The Fundamental Theorem of Calculus

The definite integral was built as a limit of Riemann sums, and evaluating even ${\int }_0^1{x}^2\mathbf{d}x$ that way took a summation identity and a page of algebra. The derivative came from a seemingly unrelated question — the slope of a tangent. The Fundamental Theorem of Calculus states that these two constructions are inverse to each other, and that single fact converts most integral evaluations into antiderivative lookups. Isaac Barrow saw the inverse relationship; Newton and Leibniz turned it into a method.

The theorem has two parts. Part 1 differentiates an integral and recovers the integrand. Part 2 evaluates an integral by subtracting the values of an antiderivative. Both hinge on a single object: the function that accumulates area as its upper limit moves.

The area accumulation function

Fix a continuous $f$ on $[a,b]$ and define a new function by letting the upper limit of integration vary:

For each fixed $x$ this is a definite integral, hence a number; letting $x$ move makes $g$ a function of $x$. The dummy variable is written $t$ so it is not confused with the limit $x$. When $f\ge 0$, the value $g(x)$ is the area under the graph from $a$ out to $x$ — the area so far.

Compute $g$ for a concrete case, $f(t)=t$ and $a=0$. The region under $y=t$ from $0$ to $x$ is a triangle of base and height $x$, so

The derivative of the accumulation function returned the integrand — a fact that holds for every continuous $f$.

Part 1: differentiating an integral

Why should $g^{\prime }(x)=f(x)$ in general? Increase the upper limit by a small $h$. The difference $g(x+h)-g(x)$ is the area of the thin sliver under $f$ between $x$ and $x+h$. For small $h$ that sliver is nearly a rectangle of width $h$ and height $f(x)$, so

Made rigorous, the approximation becomes an equality in the limit. On $[x,x+h]$ the Extreme Value Theorem gives points $u,v$ where $f$ attains its minimum $m=f(u)$ and maximum $M=f(v)$; the bounding property of integrals then sandwiches the difference quotient,

As $h\to 0$ both $u$ and $v$ are squeezed toward $x$, and continuity forces $f(u)\to f(x)$ and $f(v)\to f(x)$. By the Squeeze Theorem the middle term has the same limit.

In words: differentiating a definite integral with respect to its upper limit returns the integrand evaluated at that limit. Every continuous function therefore has an antiderivative, namely its own accumulation function — even when no formula for it exists in elementary terms.

Reading Part 1

Two consequences of Part 1 are worth stating on their own.

• Antiderivatives always exist. Many integrals cannot be expressed with elementary functions. The Fresnel function $S(x)={\int }_0^x\sin (t^2/2)\mathbf{d}t$ has no closed form, yet Part 1 gives its derivative immediately: $S^{\prime }(x)=\sin (x^2/2)$.
• A moving limit needs the Chain Rule. When the upper limit is itself a function $u=u(x)$, compose. For $\frac{\mathbf{d}}{\mathbf{d}x}{\int }_1^{x^4}\sec t\mathbf{d}t$, set $u=x^4$; then by Part 1 and the Chain Rule, $$\frac{\mathbf{d}}{\mathbf{d}x}\underset{1}{\overset{x^4}{\int }}\sec t\mathbf{d}t=\sec (u)\frac{\mathbf{d}u}{\mathbf{d}x}=\sec (x^4)\cdot 4x^3.$$

Several functions of applied mathematics are defined by exactly this kind of integral, their integrands having no elementary antiderivative. Part 1 differentiates each at sight:

Each accumulation function inherits its shape from the sign of its integrand: it rises where the integrand is positive, falls where the integrand is negative, and turns at the integrand's zeros.

Part 2: evaluating an integral

Part 1 makes $g(x)={\int }_a^{x}f(t)\mathbf{d}t$ an antiderivative of $f$. Any other antiderivative $F$ differs from $g$ by a constant, because two functions with the same derivative on an interval differ by a constant. Writing $F(x)=g(x)+C$ and evaluating at the endpoints, the constant cancels in the difference:

since $g(a)=0$ and $g(b)={\int }_a^{b}f$. That is the second part.

The definite integral, defined through all the values of $f$ on $[a,b]$, is found from the values of one antiderivative at just the two endpoints. The standard notation for the difference is a bracket:

Return to ${\int }_0^1{x}^2\mathbf{d}x$, which cost a summation identity before. An antiderivative of $x^2$ is $F(x)=\frac{1}{3}{x}^3$, so

The choice of antiderivative does not matter, since any two differ by a constant that cancels; use the simplest, dropping $+C$ entirely.

One caution: Part 2 requires $f$ continuous on all of $[a,b]$. Applying it blindly across a discontinuity produces nonsense, such as the false ${\int }_{-1}^3{x}^{-2}\mathbf{d}x=-\frac{4}{3}$ for an integrand that is positive everywhere it is defined and in fact has an infinite discontinuity at $x=0$, so the integral does not exist as an ordinary one.

The two parts together

Stated side by side, the two parts express one relationship.

Each undoes the other. Integrating $f$ and differentiating the result gives $f$ back; differentiating $F$ and integrating the result gives back the change in $F$. Differentiation and integration are inverse processes, and the challenging area, volume, and length problems that occupied mathematicians from Archimedes to Fermat become routine once an antiderivative is in hand.

Indefinite integrals

Because Part 2 turns evaluation into antidifferentiation, antiderivatives deserve a notation of their own. The symbol $\int f(x)\mathbf{d}x$ without limits denotes the general antiderivative — a whole family of functions differing by a constant.

The connection is Part 2: if $F$ is an antiderivative, ${\int }_a^{b}f(x)\mathbf{d}x=[\int f(x)\mathbf{d}x{]}_a^b$. Every differentiation rule, read backwards, is an antidifferentiation formula. The basic table:

Each entry is checked by differentiating the right side. Sums and constant multiples distribute across the integral sign exactly as they do for definite integrals, so polynomials integrate term by term. For a general indefinite integral,

A definite version, applying Part 2 to a polynomial:

which matches the Riemann-sum value computed the long way from the definition. The table is deliberately short; it grows as new derivatives are learned, and the exponential and logarithmic functions add the entries for $e^x$ and $1/x$. Antiderivatives that are not in the table need a technique, beginning with the Substitution Rule.

The Net Change Theorem

Rewriting Part 2 with $F^{\prime }=f$ named as a rate produces a statement with a direct physical reading. Since $F^{\prime }(x)$ is the rate of change of $F$, and $F(b)-F(a)$ is the total change of $F$ across the interval:

Whatever $F$ measures, integrating its rate over $[a,b]$ recovers how much $F$ changed. A few instances:

• Volume from flow. If $V^{\prime }(t)$ is the rate water flows into a reservoir, then ${\int }_{t_1}^{t_2}{V}^{\prime }(t)\mathbf{d}t=V(t_2)-V(t_1)$ is the change in stored water.
• Population from growth rate. ${\int }_{t_1}^{t_2}{n}^{\prime }(t)\mathbf{d}t=n(t_2)-n(t_1)$, the net change in population, births and deaths both folded in.
• Cost from marginal cost. ${\int }_{x_1}^{x_2}{C}^{\prime }(x)\mathbf{d}x=C(x_2)-C(x_1)$, the added cost of raising production from $x_1$ to $x_2$ units.
• Mass from density. For a rod of linear density $\rho (x)=m^{\prime }(x)$, ${\int }_a^b\rho (x)\mathbf{d}x=m(b)-m(a)$ is the mass of the segment.

The unit of ${\int }_a^{b}f(x)\mathbf{d}x$ is the product of the unit of $f$ and the unit of $x$. Integrating megawatts against hours gives megawatt-hours of energy; integrating a density in kg/m against meters gives kilograms.

Displacement versus distance

Motion along a line makes the sign convention concrete. If $v(t)=s^{\prime }(t)$ is velocity, the Net Change Theorem gives displacement,

the net change in position — with leftward motion ($v<0$) subtracting. Total distance traveled counts every stretch as positive, so it integrates the speed $|v(t)|$: