(x) = \Phi'(u(x))\cdot u'(x) = f(u(x))\,u'(x).$$

Example. $\dfrac{d}{dx}\displaystyle\int_0^{x^2}\sin t\,dt = \sin(x^2)\cdot 2x.$ You never compute the integral — you just evaluate the integrand at the top limit and multiply by the derivative of that limit.

When both limits vary, split at any convenient constant and apply the rule to each piece. The general form (the Leibniz integral rule in its simplest case) is

$$\frac{d}{dx}\int_{u(x)}^{v(x)} f(t)\,dt = f(v(x))\,v'(x) - f(u(x))\,u'(x).$$

Common Pitfall. Forgetting the chain-rule factor is the single most common error here. $\frac{d}{dx}\int_1^{x^3} e^{-t^2}\,dt$ is not $e^{-x^2}$ and not $e^{-(x^3)^2}$ alone — it is $e^{-(x^3)^2}\cdot 3x^2 = 3x^2 e^{-x^6}$. The derivative of the upper limit, $3x^2$, must come along.

14.9 Average Value and the Mean Value Theorem for Integrals

FTC also pins down what the average value of a continuous function means. The average of finitely many numbers is their sum divided by the count; the continuous analog replaces the sum by an integral:

$$\bar f = \frac{1}{b-a}\int_a^b f(x)\,dx.$$

Geometrically, $\bar f$ is the height of the rectangle on base $[a,b]$ whose area equals the area under $f$. The Mean Value Theorem for Integrals then guarantees this height is actually achieved: there is a point $c \in [a,b]$ with $f(c) = \bar f$. Using FTC, with $F' = f$,

$$\bar f = \frac{F(b) - F(a)}{b - a},$$

which is exactly the average slope of $F$ across $[a,b]$ — and the Mean Value Theorem for derivatives says some interior slope $F'(c) = f(c)$ equals it. The two mean value theorems are the same statement viewed through FTC.

Example. The average value of $f(x) = x^2$ on $[0,3]$ is $\frac{1}{3}\int_0^3 x^2\,dx = \frac{1}{3}\cdot\frac{27}{3} = 3$, attained at $c = \sqrt 3 \in [0,3]$.

14.10 Applications Across Every Field

FTC is not a physics theorem with applications elsewhere; it is field-agnostic. Here are four readings of the same mathematics.

Real-World Application — Pharmacokinetics (biology/medicine). If a drug is eliminated from the bloodstream at rate $r(t)$ (mg/hr), then the total amount cleared between hours $a$ and $b$ is $\int_a^b r(t)\,dt$. The "area under the concentration curve" (AUC) — literally a definite integral evaluated by FTC — is the single most important quantity in dosing: it determines total drug exposure and is what regulators require for proving two drug formulations are bioequivalent.

Economics — consumer cost from marginal cost. A firm's marginal cost $C'(q)$ describes the cost of the next unit. The added cost of scaling production from $q_1$ to $q_2$ units is the net change $\int_{q_1}^{q_2} C'(q)\,dq = C(q_2) - C(q_1)$ — no need to know fixed costs, which cancel in the subtraction.

Physics — work from a variable force. Stretching a spring with force $F(x) = kx$ from rest to extension $a$ requires work $W = \int_0^a kx\,dx = \tfrac{1}{2}ka^2$. We guessed this in Chapter 12 from energy arguments; FTC derives it.

Data science / statistics — the area under the normal curve. Recall the anchor example introduced in Chapter 13: the bell curve $\phi(x) = \frac{1}{\sqrt{2\pi}}e^{-x^2/2}$. A probability is an area: $P(a \le X \le b) = \int_a^b \phi(x)\,dx = \Phi(b) - \Phi(a)$, where $\Phi$ is the cumulative distribution function — the accumulation function of $\phi$. FTC tells us $\Phi' = \phi$ and that probabilities are net changes in $\Phi$. The catch, which we return to in §14.12 and resolve with Taylor series in Chapter 23, is that $\Phi$ has no elementary formula. FTC still applies perfectly; we just cannot write the antiderivative with the functions we know.

14.11 Computation: Verifying and Visualizing FTC

Python lets you see the accumulation function being born and confirm your hand answers. The three-tier pattern — analytic, by hand, then machine check — is our standard.

import numpy as np
from scipy.integrate import quad

# Hand result (Example 1): ∫₀¹ x² dx = 1/3 via FTC, antiderivative x³/3
numerical, _ = quad(lambda x: x**2, 0, 1)   # numerical integral, no antiderivative used
print(f"FTC by hand : {1/3:.10f}")
print(f"scipy quad  : {numerical:.10f}")     # both print 0.3333333333

The point of the check is conceptual: quad approximates the integral numerically (a sophisticated Riemann-type sum), while FTC gives it exactly. Agreement to ten digits is FTC and numerical analysis shaking hands.

We can also watch FTC Part 1 directly — the accumulation function $F(x)=\int_0^x f$ should have derivative equal to $f$:

import numpy as np

f = np.sin                       # integrand
x = np.linspace(0, 2*np.pi, 1000)
dx = x[1] - x[0]

# Build the accumulation function F(x) = ∫₀ˣ sin t dt by cumulative summation
F = np.concatenate([[0], np.cumsum(f(x[:-1]) * dx)])   # left Riemann accumulation

# FTC Part 1 prediction: dF/dx should equal f(x) = sin(x)
dF = np.gradient(F, dx)
print("max |dF/dx - sin x| =", np.max(np.abs(dF - f(x))))   # ~1e-3, shrinks with finer mesh

Computational Note. np.cumsum is the discrete accumulation function: each entry is the running total of everything before it — exactly $\int_a^x$ discretized. Differentiating that running total with np.gradient recovers the original integrand to within the discretization error. You have just demonstrated FTC Part 1 numerically: the derivative of the accumulation is the integrand. Refine linspace and the error falls toward zero.

14.12 What FTC Does Not Promise

FTC Part 2 says: if you can find an antiderivative, the integral is a subtraction. It does not promise that an antiderivative can be written using elementary functions (polynomials, roots, exponentials, logarithms, trig, and their finite combinations). Astonishingly, most integrands have no elementary antiderivative:

$$\int e^{-x^2}\,dx \ \ (\text{the error function, } \mathrm{erf}), \qquad \int \sin(x^2)\,dx\ \ (\text{Fresnel}), \qquad \int \frac{1}{\ln x}\,dx\ \ (\text{logarithmic integral}).$$

This is not a failure of cleverness — it is a theorem (Liouville, 1830s) that these antiderivatives provably escape the elementary functions. FTC still holds: each integral equals "antiderivative at $b$ minus at $a$." We simply evaluate that antiderivative numerically or through a series instead of a formula. The normal-curve integral $\int \phi$ from §14.10 is the most important member of this club, and Chapter 23 shows how Taylor series compute it to any precision.

Historical Note. Newton (working ~1665–66) and Leibniz (~1675–84) discovered FTC independently, igniting one of the bitterest priority disputes in the history of science. Newton's notation — the fluxion dot $\dot{y}$ — served physics; Leibniz's notation — the ratio $dy/dx$ and the elongated-S integral sign $\int$ (an "S" for summa, sum) — was so much better for calculation that continental Europe outpaced England for a century. We use Leibniz's symbols to this day. Both men deserve full credit; the mathematics was bigger than either.

14.13 Why FTC Is Called "Fundamental"

Three independent claims justify the name, and FTC satisfies all three:

1. It is foundational. It rests directly on the definitions of derivative (Chapter 6) and definite integral (Chapter 13) and binds them together.
2. It is unifying. The tangent problem and the area problem — separate for two millennia — are revealed as one problem with two directions. FTC Part 1 reads "differentiation undoes integration"; FTC Part 2 reads "integration undoes differentiation."
3. It is computational. It converts the intractable (limits of Riemann sums) into the routine (find an antiderivative, subtract). Essentially all of applied integral calculus runs on this conversion.

And it is the seed of everything that follows. Every major theorem of vector calculus is a higher-dimensional FTC, all sharing one slogan — the integral of a derivative over a region equals the values on the boundary:

• the Fundamental Theorem for Line Integrals, $\int_C \nabla f\cdot d\mathbf{r} = f(B) - f(A)$ (Chapter 35);
• Green's Theorem relating a boundary line integral to a region double integral (Chapter 35);
• Stokes' Theorem and the Divergence Theorem (Chapter 37);
• and their grand unification $\int_{\partial M}\omega = \int_M d\omega$ (Chapter 38).

Hold onto the slogan; you will meet it again and again, each time in one more dimension.

Add to Your Modeling Portfolio. Add a "net change" computation to your model, using FTC to recover a total from a rate. Biology: integrate a population growth rate $P'(t)$ to get net population change over a season. Economics: integrate marginal revenue $R'(q)$ to recover total revenue from selling $q_1$ to $q_2$ units. Physics: integrate force $F(x)$ along a displacement to compute work done. Data Science: integrate a probability density $p(x)$ to get the cumulative probability $\int_a^b p$, the foundation of every CDF you will build.

14.14 A Common-Error Gallery

FTC is simple to state and easy to misapply. Here are the five mistakes that cost the most points, each with the correction.

Error 1 — Ignoring a discontinuity inside the interval. Applying the evaluation bar across a vertical asymptote (the §14.4 pitfall). Correction: confirm $f$ is continuous on the entire closed interval first; if not, it is an improper integral (Chapter 17), not a routine FTC computation.

Error 2 — Dropping the chain-rule factor when limits are functions. Writing $\frac{d}{dx}\int_0^{x^2}\sin t\,dt = \sin(x^2)$ instead of $\sin(x^2)\cdot 2x$. Correction: the derivative of the upper limit always multiplies in (§14.8).

Error 3 — Confusing net displacement with total distance. Reporting $\int_a^b v\,dt$ as "distance traveled." Correction: that integral is net change; total distance is $\int_a^b |v|\,dt$, which requires splitting at the sign changes of $v$.

Error 4 — Reversing the subtraction. Computing $F(a) - F(b)$ instead of $F(b) - F(a)$, flipping the sign of every answer. Correction: top limit first, always: $F(\text{top}) - F(\text{bottom})$.

Error 5 — Treating "$+C$" as if it mattered in a definite integral. Carrying the constant of integration through to a numerical answer. Correction: in $\int_a^b$, the constant cancels (§14.4) — never write it.

Check Your Understanding. A particle has velocity $v(t) = t - 2$ (m/s) on $[0,3]$. Find (a) its net displacement and (b) the total distance it travels.

Answer

(a) Net displacement $= \int_0^3 (t-2)\,dt = \big[\tfrac{t^2}{2} - 2t\big]_0^3 = (\tfrac92 - 6) - 0 = -\tfrac32$ m. (b) $v < 0$ on $[0,2)$ and $v>0$ on $(2,3]$, so total distance $= \int_0^2 (2-t)\,dt + \int_2^3 (t-2)\,dt = \big[2t-\tfrac{t^2}{2}\big]_0^2 + \big[\tfrac{t^2}{2}-2t\big]_2^3 = 2 + \tfrac12 = \tfrac52$ m. The two answers differ precisely because of the backward leg — Error 3 in action.

Real-World Application — Cardiac output (medicine). Clinicians measure how much blood the heart pumps per minute by injecting a known mass of dye into the bloodstream and recording its downstream concentration $c(t)$. The cardiac output is $\frac{\text{dye mass}}{\int_0^T c(t)\,dt}$ — a definite integral of the concentration curve, evaluated by FTC (or numerically when no formula fits the data). The "area under the curve" is, once again, the quantity that carries the meaning.

Looking Ahead

You can now evaluate any definite integral whose antiderivative you can find. The obvious next question is: how do you find antiderivatives that are not obvious? Chapter 15 develops the two great techniques — $u$-substitution (the reverse chain rule) and integration by parts (the reverse product rule). Chapter 16 adds trigonometric integrals, trigonometric substitution, partial fractions, and numerical integration for when no formula exists. Chapter 17 handles improper integrals — the kind we mishandled in the §14.4 pitfall. Chapter 18 unleashes FTC on areas, volumes, arc length, and work, and Chapter 19 turns it loose on differential equations. From here on, FTC is simply assumed; it is the floor you stand on.

Reflection

For two thousand years, areas and tangents were different subjects studied by different people. Newton and Leibniz saw — almost simultaneously, and to their lasting mutual fury — that they are inverse operations. That single realization reorganized mathematics: suddenly every area problem became an antiderivative problem, every rate became a total, and the language of change became computable. From Newtonian mechanics to Maxwell's equations to the training of neural networks, the Fundamental Theorem of Calculus is quietly doing the arithmetic. You have just reached the center of the subject. Take a moment to appreciate it — then turn the page and learn to integrate anything.