The First Fundamental Theorem of Calculus

The theorem is the computational engine behind essentially all elementary techniques for evaluating integrals, most directly the Second Fundamental Theorem of Calculus, which uses the existence of an antiderivative (guaranteed here) to reduce $\int_a^b f$ to evaluating any antiderivative at the endpoints. It also underlies Integration by Substitution, where differentiating a composed accumulation function via the chain rule and this theorem yields the change-of-variables formula. In physics, it justifies computing displacement from velocity, work from force, and charge from current, by treating physical accumulation functions as literal Riemann integrals whose derivative recovers the instantaneous rate. In probability and statistics, the cumulative distribution function $F(x) = \int_{-\infty}^x f(t)\,dt$ of a continuous random variable is differentiable with $F' = f$ (the density) at every point of continuity of the density — a direct instance of this theorem, essential for deriving likelihoods and for change-of-variables formulas for random variables. In numerical analysis and control theory, the theorem justifies why numerically accumulating a signal (e.g. via a running sum or an analog integrator circuit) and then differentiating recovers the original signal, up to the smoothness constraints on the input. In complex analysis, the analogous statement for contour integrals of holomorphic functions is a cornerstone in proving that holomorphic functions have primitives locally, feeding into Cauchy's theorem.

:::quiz[Which prerequisite fact is used to guarantee $F(x)=\int_a^x f$ is well-defined for every $x\in[a,b]$ when $f$ is Riemann integrable on $[a,b]$?]

• The Mean Value Theorem for derivatives
• Restriction of an integrable function to a subinterval remains integrable ✓
• The chain rule
• L'Hôpital's rule

Ask the tutor: Can you show me why $F$ being Lipschitz (not just continuous) matters in applications?

Think of $f(t)$ as a velocity and $F(x)$ as the total signed distance traveled from time $a$ to time $x$. The derivative $F'(c)$ asks: as we nudge the endpoint $x$ away from $c$, how fast does accumulated distance change per unit of $x$? Intuitively this should be exactly the instantaneous velocity $f(c)$ — but this intuition silently assumes $f$ doesn't jump around near $c$. The key algebraic move is additivity: $F(x) - F(c) = \int_c^x f(t)\,dt$. Dividing by $x - c$ turns this into the average value of $f$ over the interval from $c$ to $x$. As $x \to c$, this shrinking window's average value should converge to $f(c)$ itself — precisely because $f$ is continuous at $c$, so nearby values of $f$ cluster around $f(c)$. If $f$ jumps at $c$, the average taken from one side can differ from $f(c)$, and the derivative can fail to exist or match.

The shaded region's area is $F(x) - F(c)$; dividing that area by the base length $x-c$ gives the average height of the curve over $[c,x]$, which visibly hugs the red line $f(c)$ as the window narrows — provided $f$ doesn't jump near $c$.

Example 1: An asymmetric, non-symmetric piecewise integrand. Let $f : [0,3] \to \mathbb{R}$ be defined by

Note $f$ is continuous at $t=1$ from the left ($1^2=1$) and from the right ($2(1)-1=1$), so in fact $f$ is continuous on all of $[0,3]$ (this is deliberately not a symmetric or "nice round number" example — check it doesn't accidentally hide a jump). Compute $F(x) = \int_0^x f(t)\,dt$ for $x \in [0,3]$. For $0 \le x \le 1$:

For $1 < x \le 3$: using additivity,

Check continuity of $F$ at $x=1$: from the first branch, $F(1) = 1/3$; from the second branch as $x\to 1^+$, $\frac13 + 1 - 1 = \frac13$. They match, confirming Part 1 of the theorem. Now differentiate. For $x<1$: $F'(x) = x^2 = f(x)$. For $x>1$: $F'(x) = 2x - 1 = f(x)$. At $x=1$ itself, since $f$ is continuous there, the theorem guarantees $F'(1)$ exists and equals $f(1)=1$; indeed the one-sided derivatives from the two formulas are $F'(1^-) = 1^2 = 1$ and $F'(1^+) = 2(1)-1=1$, agreeing. So $F$ is differentiable everywhere on $[0,3]$ with $F' = f$, exactly as guaranteed — and note we verified it both via the theorem and via brute-force computation, confirming the theorem's prediction.

Example 2: A jump discontinuity — where the theorem correctly stays silent. Let $g : [0,2] \to \mathbb{R}$ be

This is a step function, integrable (bounded, one discontinuity), continuous everywhere on $[0,2]$ except at $t=1$. Let $G(x) = \int_0^x g(t)\,dt$. For $0 \le x \le 1$: $G(x) = \int_0^x 1\,dt = x$. For $1 < x \le 2$: $G(x) = \int_0^1 1\,dt + \int_1^x 5\,dt = 1 + 5(x-1) = 5x - 4$. Check continuity at $x=1$: left limit gives $G(1)=1$; right-hand formula as $x\to1^+$ gives $5(1)-4=1$. Continuous, as Part 1 guarantees regardless of $g$'s discontinuity. Now examine differentiability at $x=1$. The left-hand derivative:

The right-hand derivative:

The one-sided derivatives disagree ($1 \ne 5$), so $G'(1)$ does not exist. This is completely consistent with the theorem: $g$ is discontinuous at $t=1$, so the theorem simply makes no claim there — and indeed the claim would be false if extended blindly, since $G'(1)$ doesn't even exist, let alone equal some single value $g(1)=5$. Away from $t=1$, $g$ is continuous, and correspondingly $G'(x) = g(x)$ for all $x \ne 1$ in $[0,2]$, exactly matching the theorem's guarantee at continuity points.